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TRANSCRIPT
Proceedings of the 43rd International Physics Olympiad15th – 24th of July 2012 • Tallinn and Tartu, Estonia
PhysIcsIs LOvE
4
Estonian Ministry of Education and Research
The Estonian Information Technology Foundation
Designed by loremipsum.ee
Edited by OÜ Komadisain
Published in 2012
www.ipho2012.ee
IPhO illustrations by Toom Tragel
Cover photo by Henry Teigar
Proceedings of the 43rd International Physics Olympiad is licensed under a Creative
Commons Attribution 3.0 Estonia License.
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ContentssPEEchEs 13Opening Ceremony 13
Closing Ceremony 19
PEOPLE 25Participants 25
Organizers 54
PrOgrams 63Students 63
Leaders and Observers 67
PrObLEms and sOLuTIOns 73Theoretical Competition 73
Experimental Competition 107
rEsuLTs 131Gold Medalists 131
Silver Medalists 133
Bronze Medalists 135
Honorable Mentions 137
Special Prizes 139
Statistics of the competition results 140
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InTErnaTIOnaL bOard 151Minutes 151
Statutes 157
Syllabus 168
aPPEndIcEs 175Tartu – The World Capital of Physics 175
Circulars 183
Newsletter 197
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ForewordEstonia had the honour of hosting the 43rd International Physics Olympiad (IPhO),
which took place from the 15th to the 24th of July 2012. On this occasion the dele-
gations of 80 countries came together, making it one of the largest international
events ever hosted by Estonia. The success of the venue became possible owing to
the long-term efforts of a large number of people – the devoted members of the
Steering Committee, Organizing Committee, Markers, Academic Committee, and
the numerous volunteers – the Guides, Media Team, etc. Of equal importance was
the comprehensive support of the Estonian Ministry of Education and Research,
as well as the aid of the University of Tartu, Tallinn University of Technology,
and the main sponsors. Last but not least, the success of the Olympiad stems
from the support and cooperation of all the participants – contestants, leaders,
observers, and visitors! Organizing such a huge event was a challenging but highly
rewarding task!
The International Physics Olympiad has a long history, and many of its tradi-
tions originate from an era which was very different from the current one; perhaps
the most important difference lies in the communication and information tech-
nologies. A dozen years ago at an IPhO, a group of team leaders had a discussion
about the prospects of the IPhO in the internet era. Will it remain, or will it be
superceded by online competitions? During the subsequent years, the IPhO has not
only stayed but also grown. The fun of face-to-face discussions with new friends
and time spent together can never be substituted by meetings in virtual space.
However, the advent of new technologies requires and also makes possible certain
changes in the organization of the IPhO. The organizers of the 43rd IPhO ventured
to introduce several innovations, most of which received praising feedback from
the participants. In particular, it was the first Olympiad where (a) the students
and team leaders stayed in different cities; (b) the team leaders were asked to
provide only digital copies of the problem translations (thus bypassing the hard
copies); (c) there was a fully automated scanning of the students’ solutions, which
made use of barcodes; (d) the distribution of the graders’ marks and submitting of
the leaders’ marks was made digitally online; (e) the competition city became the
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World Capital of Physics; (d) there was a “career day” during which the students
had an opportunity to visit the information booths of some of the leading uni-
versities in the world; (f) the Olympiad was preceded by a 10-month-long online
competition “Physics Cup – IPhO2012”.
Another innovation of this Olympiad – perhaps not as prominent but by no
means less important than the ones listed above – lies in the style of the problems.
To be more specific, it was not so much an innovation, but rather reverting back to
the style of the problems used 20 or more years ago. During the last two decades,
the problem texts had become longer and longer, with numerous sub-questions,
often with multiple nested structures. The reasoning behind such a trend was
simple: the number of contestants (and the number of languages in which the
solutions were written) became bigger and bigger, making the grading process
increasingly difficult. A larger number of smaller tasks make it easier to achieve a
fair grading. However, there were also serious implications: the students needed
to waste a lot of time on reading the texts; there was less room for creativity – most
often, the approach to the solution was already written into the problem text.
It has been argued that the style of the IPhO problems from the last decade
was closer to the real tasks of the science of physics. This claim, however, is not
entirely correct. Indeed, a good physical research involves several stages: (a) find-
ing a challenging and important topic; (b) making a solvable model – neglecting
marginal effects and keeping only the qualitatively important components (d)
solving the problem which was formulated at the previous stage; (e) analyzing
the implications of the solution. So, it is true that an immediate problem solving
makes up only a small part of a physical study. Nevertheless, a good insight into
physical phenomena, which is developed by solving creative physical problems,
is also very important during stages (a) and (b). Furthermore, in order to advance
with really important and innovative topics where a new physics is to be devel-
oped, creativity is unavoidable. While the technical skills can be easily developed
later, typically during university studies, creativity needs to be developed from
the very beginning of physics studies. Creative problems also make physics fun,
which is what attracts talented young people. This was the reasoning behind the
decision of the Academic Committee of the 43rd IPhO to take the risk of making
shorter and more creative problems, the solutions of which were more difficult to
grade. Now, while looking back, one can say that the risk paid off: the feedback
was nothing but positive. Let us cite here a contestant from China, Hengyun Zhou:
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“I liked this year’s IPhO problems very much. Having
gone through most of the past papers, I think that this
year’s problems are the best to date. First, they consisted
of difficult enough problems, and left most of the think-
ing process for the students so that we had to use all our
knowledge and skills to figure out the correct approach
to the problems. Many of the problems in the past paved
the entire way for the students, so all students had to
do was follow the instructions, but this year we had to
come up with a method of our own. Additionally, this
year’s problems emphasised the physics rather than
mathematic skills. The most difficult part in the prob-
lems was building an appropriate model, and that part
really intrigued me, although I failed to build a correct
model in many problems in the end.”
Jaak kIkas and Jaan kaLdaAcademic Committee of the 43rd IPhO
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Sir Harold Kroto1996 Nobel Prize winner,
the honorary guest of
the IPhO2012
“The job of the scien-
tists is to check, not
to believe everything.
The freedom to doubt
is a privilege. If you
know you are unsure,
you have a chance to
change the situation.”
From the academic lecture by Sir Harold Kroto July 20th Tartu – the World Capital of Physics
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SpeechesOpening CeremonyEnE ErgmaPresident of the Parliament of Estonia
Dear participants of the International Physics Olympiad!
Dear delegation leaders, guests, ladies and gentlemen!
It is a great pleasure for me to greet you in Tallinn, the
capital of the Republic of Estonia.
Welcome!
Already in 1996, physicist and Member of the
Academy of Sciences Jaak Aaviksoo, who was then the
Minister of Education of the Republic of Estonia, submit-
ted the application to hold the IPhO in Estonia in 2012.
This took place only five years after Estonia had restored
its independence, but already then we were convinced
that the motor of progress in the 21st century would be an
economy that is based on science and high technologies
and needs specialists with a strong science education.
Since then, young people of Estonia have success-
fully participated in several Olympiads, and have not
returned home from them empty-handed. The fact that
the International Physics Olympiad is held in Estonia
became the reason why it was decided to declare this aca-
demic year the Year of Science in Estonia, and why Tartu
becomes the World Capital of Physics on the 20th of July.
But why is the year 2012 so special for physicists?
I believe there is no sense in asking this question in
this hall, because everybody knows the answer – Higgs
Photo by Henry Teigar
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boson does exist! The almost half-a-century-long saga of experimental discovery
of this particle shows how complicated it is to get nature to reveal its secrets. But
physicists are purposeful people and they are not deterred by difficulties.
Dear future young colleagues!
I would like to tell you that an education in physics is the best thing in the
world. Having studied physics at Moscow University and worked for many years
as an astrophysicist both in science institutions and as a university lecturer, I can
tell you that dealing with physics and science is really fun. And if the laboratories
on Earth become too narrow for you, turn your glance towards the Universe, where
the great Creator, Nature, has built the most perfect physics laboratory, the con-
ditions of which scientists will never be able to reproduce on Earth.
I wish you all a successful Olympiad, a nice stay in Estonia and a successful
future career in science.
And many thanks to the instructors of the young people and to the organizers
of the Olympiad!
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hans JOrdEnsThe President of the International Physics Olympiad
Your Excellency,
Distinguished guests,
Dear participants,
For all those who love physics, participation in the
International Physics Olympiad is an exciting event that
cannot be overestimated for its importance. Large-scale
events like this one will have a profound impact on the
lives of all participants whatever they will do later on.
The IPhO is like a watershed: there is one life before the
Olympiad and another one afterwards. And the two are
very different.
For that reason, we are grateful to Estonia as the
country that hosts the Olympiad today. I had the priv-
ilege to visit Estonia one month ago to confer with the
organizers about the progress of the organization and
I was really impressed by the work they had done. You
must be aware that Estonia is a young and rather small
country with an even smaller population. To organize a
large-scale event like the Physics Olympiad is something
that is not done overnight. On top of this relatively dif-
ficult job Estonia wanted to have the competition take
place in two cities almost 200 km apart. The solution to
that was the introduction of modern techniques, much
more than we have ever used before in the IPhO. You
should not be surprised that Skype, being developed in
Estonia, will be one of them.
I hope you will enjoy your stay in Estonia. Estonia is
Photo by Siim Pille
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a lovely country with very interesting cities, with ancient histories the traces of
which are still very visible. But most of the country is covered by a pastoral and in
some places also rough and wild nature. You will have the opportunity to enjoy
all that during the excursions.
The students stay in the city of Tartu which is situated more to the south-
east. You will like this small town for her nice old centre. But you will most of
all remember Tartu as the city where you took the theoretical and experimental
tests, which, I can assure you, pose interesting physics problems that demand all
of your skills to solve them.
It is a privilege to take part in the Physics Olympiad. It is our aim to have as
many countries as possible participating in the competition. I, therefore, regret
that from the 104 invited countries 24 could not send a team for financial, organ-
izational or other reasons. In the world today we should be able to provide free
access for someone who wants to participate in this event, no matter where he or
she lives. As long as I am president of the International Physics Olympiad I take
it as my duty to promote that.
But still, 80 teams are present. You find yourself amongst some 400 others who
have the same fascination for physics as you have. That already by itself is a great
experience. I hope you take the opportunity to make friends. In this generation
more than ever before it is easy to stay in touch. Science and especially physics is
an international activity. And due to that physicists are able to tackle and solve
problems like the proof of the existence of the Higgs boson as they just did in
CERN. But while some questions are solved, new ones pop up. We are truly living
in exciting times.
I, therefore, wish you all the best in this competition and I hope that it may
be your first step towards a career in Physics.
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Jaak kIkasChairman of the Academic Committee
Honorable President of the Estonian Parliament,
Honorable President of the International Physics
Olympiad,
Dear participants of the 43rd International Physics
Olympiad,
Honorable guests, Ladies and gentlemen,
On behalf of the Academic Committee of the 43rd
International Physics Olympiad it is my pleasure to
welcome you to Estonia. The IPhO is a big event for
Estonia – never before have we hosted an international
gathering with so many nations participating, not to
mention the number of brilliant young minds – the
future of world physics.
Estonia has been participating in the IPhOs since
1992, and since then we have enjoyed the hospitality of
21 countries. We very much hope that we can, in turn,
provide a bit of the warmth that we ourselves have
enjoyed all over the world. The Academic Committee
has been working hard to make the coming Olympiad
a memorable event for you. First of all, we have tried
to prepare a balanced set of problems, including both
simple and truly challenging questions, rich in physical
content and relatively simple mathematically. We also
hope that beside the examinations the Olympiad is a
chance for you to find new friends, have a good time,
and learn a bit about our country and the people living
here. Well, yesterday you already learned something
Photo by Siim Pille
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about the Estonian weather. To be honest, it can be better. Sometimes J
And don’t worry that after they have discovered the Higgs there’s nothing left
for you to discover. Because if you raise your eyes to the skies, as academician
Ergma recommended, what do you see there? Almost nothing! Because about
95% of what is up there is hidden from our sight – I mean the dark matter and
dark energy. What these really are – it may well be your chance to find out. And
on your road to big discoveries, participation in the IPhO is a pretty big step.
On behalf of the Academic Committee I wish you every success in the forth-
coming competition.
Enjoy the Olympiad!
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Jaak aavIksOOMinister of Education and Research
Dear friends of physics, I would like to start with thank-
ing you, all of you. Firstly because of your love towards
physics. As a physicist I know what it means. It is a
personal pleasure, it is a contribution to technological
development, and I hope very much it is also a contri-
bution to a better world.
I would also like to thank you for making Estonia,
Tallinn and Tartu the Capital of Physics for at least 7 days.
It is your work, it is your love towards physics that has
made it possible. Thank you for coming! Thank you for
your efforts, solving complicated theoretical problems,
being skilled in experiments. And last but not least,
thank you for joining the physics family here in Estonia.
Of course I would like to congratulate all of you who
have made their way to the prizes, be it silver, bronze or
gold. I would also like to thank all of your teachers. All of
those people who have helped you to achieve what you have
achieved so far. I would like to thank your families. And I
would like to wish you, every one of you, success in your
future endeavours, in physics and in your personal lives.
And last but not least, try to make this world a bet-
ter place through your love towards physics, through
your love towards friends, different people, developing
ways and means to contribute to social progress and a
peaceful world.
Thank you for coming and congratulations on your
achievements!
Photo by Karl Veskus
Closing Ceremony
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hans JOrdEnsThe President of the International Physics Olympiad
Your Excellency,
Distinguished guests,
Dear participants.
First of all, I want to thank our hosts from Estonia
who made this Olympiad a great success. Much work is
needed to have an Olympiad run well, especially when
you take into account the relatively large burden on the
shoulders of a small country like Estonia. Hundreds of
people, staff, guides and volunteers, were involved in
the organization to make the competition work for some
400 participants.
So please join me in a big round of applause for the
organizers and all those who made this Olympiad possible!
The Physics Olympiad is, first and foremost, a com-
petition and I’d, therefore, like to say a few words,
especially to those who participated: the competitors.
I really hope you enjoyed your stay here in Estonia. I
hope you are satisfied with your results in the competi-
tion. I hope you enjoyed the company of your peers from
all over the world and I hope you could make friends
with some of them. All this is important. But I also hope
that you feel privileged that you could take part in the
International Physics Olympiad. Taking into account
the millions of young people of your age, you now belong
to a very select group. The French have a nice phrase for
that: “Noblesse oblige”, meaning “Nobility obliges”, or
in other words, privilege entails responsibility.
Photo by Karl Veskus
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There is an aspect of being excellent of which you must be aware. From now on
you will be regarded as a role model, and that gives you a certain responsibilities. Not
only your peers but most of all those who are younger than you will look up to you.
You have the possibility to inspire youngsters much more than anyone else, simply
because you are young yourself and because you have already achieved so much.
So don’t spend all your time just studying science. Participate in outreach
activities as well and use the abilities you have obtained from being here and the
position of the role model you have become. I would say: show the world that
science is fascinating and exciting. That it can be understood and that it should
be understood in order to make proper decisions. There is a significant amount of
scientific illiteracy amongst people who rule the world. Science not only teaches
you about the laws of nature but it stimulates a critical attitude towards what
you observe. It teaches you to distinguish between facts and fiction. It is that
critical attitude that keeps people’s eyes open in the quest for truth. And you are
the ambassadors to advocate that.
I’d like to conclude by wishing you all the best in your future lives and let us
hear from you, if not as a Nobel Prize Laureate then at least as someone who made
a difference. Try to become a happy person and let physics help you to achieve that.
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nIELs chrIsTIan harTLIngPresident of Organising Committee IPhO 2013
Distinguished guests, ladies and gentlemen,
Denmark has participated in the International
Physics Olympiad since Oslo 1996. A few years later,
in 1999, it was decided that Denmark should host the
International Physics Olympiad in 2013.
The year 2013 was mainly chosen in memory of the
100th anniversary of Niels Bohr’s theory of the hydro-
gen atom. One may say that this theory marks the very
beginning of quantum mechanics. Therefore, it seemed
natural to celebrate this year with young physics stu-
dents from all over the world. And to be quite honest:
We also hoped that the anniversary would help us
gather the necessary funds.
Back in 1999, we had a feeling that the year 2013 was
a distant future. But as time flies, we now realize with
some surprise that there is only one year left.
Estonia and Denmark have a lot in common. We are
both small nations, more or less the same size, with
a coastline on the Baltic Sea, and we have the same
weather. What you may not know is that, according to
legend, Denmark got its flag in Estonia. During a bat-
tle against Estonia on the 15th of June 1219, the Danish
army was about to lose. Then suddenly a flag fell from
the sky, and a voice said, “Under this flag you will win”!
The Danish Army did win, and we got our red and white
flag, which is the oldest in the world. So we may say
that we got our national flag from Estonia! [And today,
Photo by Merily Salura
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once more, we received a flag in Estonia.]
This year we’ve had a wonderful time here in Estonia, and finally I want to
thank our hosts for these amazing days, which we will never forget. You have
made a fantastic arrangement. Now Estonia hands over the baton to Denmark,
but it will not be easy to match this marvelous organization.
We look very much forward to welcoming you in Copenhagen, Denmark, to
the 44th International Physics Olympiad in 2013.
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PeopleParticipantsInTErnaTIOnaL PhysIcs OLymPIad cOmmITTEEPresident Hans Jordens
Executive Secretary Ming Juey Lin
aLbanIaStudent Geri Emiri
Arled Papa
Leader Antoneta Deda
armEnIaStudent Vardan Avetisyan
Aram Mkrtchyan
Virab Gevorgyan
Aleksandr Petrosyan
Razmik Hovhannisyan
Leader Gagik Grigoryan
Bilor Kurghinyan
azErbaIJanStudent Haji Piriyev
Valeh Farzaliyev
Ramazan Ramazanov
Farid Mammadov
Ahmad Mehribanli
Leader Mirzali Murguzov
Rana Mammadova
Observer Rashadat Gadmaliyev
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ausTraLIaStudent Eric Huang
Jonathan Lay
Nicholas Salmon
Siobhan Tobin
Christopher Whittle
Leader Matthew Verdon
Bonnie Zhang
Observer Alix Verdon
ausTrIaStudent Oliver Edtmair
Christoph Weis
Tobias Karg
Maximilian Ruep
Martin Stadler
Leader Helmuth Mayr
Engelbert Stuetz
bangLadEshStudent Kinjol Barua
Wasif Ahmed
Shinjini Saha
Shovon Biswas
Ahmed Maksud
Leader Fayez Ahmed Jahangir Masud
M.Arshad Momen
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bELarusStudent Ihar Lobach
Albert Samoilenka
Aliaksey Khatskevich
Aliaksandr Yankouski
Vadzim Reut
Leader Anatoli Slabadzianiuk
Anton Mishchuk
Observer Leonid Markovich
bELgIumStudent Romain Falla
Leandro Salemi
Basile Rosen
Basile Vermassen
Mathias Stichelbaut
Leader Bernadette Hendrickx
Philippe Leonard
Observer Sophie Houard
Matthieu Dontaine
bOLIvIaStudent Cesar Tapia
Leader Veronica Subieta
bOsnIa and hErzEgOvInaStudent Selver Pepic
Amer Ajanovic
Nudzeim Selimovic
Sladjan Veselinovic
Stefan Gvozdenovic
Leader Rajfa Musemic
Rodoljub Bavrlic
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brazILStudent Luis Gustavo Lapinha Dalla Stella
Guilherme Renato Martins Unzer
Lara Timbo Araujo
Ivan Tadeu Ferreira Antunes Filho
Jose Luciano De Morais Neto
Leader Euclydes Marega Junior
Fernando Wellysson de Alencar Sobreira
Observer Leonardo Bruno Pedroza Pontes Lima
Ronaldo Fogo
Antonio Giacomo Pedrine
buLgarIaStudent Katerina Naydenova
Yordan Yordanov
Veselin Karadzhov
Kaloyan Darmonev
Konstantin Gundev
Leader Victor Ivanov
Miroslav Abrashev
canadaStudent Sepehr Ebadi
Yun Jia (Melody) Guan
Tristan Downing
Henry Honglei Wu
Simon Blouin
Leader Andrzej Kotlicki
Jean-François Caron
Visitor Chantal Haussmann
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cOLOmbIaStudent Daniel Eduardo Fajardo Fajardo
Andres Rios Tascon
Andres Zorrilla Vaca
Leader Fernando Vega Salamanca
Eduardo Zalamea Godoy
crOaTIaStudent Samuel Bosch
Bruno Buljan
Luka Skoric
Karlo Sepetanc
Grgur Simunic
Leader Nikolina Novosel
Ticijana Ban
czEch rEPubLIcStudent Ondřej Bartoš
Jakub Vošmera
Stanislav Fořt
Martin Raszyk
Lubomír Grund
Leader Jan Kříž
Bohumil Vybíral
cyPrusStudent Nicolas Shiaelis
Anastasios Stylianou
Marios Ioannou
Marios Maimaris
Fidias Ieridis
Leader Demetrios Philippou
Emmanouil Lioudakis
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dEnmarkStudent Molte Emil Strange Andersen
Jakob Lass
Christian Aamand Witting
Kasper Tolborg
Nikolaj Theodor Thams
Leader Jens Ulrik Lefmann
Christian Thune Jacobsen
Observer Niels Christian Hartling
Marianne Hartling
Niels Østergård
Maja Lehmann Jacobsen
Henrik Bruus
EL saLvadOrStudent Bryan Alexander Escalante Castro
Valerie Argentina Dominguez Rivera
Julio Carlos Chorro Huezo
Leader Jose Roberto Dimas Valle
Raúl Alvarenga
EsTOnIaStudent Jaan Toots
Tanel Kiis
Kaur Aare Saar
Kristjan Kongas
Andres Erbsen
Leader Mihkel Kree
Taavi Pungas
31
FInLandStudent Iiro Lehto
Matias Mannerkoski
Jyri Maanpää
Arttu Yli-Sorvari
Tapio Hautamäki
Leader Heikki Mäntysaari
Lasse Franti
FrancEStudent Jonathan Dong
Jean Douçot
Paul Kirchner
Theodor Misiakiewicz
Simon Pirmet
Leader Christian Brunel
Nicolas Billy
Observer Solene Chevalier-Thery
gEOrgIaStudent Jemal Shengelia
Giorgi Kobakhidze
Sergi Chalauri
Saba Kharabadze
Sandro Maludze
Leader Vladimir Paverman
Kakhaber Tavzarashvili
Observer Mariami Rusishvili
32
gErmanyStudent Qiao Gu
Sebastian Linß
Vu Phan Thanh
Lorenz Eberhardt
Georg Krause
Leader Stefan Petersen
Gunnar Friege
Observer Jochen Kröger
grEEcEStudent Stavros Efthymiou
Fotios Vogias
Emmanouil Sakaridis
Emmanouil Vourliotis
Michalis Halkiopoulos
Leader George Kalkanis
Panagiotis Tsakonas
hOng kOngStudent Lam Ho Tat
Lai Kwun Hang
Chan Cheuk Lun
Fung Tsz Chai
Lo Hei Chun
Leader Dik Wai Yin
Wong Kwok Yee
Observer Ng Siu Cho
Sun Ke
33
hungaryStudent Péter Juhász
Áron Dániel Kovács
Zoltán Laczkó
Roland Papp
Attila Szabó
Leader Péter Vankó
Máté Vigh
Observer Ferenc Sarlós
IcELandStudent Hólmfríður Hannesdóttir
Atli Thor Sveinbjarnarson
Freyr Sverrisson
Pétur Rafn Bryde
Stefan Alexis Sigurðsson
Leader Ingibjörg Haraldsdóttir
Martin Swift
IndIaStudent Bijoy Singh Kochar
Jeevana Priya Inala
Kunal Singhal
Pulkit Tandon
Rahul Trivedi
Leader Patrick Dasgupta
Raghavendra Maigur Krishna
Observer Shirish Pathare
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IndOnEsIaStudent I Made G.N. Kumara
Luqman Fathurrohim
Ramadhiansyah Ramadhiansyah
Werdi Wedana Gunawan
Adrian Nugraha Utama
Leader Mohammad S. Rosid
Kamsul Abraha
Observer Bobby Eka Gunara
Visitor Bambang Hartono
IrELandStudent John Cristopher Horatio Mulholland
Dale Alexander Hughes
Liam Tomas Mulcahy
Thomas Sherlock Wyse Jackson
Leader Eamonn Cunningham
David Rea
IsLamIc rEPubLIc OF IranStudent Mohamad Ansarifard
Mehrdad Malak Mohammadi
Amir Yousefi
Ramtin Yazdanian
Sajad Khodadadian
Leader Mehdi Saadat
Ayoub Esmailpour
Observer Seyyed Nader Rasuli
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IsraELStudent Itay Knaan Harpaz
Chen Solomon
Yigal Zegelman
Ittai Rubinstein
Eden Segal
Leader Eli Raz Somech
Igor Lisenker
Observer Yoav Merhav
ITaLyStudent Federica Maria Surace
Roberto Albesiano
Michele Fava
Martin Vlashi
Federico Re
Leader Dennis Luigi Censi
Paolo Violino
Observer Giorgio Busoni
Francesco Minosso
JaPanStudent Yuichi Enoki
Tasuku Omori
Kazumi Kasaura
Kohei Kawabata
Hiromasa Nakatsuka
Leader Fumiko Okiharu
Kazuo Kitahara
Observer Tadao Sugiyama
Masashi Mukaida
Yuto Murashita
Katsuhiko Shinkaji
Masao Ninomiya
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kazakhsTanStudent Ivan Senyushkin
Ilya Vilkoviskiy
Kaisarbek Omirzakhov
Nurzhas Aidynov
Mussa Rajamov
Leader Askar Davletov
Guliya Nurbakova
Observer Yernur Rysmagambetov
kuwaITStudent Nour Alsajari
Rawan Alsenin
Suad Alasfoor
Maryam Ramadan
Zainab Busalehhah
Leader Tareq Abdullah
Eman Hamad
Observer Anoud Alkandari
Visitor Yousef Alsenin
Muhammad Alsajari
Huda Esmail
Khaled Malak
Nozhah Almatrood
kyrgyzsTanStudent Fedor Ignatov
Salizhan Kylychbekov
Zakirbek Mamatayir uulu
Ermek Belekov
Meder Kutbidin uulu
Leader Raia Sultanalieva
Abdymanap Tashmamatov
Observer Ainagyl Osmonalieva
37
LaTvIaStudent Luka Ivanovskis
Georgijs Trenins
Kristaps Znotins
Andris Gerasimovics
Maris Serzans
Leader Vyacheslavs Kashcheyevs
Andris Muznieks
LIEchTEnsTEInStudent Benedikt Kratochwil
Lukas Lang
David Hälg
Leader Fritz Epple
Daniel Oehry
LIThuanIaStudent Mantas Abazorius
Tomas Čerškus
Daumantas Kavolis
Marius Kerys
Žygimantas Stražnickas
Leader Pavelas Bogdanovičius
Edmundas Kuokštis
Observer Indrė Grigaitytė
macaO, chInaStudent Chan Lon Wu
Wai Pan Si
Ka Fai Chan
Wai Hong Lei
Wai Hei Hoi
Leader Iat Neng Chan
38
macEdOnIaStudent Vesna Bacheva
Filip Simeski
Biljana Mitreska
Ljupcho Petrov
Leader Stanisha Veljkovikj
maLaysIaStudent Lee Yuan Zhe
Yeoh Chin Vern
Ooi Chun Yeang
Koay Hui Wen
Imran Ariffin
Leader Wan Mohd Aimran Wan Mohd Kamil
Chin Mai Ying
Observer Rosman Md Ajis
mExIcOStudent Eduardo Acosta-Reynoso
Javier Mendez-Ovalle
Kevin Bustillos-Barrera
Alberto Trejo-Avila
Jorge Torres-Ramos
Leader Victor Romero-Rochin
Raul Espejel-Morales
mOLdOvaStudent Cristian Zanoci
Ion Toloaca
Nicoleta Colibaba
Dinis Cheian
Ilie Popanu
Leader Victor Paginu
Igor Evtodiev
39
mOngOLIaStudent Tsogt Baigalmaa
Battushig Myanganbayar
Munkhtsetseg Battulga
Battsooj Bayarsaikhan
Bilguun Batjargal
Leader Batsukh Garmaa
Baatarchuluun Tsermaa
Observer Sharavsuren Byamba
Soyolmaa Dorjyanjmaa
mOnTEnEgrOStudent Petar Tadic
Marko Petric
Nikola Potpara
Vladimir Pejovic
Janko Radulovic
Leader Jovan Mirkovic
Nevenka Antovic
Observer Tatijana Carapic
nEThErLandsStudent Koen Dwarshuis
Troy Figiel
Ruben Doornenbal
Thijs van der Gugten
Martijn van Kuppeveld
Leader Ad Mooldijk
Enno van der Laan
40
nIgErIaStudent Ayomide Andrae Bamidele
Musa Muhammed Damina
John Nwankwo Chijioke
Anthony Okonkwo
Michael Tari Charles Okhide
Leader Ayhan Yaman
Lewis Obagboye
Observer Alaba Aminat Agbaje
Okey Junior Chikezie
nOrwayStudent Tiantian Zhang
Håkon Tásken
Oda Lauten
Anders Strømberg
Marius Leiros
Leader Torbjørn Mehl
Joakim Bergli
PakIsTanStudent Muhammad Suhaib Qasim
Usman Ayyaz
Usman Ali Javid
Muhammad Taimoor Iftikhar
Leader Shahid Qamar
Muhammad Aftab Rafiq
41
PEOPLE’s rEPubLIc OF chInaStudent Wenzhuo Huang
Yijun Jiang
Hengyun Zhou
Siyuan Wei
Chi Shu
Leader Xiaolin Chen
Kun Xun
Observer Liangzhu Mu
Chunling Zhang
Feng Song
POLandStudent Bartlomiej Zawalski
Michal Pacholski
Kacper Oreszczuk
Filip Ficek
Jan Rydzewski
Leader Jacek Jasiak
Jan Mostowski
POrTugaLStudent Francisco Machado
Pedro Paredes
Manuel Cabral
Matheus Marreiros
Simão João
Leader Fernando Nogueira
Rui Travasso
PuErTO rIcOStudent Logan Abel
Leader Hector Jimenez
42
rEPubLIc OF kOrEaStudent Woojin Kweon
Sooshin Kim
Wonseok Lee
Jaemo Lim
Suyeon Choi
Leader Sung-Won Kim
Ki Wan Jang
Observer Chan Ju Kim
Hyun Joo Lee
Yuri Kang
Weon Kyun Mok
Kug-Hyung Lee
rEPubLIc OF sIngaPOrEStudent Ding Yue
Huan Yan Qi
Kuan Jun Jie, Joseph
Soo Wah Ming, Wayne
Ang Yu Jian
Leader Rawat Rajdeep Singh
Chung Keng Yeow
Observer Chng Chia Yi
Berthold-Georg Englert
Visitor Aleksandra Englert
rOmanIaStudent Tudor Giurgică-Tiron
Dan - Cristian Andronic
Sebastian Florin Dumitru
Tudor Ciobanu
Roberta Răileanu
Leader Delia-Constanţa Davidescu
Adrian Dafinei
Observer Victor Păunescu
43
russIaStudent Alexandra Vasilyeva
Nikita Sopenko
Ivan Ivashkovskiy
Lev Ginzburg
David Frenklakh
Leader Valery Slobodyanin
Dmitry Aleksandrov
Observer Mikhail Osin
saudI arabIaStudent Sulaiman Almatroudi
Abdullah Alsalloum
Ali Alhulaymi
Mohammad Alhejji
Homoud Alharbi
Leader Najm Al Hosiny
Sandu Golcea
Observer Abdulaziz Alharthi
Mahmoud Nagadi
Hind Aldossari
Laila Babsail
Visitor Aljoharah Almetrek
Abdulaziz Alrashed
sErbIaStudent Tamara Šumarac
Milan Kornjača
Milan Krstajić
Jovan Blanuša
Ilija Burić
Leader Aleksandar Krmpot
Mihailo Rabasović
44
sLOvakIaStudent Peter Kosec
Patrik Svancara
Patrik Turzak
Andrej Vlcek
Samuel Beznak
Leader Ivo Cap
Lubomir Mucha
Observer Lubomir Konrad
Visitor Klara Capova
sLOvEnIaStudent Domen Ipavec
Matevž Marinčič
Jan Šuntajs
Jurij Tratar
Miha Zgubič
Leader Ciril Dominko
Jurij Bajc
sOuTh aFrIcaStudent Thiolan Prevan Naidoo
Avthar Sewrathan
Xolela Jara
Lloyd Mahadeo
Shihal Menesher Sapry
Leader Mervlyn Moodley
Bagtyyar Jorayev
45
sPaInStudent Roberto Alegre
Francesc-Xavier Gispert Sánchez
David Trillo Fernández
Marc Rodà Llordés
Aitor Azemar
Leader Juan Leon
Esperanza García-Carpintero Romero
srI LankaStudent Chanaka Manoj Singhabahu
Dombagaha Gedara Prasad Randika Maithriepala
Liraj Harsha Prabath Kodithuwakku
M. Janidu Chandrashantha Gunarathna
Edurapotha Gamaralalage Inoka Amanthie Dharmasena
Leader Ramal Vernil Coorey
surInamEStudent Chaandnie Bandhoe
Raynesh Kanhai
Priya Kasimbeg
Suraj Kishoen Misier
Leader Tjien Bing Tan
Ignaas Jimidar
Visitor Chantal Hewitt
Lachman Jurgen
46
swEdEnStudent Johan Runeson
Andréas Sundström
Carl Smed
Viktor Djurberg
Simon Johansson
Leader Max Kesselberg
Bo Söderberg
Observer Anne-Sofie Mårtensson
Visitor Margareta Kesselberg
swITzErLandStudent Thanh Phong Lê
Dominic Schwarz
Sebastian Käser
Laura Gremion
Christoph Schildknecht
Leader Lionel Philippoz
Simon Birrer
Observer Johanna Nyffeler
syrIan arab rEPubLIc Student Ghadeer Shaaban
Osama Yaghi
Mohamad Nour Ahmad
Mohamed Alrazzouk
Leader Akil Salloum
Observer Farkad Alramadani
47
TaIwanStudent Kai-Chi Huang
Jun-Ting Hsieh
Wei-Jen Ko
Yu-Ting Liu
Chien-An Wang
Leader Chih-Ta Chia
Shang-Fang Tsai
Observer Chung-Yu Mou
Tzong-Jer Yang
Chon Saar Chu
Jiun-Huei Wu
Yen-Chen Yu
TaJIkIsTanStudent Abdukhomid Nurmatov
Rabboni Bafoev
Adhamzhon Shukurov
Shakhzodi Rustamdzhon
Isfandiyor Safarov
Leader Ilkhom Khotami
ThaILandStudent Pongsapuk Sawaddirak
Puthipong Worasaran
Paphop Sawasdee
Supanut Thanasilp
Nathanan Tantivasadakarn
Leader Sirapat Pratontep
Phichet Kittara
Observer Suwan Kusamran
Raksapol Thananuwong
48
TurkEy Student Atinc Cagan Sengul
Oguzhan Can
Abdurrahman Akkas
Mustafa Selman Akinci
Mehmet Said Onay
Leader Ibrahim Gunal
Onur Ozcan
TurkmEnIsTanStudent Mekan Toyjanov
Meylis Malikov
Kemal Babayev
Agajan Odayev
Övezmyrat Övezmyradow
Leader Halit Coşkun
Gylychmammet Orazov
ukraInEStudent Volodymyr Sivak
Vsevolod Bykov
Vladysslav Diachenko
Volodymyr Rozsokhovatskyi
Yevgen Cherniavskyi
Leader Boris Kreminskyi
Stanislav Vilchynskyi
Observer Bushtruk Artem
49
unITEd kIngdOmStudent Adam Brown
Richard Thorburn
Peter Budden
Eric Wieser
Frank Bloomfield
Leader Robin Hughes
Paul Nicholls
Observer Sian Owen
Visitor Muriel Irene Hughes
unITEd sTaTEs OF amErIcaStudent Allan Sadun
Eric Schneider
Jeffrey Cai
Jeffrey Yan
Kevin Zhou
Leader Paul Stanley
Andrew Lin
vIETnamStudent Xuan Hien Bui
Viet Thang Dinh
Phi Long Ngo
Ngoc Hai Dinh
Huy Quang Le
Leader The Khoi Nguyen
Minh Thi Tran
Observer Van Vinh Le
Van Pham Tran
Quang Tuan Ngo
Thai Hoc Bui
Van Vu Ha
Xuan Thanh Ha
50
51Photos by Andres Mihkeson, Siim Pille and Merily Salura
52
53Photos by Andres Mihkeson, Siim Pille and Merily Salura
54
OrganizerssTEErIng cOmmITTEE
ChairmanJanar Holm Estonian Ministry of Education and Research
SecretaryViire Sepp Gifted and Talented Development Centre
of University Tartu
MembersErgo Nõmmiste University of Tartu, Institute of Physics
Jaak Kikas University of Tartu
Jaan Kalda Tallinn University of Technology, Institute of Cybernetics
Jakob Kübarsepp Tallinn University of Technology
Kaido Reivelt Estonian Physical Society
Kristjan Haller University of Tartu
Marco Kirm University of Tartu, Institute of Physics
Peeter Saari Estonian Academy of Sciences
Rait Toompere Archimedes Foundation
Raivo Stern National Institute of Chemical Physics and Biophysics
Toomas Sõmera Estonian Information Technology Foundation
Ülle Kikas Estonian Ministry of Education and Research
55
acadEmIc cOmmITTEEJaak Kikas Head of the Academic Committee
Head of Experimental Examination
Jaan Kalda Head of Theoretical Examination
Alar Ainla
Eero Uustalu
Endel Soolo
Mihkel Heidelberg
Oleg Košik
Rünno Lõhmus
Siim Ainsaar
Stanislav Zavjalov
Taavi Adamberg
OrganIzIng cOmmITTEEEne Koitla Head of the Organizing Committee
Marily Hendrikson Project Manager
Annika Vihul Head of accounting,
transportation and accommodation
Eneli Sutt Head of information technology
Kerli Kusnets Head of media
Malle Tragon Head of events and catering
Anna Gureeva Heads of group leaders
Julia Šmakova Heads of group leaders
56
markErsHelle Kaasik Head of the Markers Team
Aigar Vaigu Estonia
Alar Ainla Estonia
Aleksandr Bitjukov Estonia
Aleksandr Morozenko Estonia
Aleksandr Pištšev Estonia
Andreas Valdmann Estonia
Andres Jaanson Estonia
Anna-Stiina Suur-Uski Finland
Ants Remm Estonia
Antti Karjalainen Finland
Arvo Mere Estonia
Bahar Mehmani Germany
Christian Laut Ebbesen Denmark
Eemeli Samuel Tomberg Finland
Eero Vaher Estonia
Endel Soolo Estonia
Erik Paemurru Estonia
Filip Studnička Czech Republic
Gleb Široki Estonia
Gyula Honyek Hungary
Hannu Jaakko Lauri Siikonen Finland
Heiki Niglas Estonia
Helle Kaasik Estonia
Henri Johannes Ylitie Finland
Herry Kwee Indonesia
Jaak Jaaniste Estonia
Jaakko Uusitalo Finland
Jaan Katus Estonia
Jaanus Sepp Estonia
Juho Kahala Finland
Kadi Liis Saar Estonia
Kert Pütsepp Estonia
Klára Baranyai Hungary
57
Kristian Kuppart Estonia
Madis Ollikainen Estonia
Maksim Säkki Estonia
Markko Paas Estonia
Mihkel Pajusalu Estonia
Mihkel Rähn Estonia
Mikko Ervasti Finland
Oleg Košik Estonia
Oleksii Chechkin Ukraine
Otso Olavi Ossian Huuska Finland
Ottb Rebane Estonia
Rauno Siinmaa Estonia
Reio Põder Estonia
Riho Taba Estonia
Roland Matt Estonia
Sami Kivistö Finland
Sanli Faez Germany
Shahabedin Chatraee Islamic Republic of Iran
Stanislav Zavjalov Estonia
Zainul Abidin Indonesia
Taavi Vaikjärv Estonia
Teemu Johannes Hynninen Finland
Tiit Sepp Estonia
Timo Olli Johannes Voipio Finland
Valter Kiisk Estonia
Vasja Susič Slovenia
Ville Suur-Uski Finland
58
vOLunTEErsAgnes Vask
Airike Jõesaar
Allan-Cristjan Puks
Anastassia Samovitš
Andreas Ragen Ayal
Andres Ainelo
Andres Allik
Andres Mihkelson
Anete Merilin Leetberg
Anete Sammler
Anete Viise
Anna Dunajeva
Anna Jazõkova
Anna Krajuškina
Annemari Sepp
Anni Müüripeal
Anni Sandra Varblane
Annika Lukner
Annika Pille
Ants Johanson
Anu Viks
Artur Panov
Arvo Ehrstein
Ats Kurvet
Auli Relve
Ave-Stina Udam
Bety Mehide
Brenda Rauniste
Diana Oidingu
Dmitri Lanevski
Donatas Braziulis
Egert Vinogradov
Elina Libek
Enna Elismäe
Eno Paenurk
Erik Ilbis
Eva Mõtshärg
Evelin Pihlap
Eveli Soo
Gerli Krjukov
Gerli Vaik
Gertrud Metsa
Grete Helena Roose
Grethe Aikevitšius
Hanna Britt Soots
Hanna Kadri Metsvaht
Hanna Kivila
Hanna Moor
Hanna-Loore Hansen
Hedvig Tamman
Helbe-Laura Nikitkina
Helena Ainsoo
Helena Talimaa
Heli Aomets
Heli Pärn
Henry Teigar
Ida Rahu
Inger Kangur
Ivo Kruusamägi
Jaanika Jensen
Jane Lihtmaa
Janne Disko
Jasper Kursk
Jelizaveta Dõljova
Jelizaveta Žatkina
Johanna-Maria Muuga
Joonas Jäme
Jorma Veiderpass
59
Juhani Almers
Julia Gavrilova
Kadi Ainsaar
Kadi Külasalu
Kadri Alumets
Kadri Ann Rebane
Kadri Eek
Kadri Tinn
Kaisa Jõgi
Karl Kütt
Karl Veskus
Karl-Mattias Tepp
Karolin Rõõm
Kaspar Märtens
Kati Randmäe
Katrin Tuude
Keidi Suursaar
Ken Riisalu
Kerli Kalk
Kerstin Kivila
Krista Kallavus
Kristi Kartus
Kristiina Štõkova
Kristin Ehala
Kristin Liiksaar
Kristine Diane Liive
Kristine Leetberg
Kristjan Kalve
Ksenia Kukuskina
Laura Liisa Lankei
Laura Soon
Liina Nõmm
Liis Kass
Liis Nurmis
Liis Talimaa
Liisa Hunt
Liisa Veerus
Liisi Liivalaid
Liisi Mõtshärg
Liisi Sünd
Liisu Miller
Lisett Kiudorv
Lona-Liisa Sutt
Ly Pärnaste
Maksim Ivanov
Maksim Mišin
Marek Järvik
Maria Krajuškina
Mari-Liis Jaansalu
Mariliis Maamägi
Mari-Liis Tamm
Maris Ertmann
Maris Palo
Marit Puusepp
Mark Gimbutas
Marren Tiivits
Mart Ernits
Marta Tanaga
Marta Vihtre
Mary-Ann Kubre
Merilin Kalavus
Merilin Vesingi
Merily Salura
Merle Lust
Merlin Russak
Mette-Triin Purde
Mihkel Lepik
Mihkel Tali
Minna-Triin Kohv
Mirjam Laurimäe
60
Mirjam Mikk
Morten Piibeleht
Natalia Nekrassova
Nele Kriisa
Olga Bulgakova
Oliver Grauberg
Ott Kekišev
Paap Koemets
Paul Liias
Pille-Riin Peet
Raimo Armus
Rando Porosk
Rasmus Kuusmann
Reile Juhanson
Rene Rünt
Riinu Ansper
Rudolf Bichele
Saile Mägi
Sander Benga
Sander Kütisaar
Sander Soo
Sander Udam
Sandhra-Mirella Valdma
Sergei Jakovlev
Siim Kaspar Uustalu
Siim Pille
Siiri Mägi
Sille Hausenberg
Simona Kalatšov
Sirje Kollom
Sten Aus
Stina Avvo
Teisi Timma
Terje Kapp
Tiina Pärtel
Tiina Turban
Triin Rebane
Triin Ärm
Triinu Hordo
Uku-Kaspar Uustalu
Ulla Meeri Liivamägi
Urmet Paloveer
Üllar Kivila
61Photos by Siim Pille, Andres Mihkeson and Henry Teigar
62
63
ProgramsStudentssunday, 15Th OF JuLyArrival and registration at Sokos Hotel Viru
17:00 – 17:30 Departure from the hotel,
transportation to Open Air Museum
18:00 – 21:00 Icebreaking – Estonian Open Air Museum
21:00 – 22:00 Arrival to the hotel
mOnday, 16Th OF JuLy07:00 – 08:00 Breakfast
08:00 – 09:00 Departure from the hotel, putting luggage onto buses
Mobile phones and laptops being collected by the organisers
09:15 – 09:45 Walk to Opening Ceremony
10:00 – 12:00 Opening Ceremony – NOKIA Concert Hall
The Ceremony will be broadcasted over the Internet
and videotaped.
12:00 – 13:30 Welcome Banquet – NOKIA Concert Hall
13:30 – 17:00 Transportation to Tartu hotels
Opening ceremony Photo by Andres Mihkeson
64
17:30 – 19:00 Free time and preparation for theory
19:00 – 21:00 Dinner – Restaurant Dorpat
TuEsday, 17Th OF JuLy06:00 – 07:30 Breakfast
08:00 Transportation to Theoretical Examination
09:00 – 14:00 Theoretical Examination –
Sports building of Estonian University of Life Sciences
14:00 – 16:00 Transportation and lunch – Tartu Adventure Park
15:00 – 19:00 Games & activities – Tartu Adventure Park
19:00 – 21:00 Dinner - Restaurant Dorpat
wEdnEsday, 18Th OF JuLy07:00 – 09:00 Breakfast
09:00 – 20:00 Excursion: Rakvere Castle
20:00 – 23:00 Free time and preparation for experiment
Thursday, 19Th OF JuLy06:00 – 06:45 Students group A: Breakfast
07:00 – 09:00 Students group B: Breakfast
07:00 Students group A: Departure to Experimental Examination
08:00 – 13:00 Students group A: Experimental Examination
09:00 – 12:30 Students group B: Excursion: AHHAA Science Centre
12:30 – 13:50 Students group B: Lunch – AHHAA Science Centre
13:00 – 14:30 Students group A: Lunch – Restaurant Dorpat
students in Tartu adventure Park Photo by Siim Pille
65
13:50 Students group B: Departure to Experimental Examination
14:30 – 17:30 Students group A: Excursion: AHHAA Science Centre
15:00 – 20:00 Students group B: Experimental Examination
17:30 – 19:00 Students group A: Free time
19:00 – 21:00 Students group A: Dinner – Restaurant Volga
20:00 – 22:00 Students group B: Dinner – Restaurant Atlantis
21:30 – 22:30 Students group A: Skype meeting with Leaders
22:30 – 23:30 Students group B: Skype meeting with Leaders
FrIday, 20Th OF JuLyTartu – the World Capital of Physics
08:00 – 10:00 Breakfast
10:00 – 17:00 Tartu – the World Capital of Physics –
public science activities
13:00 – 15:00 Lunch at Tartu restaurants
17:00 – 18:00 Lecture: Sir Harold Kroto
(The 1996 Nobel Prize in Chemistry) – Vanemuise Concert Hall
18:00 – 20:00 Reception by Mayor of Tartu – Vanemuise Concert Hall
saTurday, 21sT OF JuLy08:00 – 10:00 Breakfast
10:00 – 13:00 Transportation to Tallinn
13:00 – … Free time; lunch and dinner at Tallinn restaurants
Exam Photos by Siim Pille
66
sunday, 22nd OF JuLy07:00 – 09:00 Breakfast
09:00 – 13:00 Football tournament
13:00 – 15:00 Lunch – Football Stadium
14:00 – 18:00 Football tournament continued
18:00 – 23:00 Free time and dinner at Tallinn restaurants
mOnday, 23rd OF JuLy07.00 – 09.00 Breakfast
09.00 – 13.00 Free time
13.15 – 13.45 Walk to Closing Ceremony
14.00 – 17.00 Closing Ceremony – NOKIA Concert Hall
The Ceremony will be broadcasted over the Internet
and videotaped.
17.00 – 17.30 Walk to the hotel
18.00 Transportation to the Farewall Party
18.30 – 01.00 Farewell Party – The Tallinn Song Festival Grounds
23.00 – … Round-the-clock transportation back to the hotel
TuEsday, 24Th OF JuLyDeparture
Football award. Football tournament Photos by Henry Teigar
67
Leaders and Observerssunday, 15Th OF JuLyArrival and Registration at Radisson Blu Hotel Olümpia
17:00 – 17:30 Departure from the hotel, transportation to Open Air Museum
18:00 – 21:00 Icebreaking - Estonian Open Air Museum
21:00 – 22:00 Arrival to the hotel
mOnday, 16Th OF JuLy07:00 – 09:00 Breakfast
09:15 – 09:45 Walk to Opening Ceremony
10:00 – 12:00 Opening Ceremony – NOKIA Concert Hall
The Ceremony will be broadcasted over the Internet
and videotaped.
12.00 – 13.30 Welcome Banquet – NOKIA Concert Hall
13:30 – 14:00 Walk to the hotel
14:00 – 19:00 International Board Meeting: Discussion of theoretical
problems – Radisson Blu Hotel Olümpia Conference Centre
19:00 – 21:00 Dinner – Restaurant Senso
21:00 – … International Board Meeting:
Translation of theoretical problems
students in rakvere castle Photo by Siim Pille
68
TuEsday, 17Th OF JuLy06:00 – 07:00 Breakfast
07:00 Departure from the hotel
07:00 – 10:00 Excursion: trip to Saaremaa
10:00 – 13:00 Excursion: Kaali crater and Kuressaare
13:00 – 14:30 Lunch - Mändjala Camping
14:30 – 20:00 Excursion: Saaremaa and Muhu
20:00 – 21:30 Arrival to the hotel and dinner - Restaurant Senso
20:30 – 21:30 Distribution of theory papers from the IPhO office
wEdnEsday, 18Th OF JuLy07:00 – 09:00 Breakfast
09:00 – 12:00 Free time
11:30 – 13:00 Lunch - Restaurant Senso
13:00 – 19:00 International Board Meeting:
Discussion of experimental problems
19:00 – 21:00 Dinner – Restaurant Senso
21:00 – … International Board Meeting:
Translation of experimental problems
Thursday, 19Th OF JuLy07:00 – 09:00 Breakfast
09:00 – 12:00 Free time
12:00 – 13:00 Collection of marks from Leaders (theory) – Online
13:00 – 15:00 Lunch - Restaurant Senso
Estonian Open air museum. skype meeting with students Photos by Siim Pille, Henry Teigar
69
15:00 – 19:00 Free time
19:00 – 21:00 Leaders group A: Distribution of practical papers
from IPhO office
19:00 – 21:00 Dinner – Restaurant Senso
21:30 – 22:30 Leaders group A: Skype meeting with Students
22:30 – 23:30 Leaders group B: Skype meeting with Students
01:00 – 02:00 Leaders group B: Distribution of practical papers
from IPhO office
FrIday, 20Th OF JuLyTartu – the World Capital of Physics
07:00 – 09:00 Breakfast
07:00 – 08:00 Leaders group B: Distribution of practical papers
from IPhO office
09:00 – 11:30 Transportation to Tartu
12:00 – 17:00 Tartu – the World Capital of Physics –
Public science activities
13:00 – 15:00 Lunch at Tartu restaurants
17:00 – 18:00 Lecture: Sir Harold Kroto
(The 1996 Nobel Prize in Chemistry) - Vanemuise Concert Hall
18:00 – 20:00 Reception by Mayor of Tartu – Vanemuise Concert Hall
20:00 – 22:30 Transportation to Tallinn
22:30 – 23:30 Collection of marks from Leaders (experiment) – Online
visiting ahhaa science center Photo by Siim Pille
70
saTurday, 21sT OF JuLy07:00 – 09:00 Breakfast
10:00 – 12:00 International Board Meeting
11:00 Distribution of marks (theory) - Online
12:00 – 14:00 Lunch - Restaurant Senso
14:00 – 21:00 International Board Meeting:
Moderation of theoretical papers
19:00 Distribution of marks (experiment) - Online
19:00 – 21:00 Dinner – Restaurant Senso
sunday, 22nd OF JuLy07:00 – 09:00 Breakfast
09:00 – 17:00 International Board Meeting:
Moderation of experimental papers
12:00 – 14:00 Lunch – Restaurant Senso
17:00 – 19:00 International Board Meeting:
Deciding final marks and medals
19:00 – … Free time and dinner at Tallinn restaurants
Farwell Party Photo by Merily Salura
71
mOnday, 23rd OF JuLy07:00 – 09:00 Breakfast
09:00 – 13:00 Free time
13:15 – 13:45 Walk to Closing Ceremony – NOKIA Concert Hall
14:00 – 17:00 Closing Ceremony
The Ceremony will be broadcasted over the Internet
and videotaped.
17:00 – 17:30 Walk back to the hotel
18:00 Transportation to the Farewell Party
18:30 – 01:00 Farewell Party - The Tallinn Song Festival Grounds
23:00 – … Round-the-clock transportation back to the hotel
TuEsday, 24Th OF JuLyDeparture
closing ceremony Photo by Merily Salura
72
73
Problems and solutionsThe 43rd International Physics Olympiad — Theoretical CompetitionTartu, Estonia — Thursday, July 17th 2012The 43rd International Physics Olympiad — Theoretical Competition
Tartu, Estonia — Tuesday, July 17th 2012
• The examination lasts for 5 hours. There are 3 problems
worth in total 30 points. Please note that the point
values of the three theoretical problems are not
equal.
• You must not open the envelope with the prob-
lems before the sound signal of the beginning of
competition (three short signals).
• You are not allowed to leave your working place
without permission. If you need any assistance
(broken calculator, need to visit a restroom, etc), please
raise the corresponding flag (“help” or “toilet” with a
long handle at your seat) above your seat box walls and
keep it raised until an organizer arrives.
• Your answers must be expressed in terms of those
quantities, which are highlighted in the problem text,
and can contain also fundamental constants, if needed.
So, if it is written that “the box height is a and the
width — b” then a can be used in the answer, and b
cannot be used (unless it is highlighted somewhere else,
see below). Those quantities which are highlighted in
the text of a subquestion can be used only in the answer
to that subquestion; the quantities which are highlighted
in the introductory text of the Problem (or a Part of a
Problem), i.e. outside the scope of any subquestion, can
be used for all the answers of that Problem (or of that
Problem Part).
• Use only the front side of the sheets of paper.
• For each problem, there are dedicated Solution Sheets
(see header for the number and pictogramme). Write
your solutions onto the appropriate Solution Sheets. For
each Problem, the Solution Sheets are numbered; use
the sheets according to the enumeration. Always mark
which Problem Part and Question you are deal-
ing with. Copy the final answers into the appropriate
boxes of the Answer Sheets. There are also Draft pa-
pers; use these for writing things which you don’t want
to be graded. If you have written something what you
don’t want to be graded onto the Solution Sheets (such
as initial and incorrect solutions), cross these out.
• If you need more paper for a certain problem, please raise
the flag “help” and tell an organizer the problem num-
ber; you are given two Solution sheets (you can do this
more than once).
• You should use as little text as possible: try to ex-
plain your solution mainly with equations, numbers, sym-
bols and diagrams. When textual explanation is un-
avoidable, you are encouraged to provide English
translation alongside with the text in your native
language (if you mistranslate, or don’t translate
at all, your native language text will be used dur-
ing the Moderation).
• The first single sound signal tells you that there are 30
min of solving time left; the second double sound signal
means that 5 min is left; the third triple sound signal
marks the end of solving time. After the third sound
signal you must stop writing immediately. Put all
the papers into the envelope at your desk. You are not
allowed to take any sheet of paper out of the room.
If you have finished solving before the final sound signal,
please raise your flag.
— page 1 of 5 —
74
The 43rd International Physics Olympiad — Theoretical Competition
Tartu, Estonia — Tuesday, July 17th 2012
• The examination lasts for 5 hours. There are 3 problems
worth in total 30 points. Please note that the point
values of the three theoretical problems are not
equal.
• You must not open the envelope with the prob-
lems before the sound signal of the beginning of
competition (three short signals).
• You are not allowed to leave your working place
without permission. If you need any assistance
(broken calculator, need to visit a restroom, etc), please
raise the corresponding flag (“help” or “toilet” with a
long handle at your seat) above your seat box walls and
keep it raised until an organizer arrives.
• Your answers must be expressed in terms of those
quantities, which are highlighted in the problem text,
and can contain also fundamental constants, if needed.
So, if it is written that “the box height is a and the
width — b” then a can be used in the answer, and b
cannot be used (unless it is highlighted somewhere else,
see below). Those quantities which are highlighted in
the text of a subquestion can be used only in the answer
to that subquestion; the quantities which are highlighted
in the introductory text of the Problem (or a Part of a
Problem), i.e. outside the scope of any subquestion, can
be used for all the answers of that Problem (or of that
Problem Part).
• Use only the front side of the sheets of paper.
• For each problem, there are dedicated Solution Sheets
(see header for the number and pictogramme). Write
your solutions onto the appropriate Solution Sheets. For
each Problem, the Solution Sheets are numbered; use
the sheets according to the enumeration. Always mark
which Problem Part and Question you are deal-
ing with. Copy the final answers into the appropriate
boxes of the Answer Sheets. There are also Draft pa-
pers; use these for writing things which you don’t want
to be graded. If you have written something what you
don’t want to be graded onto the Solution Sheets (such
as initial and incorrect solutions), cross these out.
• If you need more paper for a certain problem, please raise
the flag “help” and tell an organizer the problem num-
ber; you are given two Solution sheets (you can do this
more than once).
• You should use as little text as possible: try to ex-
plain your solution mainly with equations, numbers, sym-
bols and diagrams. When textual explanation is un-
avoidable, you are encouraged to provide English
translation alongside with the text in your native
language (if you mistranslate, or don’t translate
at all, your native language text will be used dur-
ing the Moderation).
• The first single sound signal tells you that there are 30
min of solving time left; the second double sound signal
means that 5 min is left; the third triple sound signal
marks the end of solving time. After the third sound
signal you must stop writing immediately. Put all
the papers into the envelope at your desk. You are not
allowed to take any sheet of paper out of the room.
If you have finished solving before the final sound signal,
please raise your flag.
— page 1 of 5 —
The 43rd International Physics Olympiad — Theoretical Competition
Tartu, Estonia — Tuesday, July 17th 2012
• The examination lasts for 5 hours. There are 3 problems
worth in total 30 points. Please note that the point
values of the three theoretical problems are not
equal.
• You must not open the envelope with the prob-
lems before the sound signal of the beginning of
competition (three short signals).
• You are not allowed to leave your working place
without permission. If you need any assistance
(broken calculator, need to visit a restroom, etc), please
raise the corresponding flag (“help” or “toilet” with a
long handle at your seat) above your seat box walls and
keep it raised until an organizer arrives.
• Your answers must be expressed in terms of those
quantities, which are highlighted in the problem text,
and can contain also fundamental constants, if needed.
So, if it is written that “the box height is a and the
width — b” then a can be used in the answer, and b
cannot be used (unless it is highlighted somewhere else,
see below). Those quantities which are highlighted in
the text of a subquestion can be used only in the answer
to that subquestion; the quantities which are highlighted
in the introductory text of the Problem (or a Part of a
Problem), i.e. outside the scope of any subquestion, can
be used for all the answers of that Problem (or of that
Problem Part).
• Use only the front side of the sheets of paper.
• For each problem, there are dedicated Solution Sheets
(see header for the number and pictogramme). Write
your solutions onto the appropriate Solution Sheets. For
each Problem, the Solution Sheets are numbered; use
the sheets according to the enumeration. Always mark
which Problem Part and Question you are deal-
ing with. Copy the final answers into the appropriate
boxes of the Answer Sheets. There are also Draft pa-
pers; use these for writing things which you don’t want
to be graded. If you have written something what you
don’t want to be graded onto the Solution Sheets (such
as initial and incorrect solutions), cross these out.
• If you need more paper for a certain problem, please raise
the flag “help” and tell an organizer the problem num-
ber; you are given two Solution sheets (you can do this
more than once).
• You should use as little text as possible: try to ex-
plain your solution mainly with equations, numbers, sym-
bols and diagrams. When textual explanation is un-
avoidable, you are encouraged to provide English
translation alongside with the text in your native
language (if you mistranslate, or don’t translate
at all, your native language text will be used dur-
ing the Moderation).
• The first single sound signal tells you that there are 30
min of solving time left; the second double sound signal
means that 5 min is left; the third triple sound signal
marks the end of solving time. After the third sound
signal you must stop writing immediately. Put all
the papers into the envelope at your desk. You are not
allowed to take any sheet of paper out of the room.
If you have finished solving before the final sound signal,
please raise your flag.
— page 1 of 5 —
75
The 43rd International Physics Olympiad — Theoretical Competition
Tartu, Estonia — Tuesday, July 17th 2012
• The examination lasts for 5 hours. There are 3 problems
worth in total 30 points. Please note that the point
values of the three theoretical problems are not
equal.
• You must not open the envelope with the prob-
lems before the sound signal of the beginning of
competition (three short signals).
• You are not allowed to leave your working place
without permission. If you need any assistance
(broken calculator, need to visit a restroom, etc), please
raise the corresponding flag (“help” or “toilet” with a
long handle at your seat) above your seat box walls and
keep it raised until an organizer arrives.
• Your answers must be expressed in terms of those
quantities, which are highlighted in the problem text,
and can contain also fundamental constants, if needed.
So, if it is written that “the box height is a and the
width — b” then a can be used in the answer, and b
cannot be used (unless it is highlighted somewhere else,
see below). Those quantities which are highlighted in
the text of a subquestion can be used only in the answer
to that subquestion; the quantities which are highlighted
in the introductory text of the Problem (or a Part of a
Problem), i.e. outside the scope of any subquestion, can
be used for all the answers of that Problem (or of that
Problem Part).
• Use only the front side of the sheets of paper.
• For each problem, there are dedicated Solution Sheets
(see header for the number and pictogramme). Write
your solutions onto the appropriate Solution Sheets. For
each Problem, the Solution Sheets are numbered; use
the sheets according to the enumeration. Always mark
which Problem Part and Question you are deal-
ing with. Copy the final answers into the appropriate
boxes of the Answer Sheets. There are also Draft pa-
pers; use these for writing things which you don’t want
to be graded. If you have written something what you
don’t want to be graded onto the Solution Sheets (such
as initial and incorrect solutions), cross these out.
• If you need more paper for a certain problem, please raise
the flag “help” and tell an organizer the problem num-
ber; you are given two Solution sheets (you can do this
more than once).
• You should use as little text as possible: try to ex-
plain your solution mainly with equations, numbers, sym-
bols and diagrams. When textual explanation is un-
avoidable, you are encouraged to provide English
translation alongside with the text in your native
language (if you mistranslate, or don’t translate
at all, your native language text will be used dur-
ing the Moderation).
• The first single sound signal tells you that there are 30
min of solving time left; the second double sound signal
means that 5 min is left; the third triple sound signal
marks the end of solving time. After the third sound
signal you must stop writing immediately. Put all
the papers into the envelope at your desk. You are not
allowed to take any sheet of paper out of the room.
If you have finished solving before the final sound signal,
please raise your flag.
— page 1 of 5 —
Problem T1. Focus on sketches (13 points)
Part A. Ballistics (4.5 points)
A ball thrown with an initial speed v0 moves in a homogeneous
gravitational field in x−z plane, where the x-axis is horizontal,
and z — vertical, antiparallel to the free fall acceleration g ;
neglect the air drag.
i. (0.8 pts) By adjusting the launching angle for a ball thrown
with a fixed initial speed v0 from the origin, targets can be
hit within the region given by
z ≤ z0 − kx2;
you can use this fact without proving it. Find the constants z0
and k.
ii. (1.2 pts) Now, the launching point can be
freely selected on the ground level z = 0, and
the launching angle can be adjusted as needed;
the aim is to hit the topmost point of a spherical
building of radius R (see fig.) with as small as
possible initial speed v0 (prior hitting the target, bouncing off
the roof is not allowed). Sketch qualitatively the shape of the
optimal trajectory of the ball (use the designated box on the
answer sheet). Note: the points are given only for the sketch.
iii. (2.5 pts) What is the minimal launching speed vmin needed
to hit the topmost point of a spherical building of radius R ?
La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part
B. Air flow around a wing (4 points)
For this Problem Part, the following information may be
useful. For a flow of liquid or gas in a tube, along a stream-
line p + ρgh + 1
2ρv2 = const., assuming that the velocity v
is much smaller than the sound speed. Here ρ is the density,
h — height, g — free fall acceleration, and p — hydrostatic
pressure. Streamlines are defined as the trajectories of fluid
particles (assuming that the flow pattern is stationary). Note
that the term 1
2ρv2 is called the dynamic pressure.
In the fig. below, a cross-section of an aircraft wing is de-
picted together with streamlines of the air flow around the
wing, as seen in the wing’s reference frame. Assume that
(a) the air flow is purely two-dimensional (i.e. that the velo-
city vectors of air lie in the figure plane); (b) the streamline
pattern is independent of the aircraft speed; (c) there is no
wind; (d) the dynamic pressure is much smaller than the at-
mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to
take measurements from the fig. on the answer sheet .
i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,
what is the speed of the air vP at the point P (marked in fig.)
with respect to the ground?
ii. (1.2 pts) In the case of high relative humidity, as the ground
speed of the aircraft increases over a critical value vcrit, a stream
of water droplets is created behind the wing. The droplets
emerge at a certain point Q. Mark the point Q in fig. on the
answer sheet. Explain qualitatively (using formulae and as few
text as possible) how you determined its position.
iii. (2.0 pts) Estimate the critical speed vcrit using the follow-
ing data: relative humidity of the air is r = 90% , specific heat
of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure
of saturated water vapour: psa = 2.31 kPa at the temperat-
ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa
at Tb = 294 K . Depending on your approximations you
may also need the specific heat of air at constant volume
cV = 0.717 × 103 J/kg · K . Note that the relative humidity is
defined as the ratio of the vapor pressure to the saturated vapor
pressure at the given temperature. Saturated vapor pressure is
defined as the vapor pressure by which vapor is in equilibrium
with the liquid.
— page 2 of 5 —
Problem T1. Focus on sketches (13 points)
Part A. Ballistics (4.5 points)
A ball thrown with an initial speed v0 moves in a homogeneous
gravitational field in x−z plane, where the x-axis is horizontal,
and z — vertical, antiparallel to the free fall acceleration g ;
neglect the air drag.
i. (0.8 pts) By adjusting the launching angle for a ball thrown
with a fixed initial speed v0 from the origin, targets can be
hit within the region given by
z ≤ z0 − kx2;
you can use this fact without proving it. Find the constants z0
and k.
ii. (1.2 pts) Now, the launching point can be
freely selected on the ground level z = 0, and
the launching angle can be adjusted as needed;
the aim is to hit the topmost point of a spherical
building of radius R (see fig.) with as small as
possible initial speed v0 (prior hitting the target, bouncing off
the roof is not allowed). Sketch qualitatively the shape of the
optimal trajectory of the ball (use the designated box on the
answer sheet). Note: the points are given only for the sketch.
iii. (2.5 pts) What is the minimal launching speed vmin needed
to hit the topmost point of a spherical building of radius R ?
La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part
B. Air flow around a wing (4 points)
For this Problem Part, the following information may be
useful. For a flow of liquid or gas in a tube, along a stream-
line p + ρgh + 1
2ρv2 = const., assuming that the velocity v
is much smaller than the sound speed. Here ρ is the density,
h — height, g — free fall acceleration, and p — hydrostatic
pressure. Streamlines are defined as the trajectories of fluid
particles (assuming that the flow pattern is stationary). Note
that the term 1
2ρv2 is called the dynamic pressure.
In the fig. below, a cross-section of an aircraft wing is de-
picted together with streamlines of the air flow around the
wing, as seen in the wing’s reference frame. Assume that
(a) the air flow is purely two-dimensional (i.e. that the velo-
city vectors of air lie in the figure plane); (b) the streamline
pattern is independent of the aircraft speed; (c) there is no
wind; (d) the dynamic pressure is much smaller than the at-
mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to
take measurements from the fig. on the answer sheet .
i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,
what is the speed of the air vP at the point P (marked in fig.)
with respect to the ground?
ii. (1.2 pts) In the case of high relative humidity, as the ground
speed of the aircraft increases over a critical value vcrit, a stream
of water droplets is created behind the wing. The droplets
emerge at a certain point Q. Mark the point Q in fig. on the
answer sheet. Explain qualitatively (using formulae and as few
text as possible) how you determined its position.
iii. (2.0 pts) Estimate the critical speed vcrit using the follow-
ing data: relative humidity of the air is r = 90% , specific heat
of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure
of saturated water vapour: psa = 2.31 kPa at the temperat-
ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa
at Tb = 294 K . Depending on your approximations you
may also need the specific heat of air at constant volume
cV = 0.717 × 103 J/kg · K . Note that the relative humidity is
defined as the ratio of the vapor pressure to the saturated vapor
pressure at the given temperature. Saturated vapor pressure is
defined as the vapor pressure by which vapor is in equilibrium
with the liquid.
— page 2 of 5 —
PrObLEm T1. FOcus On skETchEs (13 POInTs)
76
Problem T1. Focus on sketches (13 points)
Part A. Ballistics (4.5 points)
A ball thrown with an initial speed v0 moves in a homogeneous
gravitational field in x−z plane, where the x-axis is horizontal,
and z — vertical, antiparallel to the free fall acceleration g ;
neglect the air drag.
i. (0.8 pts) By adjusting the launching angle for a ball thrown
with a fixed initial speed v0 from the origin, targets can be
hit within the region given by
z ≤ z0 − kx2;
you can use this fact without proving it. Find the constants z0
and k.
ii. (1.2 pts) Now, the launching point can be
freely selected on the ground level z = 0, and
the launching angle can be adjusted as needed;
the aim is to hit the topmost point of a spherical
building of radius R (see fig.) with as small as
possible initial speed v0 (prior hitting the target, bouncing off
the roof is not allowed). Sketch qualitatively the shape of the
optimal trajectory of the ball (use the designated box on the
answer sheet). Note: the points are given only for the sketch.
iii. (2.5 pts) What is the minimal launching speed vmin needed
to hit the topmost point of a spherical building of radius R ?
La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part
B. Air flow around a wing (4 points)
For this Problem Part, the following information may be
useful. For a flow of liquid or gas in a tube, along a stream-
line p + ρgh + 1
2ρv2 = const., assuming that the velocity v
is much smaller than the sound speed. Here ρ is the density,
h — height, g — free fall acceleration, and p — hydrostatic
pressure. Streamlines are defined as the trajectories of fluid
particles (assuming that the flow pattern is stationary). Note
that the term 1
2ρv2 is called the dynamic pressure.
In the fig. below, a cross-section of an aircraft wing is de-
picted together with streamlines of the air flow around the
wing, as seen in the wing’s reference frame. Assume that
(a) the air flow is purely two-dimensional (i.e. that the velo-
city vectors of air lie in the figure plane); (b) the streamline
pattern is independent of the aircraft speed; (c) there is no
wind; (d) the dynamic pressure is much smaller than the at-
mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to
take measurements from the fig. on the answer sheet .
i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,
what is the speed of the air vP at the point P (marked in fig.)
with respect to the ground?
ii. (1.2 pts) In the case of high relative humidity, as the ground
speed of the aircraft increases over a critical value vcrit, a stream
of water droplets is created behind the wing. The droplets
emerge at a certain point Q. Mark the point Q in fig. on the
answer sheet. Explain qualitatively (using formulae and as few
text as possible) how you determined its position.
iii. (2.0 pts) Estimate the critical speed vcrit using the follow-
ing data: relative humidity of the air is r = 90% , specific heat
of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure
of saturated water vapour: psa = 2.31 kPa at the temperat-
ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa
at Tb = 294 K . Depending on your approximations you
may also need the specific heat of air at constant volume
cV = 0.717 × 103 J/kg · K . Note that the relative humidity is
defined as the ratio of the vapor pressure to the saturated vapor
pressure at the given temperature. Saturated vapor pressure is
defined as the vapor pressure by which vapor is in equilibrium
with the liquid.
— page 2 of 5 —
Problem T1. Focus on sketches (13 points)
Part A. Ballistics (4.5 points)
A ball thrown with an initial speed v0 moves in a homogeneous
gravitational field in x−z plane, where the x-axis is horizontal,
and z — vertical, antiparallel to the free fall acceleration g ;
neglect the air drag.
i. (0.8 pts) By adjusting the launching angle for a ball thrown
with a fixed initial speed v0 from the origin, targets can be
hit within the region given by
z ≤ z0 − kx2;
you can use this fact without proving it. Find the constants z0
and k.
ii. (1.2 pts) Now, the launching point can be
freely selected on the ground level z = 0, and
the launching angle can be adjusted as needed;
the aim is to hit the topmost point of a spherical
building of radius R (see fig.) with as small as
possible initial speed v0 (prior hitting the target, bouncing off
the roof is not allowed). Sketch qualitatively the shape of the
optimal trajectory of the ball (use the designated box on the
answer sheet). Note: the points are given only for the sketch.
iii. (2.5 pts) What is the minimal launching speed vmin needed
to hit the topmost point of a spherical building of radius R ?
La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part
B. Air flow around a wing (4 points)
For this Problem Part, the following information may be
useful. For a flow of liquid or gas in a tube, along a stream-
line p + ρgh + 1
2ρv2 = const., assuming that the velocity v
is much smaller than the sound speed. Here ρ is the density,
h — height, g — free fall acceleration, and p — hydrostatic
pressure. Streamlines are defined as the trajectories of fluid
particles (assuming that the flow pattern is stationary). Note
that the term 1
2ρv2 is called the dynamic pressure.
In the fig. below, a cross-section of an aircraft wing is de-
picted together with streamlines of the air flow around the
wing, as seen in the wing’s reference frame. Assume that
(a) the air flow is purely two-dimensional (i.e. that the velo-
city vectors of air lie in the figure plane); (b) the streamline
pattern is independent of the aircraft speed; (c) there is no
wind; (d) the dynamic pressure is much smaller than the at-
mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to
take measurements from the fig. on the answer sheet .
i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,
what is the speed of the air vP at the point P (marked in fig.)
with respect to the ground?
ii. (1.2 pts) In the case of high relative humidity, as the ground
speed of the aircraft increases over a critical value vcrit, a stream
of water droplets is created behind the wing. The droplets
emerge at a certain point Q. Mark the point Q in fig. on the
answer sheet. Explain qualitatively (using formulae and as few
text as possible) how you determined its position.
iii. (2.0 pts) Estimate the critical speed vcrit using the follow-
ing data: relative humidity of the air is r = 90% , specific heat
of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure
of saturated water vapour: psa = 2.31 kPa at the temperat-
ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa
at Tb = 294 K . Depending on your approximations you
may also need the specific heat of air at constant volume
cV = 0.717 × 103 J/kg · K . Note that the relative humidity is
defined as the ratio of the vapor pressure to the saturated vapor
pressure at the given temperature. Saturated vapor pressure is
defined as the vapor pressure by which vapor is in equilibrium
with the liquid.
— page 2 of 5 —
Problem T1. Focus on sketches (13 points)
Part A. Ballistics (4.5 points)
A ball thrown with an initial speed v0 moves in a homogeneous
gravitational field in x−z plane, where the x-axis is horizontal,
and z — vertical, antiparallel to the free fall acceleration g ;
neglect the air drag.
i. (0.8 pts) By adjusting the launching angle for a ball thrown
with a fixed initial speed v0 from the origin, targets can be
hit within the region given by
z ≤ z0 − kx2;
you can use this fact without proving it. Find the constants z0
and k.
ii. (1.2 pts) Now, the launching point can be
freely selected on the ground level z = 0, and
the launching angle can be adjusted as needed;
the aim is to hit the topmost point of a spherical
building of radius R (see fig.) with as small as
possible initial speed v0 (prior hitting the target, bouncing off
the roof is not allowed). Sketch qualitatively the shape of the
optimal trajectory of the ball (use the designated box on the
answer sheet). Note: the points are given only for the sketch.
iii. (2.5 pts) What is the minimal launching speed vmin needed
to hit the topmost point of a spherical building of radius R ?
La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part
B. Air flow around a wing (4 points)
For this Problem Part, the following information may be
useful. For a flow of liquid or gas in a tube, along a stream-
line p + ρgh + 1
2ρv2 = const., assuming that the velocity v
is much smaller than the sound speed. Here ρ is the density,
h — height, g — free fall acceleration, and p — hydrostatic
pressure. Streamlines are defined as the trajectories of fluid
particles (assuming that the flow pattern is stationary). Note
that the term 1
2ρv2 is called the dynamic pressure.
In the fig. below, a cross-section of an aircraft wing is de-
picted together with streamlines of the air flow around the
wing, as seen in the wing’s reference frame. Assume that
(a) the air flow is purely two-dimensional (i.e. that the velo-
city vectors of air lie in the figure plane); (b) the streamline
pattern is independent of the aircraft speed; (c) there is no
wind; (d) the dynamic pressure is much smaller than the at-
mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to
take measurements from the fig. on the answer sheet .
i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,
what is the speed of the air vP at the point P (marked in fig.)
with respect to the ground?
ii. (1.2 pts) In the case of high relative humidity, as the ground
speed of the aircraft increases over a critical value vcrit, a stream
of water droplets is created behind the wing. The droplets
emerge at a certain point Q. Mark the point Q in fig. on the
answer sheet. Explain qualitatively (using formulae and as few
text as possible) how you determined its position.
iii. (2.0 pts) Estimate the critical speed vcrit using the follow-
ing data: relative humidity of the air is r = 90% , specific heat
of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure
of saturated water vapour: psa = 2.31 kPa at the temperat-
ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa
at Tb = 294 K . Depending on your approximations you
may also need the specific heat of air at constant volume
cV = 0.717 × 103 J/kg · K . Note that the relative humidity is
defined as the ratio of the vapor pressure to the saturated vapor
pressure at the given temperature. Saturated vapor pressure is
defined as the vapor pressure by which vapor is in equilibrium
with the liquid.
— page 2 of 5 —Problem T1. Focus on sketches (13 points)
Part A. Ballistics (4.5 points)
A ball thrown with an initial speed v0 moves in a homogeneous
gravitational field in x−z plane, where the x-axis is horizontal,
and z — vertical, antiparallel to the free fall acceleration g ;
neglect the air drag.
i. (0.8 pts) By adjusting the launching angle for a ball thrown
with a fixed initial speed v0 from the origin, targets can be
hit within the region given by
z ≤ z0 − kx2;
you can use this fact without proving it. Find the constants z0
and k.
ii. (1.2 pts) Now, the launching point can be
freely selected on the ground level z = 0, and
the launching angle can be adjusted as needed;
the aim is to hit the topmost point of a spherical
building of radius R (see fig.) with as small as
possible initial speed v0 (prior hitting the target, bouncing off
the roof is not allowed). Sketch qualitatively the shape of the
optimal trajectory of the ball (use the designated box on the
answer sheet). Note: the points are given only for the sketch.
iii. (2.5 pts) What is the minimal launching speed vmin needed
to hit the topmost point of a spherical building of radius R ?
La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part
B. Air flow around a wing (4 points)
For this Problem Part, the following information may be
useful. For a flow of liquid or gas in a tube, along a stream-
line p + ρgh + 1
2ρv2 = const., assuming that the velocity v
is much smaller than the sound speed. Here ρ is the density,
h — height, g — free fall acceleration, and p — hydrostatic
pressure. Streamlines are defined as the trajectories of fluid
particles (assuming that the flow pattern is stationary). Note
that the term 1
2ρv2 is called the dynamic pressure.
In the fig. below, a cross-section of an aircraft wing is de-
picted together with streamlines of the air flow around the
wing, as seen in the wing’s reference frame. Assume that
(a) the air flow is purely two-dimensional (i.e. that the velo-
city vectors of air lie in the figure plane); (b) the streamline
pattern is independent of the aircraft speed; (c) there is no
wind; (d) the dynamic pressure is much smaller than the at-
mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to
take measurements from the fig. on the answer sheet .
i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,
what is the speed of the air vP at the point P (marked in fig.)
with respect to the ground?
ii. (1.2 pts) In the case of high relative humidity, as the ground
speed of the aircraft increases over a critical value vcrit, a stream
of water droplets is created behind the wing. The droplets
emerge at a certain point Q. Mark the point Q in fig. on the
answer sheet. Explain qualitatively (using formulae and as few
text as possible) how you determined its position.
iii. (2.0 pts) Estimate the critical speed vcrit using the follow-
ing data: relative humidity of the air is r = 90% , specific heat
of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure
of saturated water vapour: psa = 2.31 kPa at the temperat-
ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa
at Tb = 294 K . Depending on your approximations you
may also need the specific heat of air at constant volume
cV = 0.717 × 103 J/kg · K . Note that the relative humidity is
defined as the ratio of the vapor pressure to the saturated vapor
pressure at the given temperature. Saturated vapor pressure is
defined as the vapor pressure by which vapor is in equilibrium
with the liquid.
— page 2 of 5 —
Problem T1. Focus on sketches (13 points)
Part A. Ballistics (4.5 points)
A ball thrown with an initial speed v0 moves in a homogeneous
gravitational field in x−z plane, where the x-axis is horizontal,
and z — vertical, antiparallel to the free fall acceleration g ;
neglect the air drag.
i. (0.8 pts) By adjusting the launching angle for a ball thrown
with a fixed initial speed v0 from the origin, targets can be
hit within the region given by
z ≤ z0 − kx2;
you can use this fact without proving it. Find the constants z0
and k.
ii. (1.2 pts) Now, the launching point can be
freely selected on the ground level z = 0, and
the launching angle can be adjusted as needed;
the aim is to hit the topmost point of a spherical
building of radius R (see fig.) with as small as
possible initial speed v0 (prior hitting the target, bouncing off
the roof is not allowed). Sketch qualitatively the shape of the
optimal trajectory of the ball (use the designated box on the
answer sheet). Note: the points are given only for the sketch.
iii. (2.5 pts) What is the minimal launching speed vmin needed
to hit the topmost point of a spherical building of radius R ?
La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part
B. Air flow around a wing (4 points)
For this Problem Part, the following information may be
useful. For a flow of liquid or gas in a tube, along a stream-
line p + ρgh + 1
2ρv2 = const., assuming that the velocity v
is much smaller than the sound speed. Here ρ is the density,
h — height, g — free fall acceleration, and p — hydrostatic
pressure. Streamlines are defined as the trajectories of fluid
particles (assuming that the flow pattern is stationary). Note
that the term 1
2ρv2 is called the dynamic pressure.
In the fig. below, a cross-section of an aircraft wing is de-
picted together with streamlines of the air flow around the
wing, as seen in the wing’s reference frame. Assume that
(a) the air flow is purely two-dimensional (i.e. that the velo-
city vectors of air lie in the figure plane); (b) the streamline
pattern is independent of the aircraft speed; (c) there is no
wind; (d) the dynamic pressure is much smaller than the at-
mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to
take measurements from the fig. on the answer sheet .
i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,
what is the speed of the air vP at the point P (marked in fig.)
with respect to the ground?
ii. (1.2 pts) In the case of high relative humidity, as the ground
speed of the aircraft increases over a critical value vcrit, a stream
of water droplets is created behind the wing. The droplets
emerge at a certain point Q. Mark the point Q in fig. on the
answer sheet. Explain qualitatively (using formulae and as few
text as possible) how you determined its position.
iii. (2.0 pts) Estimate the critical speed vcrit using the follow-
ing data: relative humidity of the air is r = 90% , specific heat
of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure
of saturated water vapour: psa = 2.31 kPa at the temperat-
ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa
at Tb = 294 K . Depending on your approximations you
may also need the specific heat of air at constant volume
cV = 0.717 × 103 J/kg · K . Note that the relative humidity is
defined as the ratio of the vapor pressure to the saturated vapor
pressure at the given temperature. Saturated vapor pressure is
defined as the vapor pressure by which vapor is in equilibrium
with the liquid.
— page 2 of 5 —
77
Problem T1. Focus on sketches (13 points)
Part A. Ballistics (4.5 points)
A ball thrown with an initial speed v0 moves in a homogeneous
gravitational field in x−z plane, where the x-axis is horizontal,
and z — vertical, antiparallel to the free fall acceleration g ;
neglect the air drag.
i. (0.8 pts) By adjusting the launching angle for a ball thrown
with a fixed initial speed v0 from the origin, targets can be
hit within the region given by
z ≤ z0 − kx2;
you can use this fact without proving it. Find the constants z0
and k.
ii. (1.2 pts) Now, the launching point can be
freely selected on the ground level z = 0, and
the launching angle can be adjusted as needed;
the aim is to hit the topmost point of a spherical
building of radius R (see fig.) with as small as
possible initial speed v0 (prior hitting the target, bouncing off
the roof is not allowed). Sketch qualitatively the shape of the
optimal trajectory of the ball (use the designated box on the
answer sheet). Note: the points are given only for the sketch.
iii. (2.5 pts) What is the minimal launching speed vmin needed
to hit the topmost point of a spherical building of radius R ?
La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part
B. Air flow around a wing (4 points)
For this Problem Part, the following information may be
useful. For a flow of liquid or gas in a tube, along a stream-
line p + ρgh + 1
2ρv2 = const., assuming that the velocity v
is much smaller than the sound speed. Here ρ is the density,
h — height, g — free fall acceleration, and p — hydrostatic
pressure. Streamlines are defined as the trajectories of fluid
particles (assuming that the flow pattern is stationary). Note
that the term 1
2ρv2 is called the dynamic pressure.
In the fig. below, a cross-section of an aircraft wing is de-
picted together with streamlines of the air flow around the
wing, as seen in the wing’s reference frame. Assume that
(a) the air flow is purely two-dimensional (i.e. that the velo-
city vectors of air lie in the figure plane); (b) the streamline
pattern is independent of the aircraft speed; (c) there is no
wind; (d) the dynamic pressure is much smaller than the at-
mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to
take measurements from the fig. on the answer sheet .
i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,
what is the speed of the air vP at the point P (marked in fig.)
with respect to the ground?
ii. (1.2 pts) In the case of high relative humidity, as the ground
speed of the aircraft increases over a critical value vcrit, a stream
of water droplets is created behind the wing. The droplets
emerge at a certain point Q. Mark the point Q in fig. on the
answer sheet. Explain qualitatively (using formulae and as few
text as possible) how you determined its position.
iii. (2.0 pts) Estimate the critical speed vcrit using the follow-
ing data: relative humidity of the air is r = 90% , specific heat
of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure
of saturated water vapour: psa = 2.31 kPa at the temperat-
ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa
at Tb = 294 K . Depending on your approximations you
may also need the specific heat of air at constant volume
cV = 0.717 × 103 J/kg · K . Note that the relative humidity is
defined as the ratio of the vapor pressure to the saturated vapor
pressure at the given temperature. Saturated vapor pressure is
defined as the vapor pressure by which vapor is in equilibrium
with the liquid.
— page 2 of 5 —
78
Part C. Magnetic straws (4.5 points)
Consider a cylindrical tube made of a super-
conducting material. The length of the tube
is l and the inner radius is r ; always l ≫ r.
The centre of the tube coincides with the
origin, and its axis coincides with the z-
axis. There is a magnetic flux Φ through
the central cross-section of the tube, z = 0,
x2 + y2 < r2.
Superconductor is a material which expels any magnetic
field (field is zero inside it).
i. (0.8 pts) Sketch five such magnetic field lines onto the des-
ignated box of the answer sheet which pass through the five
red dots marked on the axial cross-section of the tube.
ii. (1.2 pts) Find the z-directional tension force T in the
middle of the tube (i.e. the force by which two halves of the
tube, z > 0 and z < 0, interact with each other).
iii. (2.5 pts) Now there is another tube,
identical and parallel to the first one. The
second tube has opposite direction of the
magnetic field, and its centre is placed at
y = l , x = z = 0 (so that the tubes form
opposite sides of a square). Determine the
magnetic interaction force F between the two tubes.
— page 3 of 5 —
Part C. Magnetic straws (4.5 points)
Consider a cylindrical tube made of a super-
conducting material. The length of the tube
is l and the inner radius is r ; always l ≫ r.
The centre of the tube coincides with the
origin, and its axis coincides with the z-
axis. There is a magnetic flux Φ through
the central cross-section of the tube, z = 0,
x2 + y2 < r2.
Superconductor is a material which expels any magnetic
field (field is zero inside it).
i. (0.8 pts) Sketch five such magnetic field lines onto the des-
ignated box of the answer sheet which pass through the five
red dots marked on the axial cross-section of the tube.
ii. (1.2 pts) Find the z-directional tension force T in the
middle of the tube (i.e. the force by which two halves of the
tube, z > 0 and z < 0, interact with each other).
iii. (2.5 pts) Now there is another tube,
identical and parallel to the first one. The
second tube has opposite direction of the
magnetic field, and its centre is placed at
y = l , x = z = 0 (so that the tubes form
opposite sides of a square). Determine the
magnetic interaction force F between the two tubes.
— page 3 of 5 —
Part C. Magnetic straws (4.5 points)
Consider a cylindrical tube made of a super-
conducting material. The length of the tube
is l and the inner radius is r ; always l ≫ r.
The centre of the tube coincides with the
origin, and its axis coincides with the z-
axis. There is a magnetic flux Φ through
the central cross-section of the tube, z = 0,
x2 + y2 < r2.
Superconductor is a material which expels any magnetic
field (field is zero inside it).
i. (0.8 pts) Sketch five such magnetic field lines onto the des-
ignated box of the answer sheet which pass through the five
red dots marked on the axial cross-section of the tube.
ii. (1.2 pts) Find the z-directional tension force T in the
middle of the tube (i.e. the force by which two halves of the
tube, z > 0 and z < 0, interact with each other).
iii. (2.5 pts) Now there is another tube,
identical and parallel to the first one. The
second tube has opposite direction of the
magnetic field, and its centre is placed at
y = l , x = z = 0 (so that the tubes form
opposite sides of a square). Determine the
magnetic interaction force F between the two tubes.
— page 3 of 5 —
Problem T2. Kelvin water dropper (8 points)
The following facts about the surface tension may turn out
to be useful for this problem. For the molecules of a liquid, the
positions at the liquid-air interface are less favourable as com-
pared with the positions in the bulk of the liquid. Therefore,
this interface is ascribed the so-called surface energy U = σS,
where S is the surface area of the interface and σ — the surface
tension coefficient of the liquid. Further, two fragments of the
liquid surface pull each other with a force F = σl, where l is
the length of a straight line separating the fragments.
A long metallic pipe with internal diameter d
is pointing directly downwards; water is slowly
dripping from a nozzle at its lower end, see fig.
Water can be considered to be electrically con-
ducting; its surface tension is σ and density —
ρ . Always assume that d ≪ r. Here, r is the
radius of the droplet hanging below the nozzle,
which grows slowly in time until the droplet sep-
arates from the nozzle due to the free fall acceleration g .
Part A. Single pipe (4 points)
i. (1.2 pts) Find the radius rmax of a drop just before it sep-
arates from the nozzle.
ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s
electrostatic potential is ϕ . Find the charge Q of a drop when
its radius is r .
iii. (1.6 pts) For this question, assume that r is kept con-
stant and ϕ is slowly increased. The droplet becomes unstable
and breaks into two pieces if the hydrostatic pressure inside
the droplet becomes smaller than the atmospheric one. Find
the critical potential ϕmax at which this will happen.
Part B. Two pipes (4 points)
An apparatus called “Kelvin water dropper” consists of two
pipes (identical to the one described in Part A), connected
via a T-junction, see fig. The ends of both pipes are at the
centres of two cylindrical electrodes (with height L and dia-
meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n
droplets per unit time. Droplets fall from height H into con-
ductive bowls underneath the nozzles, cross-connected to the
electrodes as shown in Fig; the electrodes are connected via a
capacitance C . There is no net charge on the system of bowls
and electrodes. Note that the water container is grounded.
The first droplet to fall will have microscopic charge, which
will cause an imbalance between the two sides and a small
charge separation across the capacitor.
i. (1.2 pts) Express the modulus
of the charge Q0 of the drops separ-
ating at the instant when the capa-
citor’s charge is q in terms of rmax
(from Part A-i). Neglect the effect
described in Part A-iii.
ii. (1.5 pts) Find the dependence
of q on time t by approximating it
with a continuous function q(t) and
assuming that q(0) = q0 .
iii. (1.3 pts) The dropper’s functioning can be hindered by
the effect shown in Part A-iii. Additionally, a limit Umax to
the achievable voltage between the electrodes is set by the
electrostatic push between a droplet and the bowl beneath it;
find Umax.
— page 4 of 5 —
PrObLEm T2. kELvIn waTEr drOPPEr (8 POInTs)
79
Problem T2. Kelvin water dropper (8 points)
The following facts about the surface tension may turn out
to be useful for this problem. For the molecules of a liquid, the
positions at the liquid-air interface are less favourable as com-
pared with the positions in the bulk of the liquid. Therefore,
this interface is ascribed the so-called surface energy U = σS,
where S is the surface area of the interface and σ — the surface
tension coefficient of the liquid. Further, two fragments of the
liquid surface pull each other with a force F = σl, where l is
the length of a straight line separating the fragments.
A long metallic pipe with internal diameter d
is pointing directly downwards; water is slowly
dripping from a nozzle at its lower end, see fig.
Water can be considered to be electrically con-
ducting; its surface tension is σ and density —
ρ . Always assume that d ≪ r. Here, r is the
radius of the droplet hanging below the nozzle,
which grows slowly in time until the droplet sep-
arates from the nozzle due to the free fall acceleration g .
Part A. Single pipe (4 points)
i. (1.2 pts) Find the radius rmax of a drop just before it sep-
arates from the nozzle.
ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s
electrostatic potential is ϕ . Find the charge Q of a drop when
its radius is r .
iii. (1.6 pts) For this question, assume that r is kept con-
stant and ϕ is slowly increased. The droplet becomes unstable
and breaks into two pieces if the hydrostatic pressure inside
the droplet becomes smaller than the atmospheric one. Find
the critical potential ϕmax at which this will happen.
Part B. Two pipes (4 points)
An apparatus called “Kelvin water dropper” consists of two
pipes (identical to the one described in Part A), connected
via a T-junction, see fig. The ends of both pipes are at the
centres of two cylindrical electrodes (with height L and dia-
meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n
droplets per unit time. Droplets fall from height H into con-
ductive bowls underneath the nozzles, cross-connected to the
electrodes as shown in Fig; the electrodes are connected via a
capacitance C . There is no net charge on the system of bowls
and electrodes. Note that the water container is grounded.
The first droplet to fall will have microscopic charge, which
will cause an imbalance between the two sides and a small
charge separation across the capacitor.
i. (1.2 pts) Express the modulus
of the charge Q0 of the drops separ-
ating at the instant when the capa-
citor’s charge is q in terms of rmax
(from Part A-i). Neglect the effect
described in Part A-iii.
ii. (1.5 pts) Find the dependence
of q on time t by approximating it
with a continuous function q(t) and
assuming that q(0) = q0 .
iii. (1.3 pts) The dropper’s functioning can be hindered by
the effect shown in Part A-iii. Additionally, a limit Umax to
the achievable voltage between the electrodes is set by the
electrostatic push between a droplet and the bowl beneath it;
find Umax.
— page 4 of 5 —
Problem T2. Kelvin water dropper (8 points)
The following facts about the surface tension may turn out
to be useful for this problem. For the molecules of a liquid, the
positions at the liquid-air interface are less favourable as com-
pared with the positions in the bulk of the liquid. Therefore,
this interface is ascribed the so-called surface energy U = σS,
where S is the surface area of the interface and σ — the surface
tension coefficient of the liquid. Further, two fragments of the
liquid surface pull each other with a force F = σl, where l is
the length of a straight line separating the fragments.
A long metallic pipe with internal diameter d
is pointing directly downwards; water is slowly
dripping from a nozzle at its lower end, see fig.
Water can be considered to be electrically con-
ducting; its surface tension is σ and density —
ρ . Always assume that d ≪ r. Here, r is the
radius of the droplet hanging below the nozzle,
which grows slowly in time until the droplet sep-
arates from the nozzle due to the free fall acceleration g .
Part A. Single pipe (4 points)
i. (1.2 pts) Find the radius rmax of a drop just before it sep-
arates from the nozzle.
ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s
electrostatic potential is ϕ . Find the charge Q of a drop when
its radius is r .
iii. (1.6 pts) For this question, assume that r is kept con-
stant and ϕ is slowly increased. The droplet becomes unstable
and breaks into two pieces if the hydrostatic pressure inside
the droplet becomes smaller than the atmospheric one. Find
the critical potential ϕmax at which this will happen.
Part B. Two pipes (4 points)
An apparatus called “Kelvin water dropper” consists of two
pipes (identical to the one described in Part A), connected
via a T-junction, see fig. The ends of both pipes are at the
centres of two cylindrical electrodes (with height L and dia-
meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n
droplets per unit time. Droplets fall from height H into con-
ductive bowls underneath the nozzles, cross-connected to the
electrodes as shown in Fig; the electrodes are connected via a
capacitance C . There is no net charge on the system of bowls
and electrodes. Note that the water container is grounded.
The first droplet to fall will have microscopic charge, which
will cause an imbalance between the two sides and a small
charge separation across the capacitor.
i. (1.2 pts) Express the modulus
of the charge Q0 of the drops separ-
ating at the instant when the capa-
citor’s charge is q in terms of rmax
(from Part A-i). Neglect the effect
described in Part A-iii.
ii. (1.5 pts) Find the dependence
of q on time t by approximating it
with a continuous function q(t) and
assuming that q(0) = q0 .
iii. (1.3 pts) The dropper’s functioning can be hindered by
the effect shown in Part A-iii. Additionally, a limit Umax to
the achievable voltage between the electrodes is set by the
electrostatic push between a droplet and the bowl beneath it;
find Umax.
— page 4 of 5 —
Problem T2. Kelvin water dropper (8 points)
The following facts about the surface tension may turn out
to be useful for this problem. For the molecules of a liquid, the
positions at the liquid-air interface are less favourable as com-
pared with the positions in the bulk of the liquid. Therefore,
this interface is ascribed the so-called surface energy U = σS,
where S is the surface area of the interface and σ — the surface
tension coefficient of the liquid. Further, two fragments of the
liquid surface pull each other with a force F = σl, where l is
the length of a straight line separating the fragments.
A long metallic pipe with internal diameter d
is pointing directly downwards; water is slowly
dripping from a nozzle at its lower end, see fig.
Water can be considered to be electrically con-
ducting; its surface tension is σ and density —
ρ . Always assume that d ≪ r. Here, r is the
radius of the droplet hanging below the nozzle,
which grows slowly in time until the droplet sep-
arates from the nozzle due to the free fall acceleration g .
Part A. Single pipe (4 points)
i. (1.2 pts) Find the radius rmax of a drop just before it sep-
arates from the nozzle.
ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s
electrostatic potential is ϕ . Find the charge Q of a drop when
its radius is r .
iii. (1.6 pts) For this question, assume that r is kept con-
stant and ϕ is slowly increased. The droplet becomes unstable
and breaks into two pieces if the hydrostatic pressure inside
the droplet becomes smaller than the atmospheric one. Find
the critical potential ϕmax at which this will happen.
Part B. Two pipes (4 points)
An apparatus called “Kelvin water dropper” consists of two
pipes (identical to the one described in Part A), connected
via a T-junction, see fig. The ends of both pipes are at the
centres of two cylindrical electrodes (with height L and dia-
meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n
droplets per unit time. Droplets fall from height H into con-
ductive bowls underneath the nozzles, cross-connected to the
electrodes as shown in Fig; the electrodes are connected via a
capacitance C . There is no net charge on the system of bowls
and electrodes. Note that the water container is grounded.
The first droplet to fall will have microscopic charge, which
will cause an imbalance between the two sides and a small
charge separation across the capacitor.
i. (1.2 pts) Express the modulus
of the charge Q0 of the drops separ-
ating at the instant when the capa-
citor’s charge is q in terms of rmax
(from Part A-i). Neglect the effect
described in Part A-iii.
ii. (1.5 pts) Find the dependence
of q on time t by approximating it
with a continuous function q(t) and
assuming that q(0) = q0 .
iii. (1.3 pts) The dropper’s functioning can be hindered by
the effect shown in Part A-iii. Additionally, a limit Umax to
the achievable voltage between the electrodes is set by the
electrostatic push between a droplet and the bowl beneath it;
find Umax.
— page 4 of 5 —
Problem T2. Kelvin water dropper (8 points)
The following facts about the surface tension may turn out
to be useful for this problem. For the molecules of a liquid, the
positions at the liquid-air interface are less favourable as com-
pared with the positions in the bulk of the liquid. Therefore,
this interface is ascribed the so-called surface energy U = σS,
where S is the surface area of the interface and σ — the surface
tension coefficient of the liquid. Further, two fragments of the
liquid surface pull each other with a force F = σl, where l is
the length of a straight line separating the fragments.
A long metallic pipe with internal diameter d
is pointing directly downwards; water is slowly
dripping from a nozzle at its lower end, see fig.
Water can be considered to be electrically con-
ducting; its surface tension is σ and density —
ρ . Always assume that d ≪ r. Here, r is the
radius of the droplet hanging below the nozzle,
which grows slowly in time until the droplet sep-
arates from the nozzle due to the free fall acceleration g .
Part A. Single pipe (4 points)
i. (1.2 pts) Find the radius rmax of a drop just before it sep-
arates from the nozzle.
ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s
electrostatic potential is ϕ . Find the charge Q of a drop when
its radius is r .
iii. (1.6 pts) For this question, assume that r is kept con-
stant and ϕ is slowly increased. The droplet becomes unstable
and breaks into two pieces if the hydrostatic pressure inside
the droplet becomes smaller than the atmospheric one. Find
the critical potential ϕmax at which this will happen.
Part B. Two pipes (4 points)
An apparatus called “Kelvin water dropper” consists of two
pipes (identical to the one described in Part A), connected
via a T-junction, see fig. The ends of both pipes are at the
centres of two cylindrical electrodes (with height L and dia-
meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n
droplets per unit time. Droplets fall from height H into con-
ductive bowls underneath the nozzles, cross-connected to the
electrodes as shown in Fig; the electrodes are connected via a
capacitance C . There is no net charge on the system of bowls
and electrodes. Note that the water container is grounded.
The first droplet to fall will have microscopic charge, which
will cause an imbalance between the two sides and a small
charge separation across the capacitor.
i. (1.2 pts) Express the modulus
of the charge Q0 of the drops separ-
ating at the instant when the capa-
citor’s charge is q in terms of rmax
(from Part A-i). Neglect the effect
described in Part A-iii.
ii. (1.5 pts) Find the dependence
of q on time t by approximating it
with a continuous function q(t) and
assuming that q(0) = q0 .
iii. (1.3 pts) The dropper’s functioning can be hindered by
the effect shown in Part A-iii. Additionally, a limit Umax to
the achievable voltage between the electrodes is set by the
electrostatic push between a droplet and the bowl beneath it;
find Umax.
— page 4 of 5 —
Problem T2. Kelvin water dropper (8 points)
The following facts about the surface tension may turn out
to be useful for this problem. For the molecules of a liquid, the
positions at the liquid-air interface are less favourable as com-
pared with the positions in the bulk of the liquid. Therefore,
this interface is ascribed the so-called surface energy U = σS,
where S is the surface area of the interface and σ — the surface
tension coefficient of the liquid. Further, two fragments of the
liquid surface pull each other with a force F = σl, where l is
the length of a straight line separating the fragments.
A long metallic pipe with internal diameter d
is pointing directly downwards; water is slowly
dripping from a nozzle at its lower end, see fig.
Water can be considered to be electrically con-
ducting; its surface tension is σ and density —
ρ . Always assume that d ≪ r. Here, r is the
radius of the droplet hanging below the nozzle,
which grows slowly in time until the droplet sep-
arates from the nozzle due to the free fall acceleration g .
Part A. Single pipe (4 points)
i. (1.2 pts) Find the radius rmax of a drop just before it sep-
arates from the nozzle.
ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s
electrostatic potential is ϕ . Find the charge Q of a drop when
its radius is r .
iii. (1.6 pts) For this question, assume that r is kept con-
stant and ϕ is slowly increased. The droplet becomes unstable
and breaks into two pieces if the hydrostatic pressure inside
the droplet becomes smaller than the atmospheric one. Find
the critical potential ϕmax at which this will happen.
Part B. Two pipes (4 points)
An apparatus called “Kelvin water dropper” consists of two
pipes (identical to the one described in Part A), connected
via a T-junction, see fig. The ends of both pipes are at the
centres of two cylindrical electrodes (with height L and dia-
meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n
droplets per unit time. Droplets fall from height H into con-
ductive bowls underneath the nozzles, cross-connected to the
electrodes as shown in Fig; the electrodes are connected via a
capacitance C . There is no net charge on the system of bowls
and electrodes. Note that the water container is grounded.
The first droplet to fall will have microscopic charge, which
will cause an imbalance between the two sides and a small
charge separation across the capacitor.
i. (1.2 pts) Express the modulus
of the charge Q0 of the drops separ-
ating at the instant when the capa-
citor’s charge is q in terms of rmax
(from Part A-i). Neglect the effect
described in Part A-iii.
ii. (1.5 pts) Find the dependence
of q on time t by approximating it
with a continuous function q(t) and
assuming that q(0) = q0 .
iii. (1.3 pts) The dropper’s functioning can be hindered by
the effect shown in Part A-iii. Additionally, a limit Umax to
the achievable voltage between the electrodes is set by the
electrostatic push between a droplet and the bowl beneath it;
find Umax.
— page 4 of 5 —
80
Problem T2. Kelvin water dropper (8 points)
The following facts about the surface tension may turn out
to be useful for this problem. For the molecules of a liquid, the
positions at the liquid-air interface are less favourable as com-
pared with the positions in the bulk of the liquid. Therefore,
this interface is ascribed the so-called surface energy U = σS,
where S is the surface area of the interface and σ — the surface
tension coefficient of the liquid. Further, two fragments of the
liquid surface pull each other with a force F = σl, where l is
the length of a straight line separating the fragments.
A long metallic pipe with internal diameter d
is pointing directly downwards; water is slowly
dripping from a nozzle at its lower end, see fig.
Water can be considered to be electrically con-
ducting; its surface tension is σ and density —
ρ . Always assume that d ≪ r. Here, r is the
radius of the droplet hanging below the nozzle,
which grows slowly in time until the droplet sep-
arates from the nozzle due to the free fall acceleration g .
Part A. Single pipe (4 points)
i. (1.2 pts) Find the radius rmax of a drop just before it sep-
arates from the nozzle.
ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s
electrostatic potential is ϕ . Find the charge Q of a drop when
its radius is r .
iii. (1.6 pts) For this question, assume that r is kept con-
stant and ϕ is slowly increased. The droplet becomes unstable
and breaks into two pieces if the hydrostatic pressure inside
the droplet becomes smaller than the atmospheric one. Find
the critical potential ϕmax at which this will happen.
Part B. Two pipes (4 points)
An apparatus called “Kelvin water dropper” consists of two
pipes (identical to the one described in Part A), connected
via a T-junction, see fig. The ends of both pipes are at the
centres of two cylindrical electrodes (with height L and dia-
meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n
droplets per unit time. Droplets fall from height H into con-
ductive bowls underneath the nozzles, cross-connected to the
electrodes as shown in Fig; the electrodes are connected via a
capacitance C . There is no net charge on the system of bowls
and electrodes. Note that the water container is grounded.
The first droplet to fall will have microscopic charge, which
will cause an imbalance between the two sides and a small
charge separation across the capacitor.
i. (1.2 pts) Express the modulus
of the charge Q0 of the drops separ-
ating at the instant when the capa-
citor’s charge is q in terms of rmax
(from Part A-i). Neglect the effect
described in Part A-iii.
ii. (1.5 pts) Find the dependence
of q on time t by approximating it
with a continuous function q(t) and
assuming that q(0) = q0 .
iii. (1.3 pts) The dropper’s functioning can be hindered by
the effect shown in Part A-iii. Additionally, a limit Umax to
the achievable voltage between the electrodes is set by the
electrostatic push between a droplet and the bowl beneath it;
find Umax.
— page 4 of 5 —
PrObLEm T3. PrOTOsTar FOrmaTIOn (9 POInTs)Problem T3. Protostar formation (9 points)
Let us model the formation of a star as follows. A spherical
cloud of sparse interstellar gas, initially at rest, starts to col-
lapse due to its own gravity. The initial radius of the ball is
r0 and the mass — m . The temperature of the surroundings
(much sparser than the gas) and the initial temperature of the
gas is uniformly T0 . The gas may be assumed to be ideal.
The average molar mass of the gas is µ and its adiabatic in-
dex is γ > 4
3. Assume that Gmµ
r0
≫ RT0, where R is the gas
constant and G — the gravity constant.
i. (0.8 pts) During much of the collapse, the gas is so trans-
parent that any heat generated is immediately radiated away,
i.e. the ball stays in a thermodynamic equilibrium with its sur-
roundings. How many times (n) does the pressure increase
while the radius is halved ( r1 = 0.5r0 )? Assume that the gas
density stays uniform.
ii. (1 pt) Estimate the time t2 needed for the radius to shrink
from r0 to r2 = 0.95r0 . Neglect the change of the gravity field
at the position of a falling gas particle.
iii. (2.5 pts) Assuming that the pressure stays negligible, find
the time tr→0 needed for the ball to collapse from r0 down to
a much smaller radius using Kepler’s Laws for elliptical orbits.
iv. (1.7 pts) At some radius r3 ≪ r0, the gas becomes dense
enough to be opaque to the heat radiation. Calculate the
amount of heat Q radiated away during the collapse from the
radius r0 down to r3.
v. (1 pt) For radii smaller than r3 you may neglect heat ra-
diation. Determine how the temperature T of the ball depends
on its radius r < r3.
vi. (2 pts) Eventually we cannot neglect the effect of the pres-
sure on the dynamics of the gas and the collapse stops at
r = r4 (with r4 ≪ r3 ). However, the radiation can still be
neglected and the temperature is not yet high enough to ignite
nuclear fusion. The pressure of such a protostar is not uniform
anymore, but rough estimates with inaccurate numerical pre-
factors can still be done. Estimate the final radius r4 and the
respective temperature T4.
— page 5 of 5 —
81
Problem T3. Protostar formation (9 points)
Let us model the formation of a star as follows. A spherical
cloud of sparse interstellar gas, initially at rest, starts to col-
lapse due to its own gravity. The initial radius of the ball is
r0 and the mass — m . The temperature of the surroundings
(much sparser than the gas) and the initial temperature of the
gas is uniformly T0 . The gas may be assumed to be ideal.
The average molar mass of the gas is µ and its adiabatic in-
dex is γ > 4
3. Assume that Gmµ
r0
≫ RT0, where R is the gas
constant and G — the gravity constant.
i. (0.8 pts) During much of the collapse, the gas is so trans-
parent that any heat generated is immediately radiated away,
i.e. the ball stays in a thermodynamic equilibrium with its sur-
roundings. How many times (n) does the pressure increase
while the radius is halved ( r1 = 0.5r0 )? Assume that the gas
density stays uniform.
ii. (1 pt) Estimate the time t2 needed for the radius to shrink
from r0 to r2 = 0.95r0 . Neglect the change of the gravity field
at the position of a falling gas particle.
iii. (2.5 pts) Assuming that the pressure stays negligible, find
the time tr→0 needed for the ball to collapse from r0 down to
a much smaller radius using Kepler’s Laws for elliptical orbits.
iv. (1.7 pts) At some radius r3 ≪ r0, the gas becomes dense
enough to be opaque to the heat radiation. Calculate the
amount of heat Q radiated away during the collapse from the
radius r0 down to r3.
v. (1 pt) For radii smaller than r3 you may neglect heat ra-
diation. Determine how the temperature T of the ball depends
on its radius r < r3.
vi. (2 pts) Eventually we cannot neglect the effect of the pres-
sure on the dynamics of the gas and the collapse stops at
r = r4 (with r4 ≪ r3 ). However, the radiation can still be
neglected and the temperature is not yet high enough to ignite
nuclear fusion. The pressure of such a protostar is not uniform
anymore, but rough estimates with inaccurate numerical pre-
factors can still be done. Estimate the final radius r4 and the
respective temperature T4.
— page 5 of 5 —
Problem T3. Protostar formation (9 points)
Let us model the formation of a star as follows. A spherical
cloud of sparse interstellar gas, initially at rest, starts to col-
lapse due to its own gravity. The initial radius of the ball is
r0 and the mass — m . The temperature of the surroundings
(much sparser than the gas) and the initial temperature of the
gas is uniformly T0 . The gas may be assumed to be ideal.
The average molar mass of the gas is µ and its adiabatic in-
dex is γ > 4
3. Assume that Gmµ
r0
≫ RT0, where R is the gas
constant and G — the gravity constant.
i. (0.8 pts) During much of the collapse, the gas is so trans-
parent that any heat generated is immediately radiated away,
i.e. the ball stays in a thermodynamic equilibrium with its sur-
roundings. How many times (n) does the pressure increase
while the radius is halved ( r1 = 0.5r0 )? Assume that the gas
density stays uniform.
ii. (1 pt) Estimate the time t2 needed for the radius to shrink
from r0 to r2 = 0.95r0 . Neglect the change of the gravity field
at the position of a falling gas particle.
iii. (2.5 pts) Assuming that the pressure stays negligible, find
the time tr→0 needed for the ball to collapse from r0 down to
a much smaller radius using Kepler’s Laws for elliptical orbits.
iv. (1.7 pts) At some radius r3 ≪ r0, the gas becomes dense
enough to be opaque to the heat radiation. Calculate the
amount of heat Q radiated away during the collapse from the
radius r0 down to r3.
v. (1 pt) For radii smaller than r3 you may neglect heat ra-
diation. Determine how the temperature T of the ball depends
on its radius r < r3.
vi. (2 pts) Eventually we cannot neglect the effect of the pres-
sure on the dynamics of the gas and the collapse stops at
r = r4 (with r4 ≪ r3 ). However, the radiation can still be
neglected and the temperature is not yet high enough to ignite
nuclear fusion. The pressure of such a protostar is not uniform
anymore, but rough estimates with inaccurate numerical pre-
factors can still be done. Estimate the final radius r4 and the
respective temperature T4.
— page 5 of 5 —
82
Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) When the stone is thrown vertically upwards, it
can reach the point x = 0, z = v20/2g (as it follows from the
energy conservation law). Comparing this with the inequality
z ≤ z0 − kx2 we conclude that
z0 = v20/2g. [0.3 pts]
Let us consider the asymptotics z → −∞; the trajectory of
the stone is a parabola, and at this limit, the horizontal dis-
placement (for the given z) is very sensitive with respect to the
curvature of the parabola: the flatter the parabola, the larger
the displacement. The parabola has the flattest shape when
the stone is thrown horizontally, x = v0t and z = −gt2/2, i.e.
its trajectory is given by z = −gx2/2v20. Now, let us recall
that z ≤ z0 − kx2, i.e. −gx2/2v20 ≤ z0 − kx2 ⇒ k ≤ g/2v2
0.
Note that k < g/2v20 would imply that there is a gap between
the parabolic region z ≤ z0 − kx2 and the given trajectory
z = −gx2/2v20 . This trajectory is supposed to be optimal for
hitting targets far below (z → −∞), so there should be no such
a gap, and hence, we can exclude the option k < g/2v20. This
leaves us with
k = g/2v20. [0.5 pts]
ii. (1.2 pts) Let us note that the
stone trajectory is reversible and due
to the energy conservation law, one
can equivalently ask, what is the min-
imal initial speed needed for a stone
to be thrown from the topmost point
of the spherical building down to the
ground without hitting the roof, and what is the respective tra-
jectory. It is easy to understand that the trajectory either needs
to touch the roof, or start horizontally from the topmost point
with the curvature radius equal to R. Indeed, if neither were
the case, it would be possible to keep the same throwing angle
and just reduce the speed a little bit — the stone would still
reach the ground without hitting the roof. Further, if it were
tangent at the topmost point, the trajectory wouldn’t touch
nor intersect the roof anywhere else, because the curvature of
the parabola has maximum at its topmost point. Then, it
would be possible to keep the initial speed constant, and in-
crease slightly the throwing angle (from horizontal to slightly
upwards): the new trajectory wouldn’t be neither tangent at
the top nor touch the roof at any other point; now we can re-
duce the initial speed as we argued previously. So we conclude
that the optimal trajectory needs to touch the roof somewhere,
as shown in Fig.
iii. (2.5 pts) The brute force approach would be writing down
the condition that the optimal trajectory intersects with the
building at two points and touches at one. This would be de-
scribed by a fourth order algebraic equation and therefore, it is
not realistic to accomplish such a solution within a reasonable
time frame.
Note that the interior of the building needs to lie inside the
region where the targets can be hit with a stone thrown from
the top with initial speed vmin. Indeed, if we can throw over
the building, we can hit anything inside by lowering the throw-
ing angle. On the other hand, the boundary of the targetable
region needs to touch the building. Indeed, if there were a
gap, it would be possible to hit a target just above the point
where the optimal trajectory touches the building; the traject-
ory through that target wouldn’t touch the building anywhere,
hence we arrive at a contradiction.
So, with v0 corresponding to the optimal trajectory, the tar-
getable region touches the building; due to symmetry, overall
there are two touching points (for smaller speeds, there would
be four, and for larger speeds, there would be none). With the
origin at the top of the building, the intersection points are
defined by the following system of equations:
x2 + z2 + 2zR = 0, z =v2
0
2g−
gx2
2v20
.
Upon eliminating z, this becomes a biquadratic equation for x:
x4
(
g
2v20
)2
+ x2
(
1
2−
gR
v20
)
+
(
v20
4g+ R
)
v20
g= 0.
Hence the speed by which the real-valued solutions disappear
can be found from the condition that the discriminant vanishes:(
1
2−
gR
v20
)2
=1
4+
gR
v20
=⇒gR
v20
= 2.
Bearing in mind that due to the energy conservation law, at
the ground level the squared speed is increased by 4gR. Thus
we finally obtain
vmin =√
v20 + 4gR = 3
√
gR
2.
Alternative solution using the fact that if a ball
is thrown from a point A to a point B (possibly at
different heights) with the minimal required launch-
ing speed, the initial and terminal velocities are equal.
This fact may be known to some of the contestants, but it
can be also easily derived. Indeed, suppose that when start-
ing with velocity v0 at point A, the ball will hit after time τ
the point B, and |v0| is the minimal speed by which hitting is
possible. Now, let us rotate the vector v0 slightly by adding
to it a perpendicular small vector u ⊥ v0 (|u| ≪ |v0|). With
the launching velocity v1 = v0 + u, the trajectory of the ball
will still go almost through point B: near the pont B, the
displacement of the trajectory cannot change linearly with |u|.
Indeed, a linear in |u| displacement would mean that with es-
sentially the same speed |v1| ≈ |v0|, it would be possible to
throw over point B, which is in a contradiction with the op-
timality of the original trajectory. Hence, the displacement
vector r0(τ) − r1(τ) needs to be parallel to the trajectory, i.e.
to the velocity vB of the ball at point B. Here, r0(t) [r1(t)]
is the radius vector of the ball as a function of time when it
was launched with velocity v0 [v1]. In a free-falling system of
— page 1 of 5 —
Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) When the stone is thrown vertically upwards, it
can reach the point x = 0, z = v20/2g (as it follows from the
energy conservation law). Comparing this with the inequality
z ≤ z0 − kx2 we conclude that
z0 = v20/2g. [0.3 pts]
Let us consider the asymptotics z → −∞; the trajectory of
the stone is a parabola, and at this limit, the horizontal dis-
placement (for the given z) is very sensitive with respect to the
curvature of the parabola: the flatter the parabola, the larger
the displacement. The parabola has the flattest shape when
the stone is thrown horizontally, x = v0t and z = −gt2/2, i.e.
its trajectory is given by z = −gx2/2v20. Now, let us recall
that z ≤ z0 − kx2, i.e. −gx2/2v20 ≤ z0 − kx2 ⇒ k ≤ g/2v2
0.
Note that k < g/2v20 would imply that there is a gap between
the parabolic region z ≤ z0 − kx2 and the given trajectory
z = −gx2/2v20 . This trajectory is supposed to be optimal for
hitting targets far below (z → −∞), so there should be no such
a gap, and hence, we can exclude the option k < g/2v20. This
leaves us with
k = g/2v20. [0.5 pts]
ii. (1.2 pts) Let us note that the
stone trajectory is reversible and due
to the energy conservation law, one
can equivalently ask, what is the min-
imal initial speed needed for a stone
to be thrown from the topmost point
of the spherical building down to the
ground without hitting the roof, and what is the respective tra-
jectory. It is easy to understand that the trajectory either needs
to touch the roof, or start horizontally from the topmost point
with the curvature radius equal to R. Indeed, if neither were
the case, it would be possible to keep the same throwing angle
and just reduce the speed a little bit — the stone would still
reach the ground without hitting the roof. Further, if it were
tangent at the topmost point, the trajectory wouldn’t touch
nor intersect the roof anywhere else, because the curvature of
the parabola has maximum at its topmost point. Then, it
would be possible to keep the initial speed constant, and in-
crease slightly the throwing angle (from horizontal to slightly
upwards): the new trajectory wouldn’t be neither tangent at
the top nor touch the roof at any other point; now we can re-
duce the initial speed as we argued previously. So we conclude
that the optimal trajectory needs to touch the roof somewhere,
as shown in Fig.
iii. (2.5 pts) The brute force approach would be writing down
the condition that the optimal trajectory intersects with the
building at two points and touches at one. This would be de-
scribed by a fourth order algebraic equation and therefore, it is
not realistic to accomplish such a solution within a reasonable
time frame.
Note that the interior of the building needs to lie inside the
region where the targets can be hit with a stone thrown from
the top with initial speed vmin. Indeed, if we can throw over
the building, we can hit anything inside by lowering the throw-
ing angle. On the other hand, the boundary of the targetable
region needs to touch the building. Indeed, if there were a
gap, it would be possible to hit a target just above the point
where the optimal trajectory touches the building; the traject-
ory through that target wouldn’t touch the building anywhere,
hence we arrive at a contradiction.
So, with v0 corresponding to the optimal trajectory, the tar-
getable region touches the building; due to symmetry, overall
there are two touching points (for smaller speeds, there would
be four, and for larger speeds, there would be none). With the
origin at the top of the building, the intersection points are
defined by the following system of equations:
x2 + z2 + 2zR = 0, z =v2
0
2g−
gx2
2v20
.
Upon eliminating z, this becomes a biquadratic equation for x:
x4
(
g
2v20
)2
+ x2
(
1
2−
gR
v20
)
+
(
v20
4g+ R
)
v20
g= 0.
Hence the speed by which the real-valued solutions disappear
can be found from the condition that the discriminant vanishes:(
1
2−
gR
v20
)2
=1
4+
gR
v20
=⇒gR
v20
= 2.
Bearing in mind that due to the energy conservation law, at
the ground level the squared speed is increased by 4gR. Thus
we finally obtain
vmin =√
v20 + 4gR = 3
√
gR
2.
Alternative solution using the fact that if a ball
is thrown from a point A to a point B (possibly at
different heights) with the minimal required launch-
ing speed, the initial and terminal velocities are equal.
This fact may be known to some of the contestants, but it
can be also easily derived. Indeed, suppose that when start-
ing with velocity v0 at point A, the ball will hit after time τ
the point B, and |v0| is the minimal speed by which hitting is
possible. Now, let us rotate the vector v0 slightly by adding
to it a perpendicular small vector u ⊥ v0 (|u| ≪ |v0|). With
the launching velocity v1 = v0 + u, the trajectory of the ball
will still go almost through point B: near the pont B, the
displacement of the trajectory cannot change linearly with |u|.
Indeed, a linear in |u| displacement would mean that with es-
sentially the same speed |v1| ≈ |v0|, it would be possible to
throw over point B, which is in a contradiction with the op-
timality of the original trajectory. Hence, the displacement
vector r0(τ) − r1(τ) needs to be parallel to the trajectory, i.e.
to the velocity vB of the ball at point B. Here, r0(t) [r1(t)]
is the radius vector of the ball as a function of time when it
was launched with velocity v0 [v1]. In a free-falling system of
— page 1 of 5 —
SolutionsPrObLEm T1. FOcus On skETchEs (13 POInTs)
83
Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) When the stone is thrown vertically upwards, it
can reach the point x = 0, z = v20/2g (as it follows from the
energy conservation law). Comparing this with the inequality
z ≤ z0 − kx2 we conclude that
z0 = v20/2g. [0.3 pts]
Let us consider the asymptotics z → −∞; the trajectory of
the stone is a parabola, and at this limit, the horizontal dis-
placement (for the given z) is very sensitive with respect to the
curvature of the parabola: the flatter the parabola, the larger
the displacement. The parabola has the flattest shape when
the stone is thrown horizontally, x = v0t and z = −gt2/2, i.e.
its trajectory is given by z = −gx2/2v20. Now, let us recall
that z ≤ z0 − kx2, i.e. −gx2/2v20 ≤ z0 − kx2 ⇒ k ≤ g/2v2
0.
Note that k < g/2v20 would imply that there is a gap between
the parabolic region z ≤ z0 − kx2 and the given trajectory
z = −gx2/2v20 . This trajectory is supposed to be optimal for
hitting targets far below (z → −∞), so there should be no such
a gap, and hence, we can exclude the option k < g/2v20. This
leaves us with
k = g/2v20. [0.5 pts]
ii. (1.2 pts) Let us note that the
stone trajectory is reversible and due
to the energy conservation law, one
can equivalently ask, what is the min-
imal initial speed needed for a stone
to be thrown from the topmost point
of the spherical building down to the
ground without hitting the roof, and what is the respective tra-
jectory. It is easy to understand that the trajectory either needs
to touch the roof, or start horizontally from the topmost point
with the curvature radius equal to R. Indeed, if neither were
the case, it would be possible to keep the same throwing angle
and just reduce the speed a little bit — the stone would still
reach the ground without hitting the roof. Further, if it were
tangent at the topmost point, the trajectory wouldn’t touch
nor intersect the roof anywhere else, because the curvature of
the parabola has maximum at its topmost point. Then, it
would be possible to keep the initial speed constant, and in-
crease slightly the throwing angle (from horizontal to slightly
upwards): the new trajectory wouldn’t be neither tangent at
the top nor touch the roof at any other point; now we can re-
duce the initial speed as we argued previously. So we conclude
that the optimal trajectory needs to touch the roof somewhere,
as shown in Fig.
iii. (2.5 pts) The brute force approach would be writing down
the condition that the optimal trajectory intersects with the
building at two points and touches at one. This would be de-
scribed by a fourth order algebraic equation and therefore, it is
not realistic to accomplish such a solution within a reasonable
time frame.
Note that the interior of the building needs to lie inside the
region where the targets can be hit with a stone thrown from
the top with initial speed vmin. Indeed, if we can throw over
the building, we can hit anything inside by lowering the throw-
ing angle. On the other hand, the boundary of the targetable
region needs to touch the building. Indeed, if there were a
gap, it would be possible to hit a target just above the point
where the optimal trajectory touches the building; the traject-
ory through that target wouldn’t touch the building anywhere,
hence we arrive at a contradiction.
So, with v0 corresponding to the optimal trajectory, the tar-
getable region touches the building; due to symmetry, overall
there are two touching points (for smaller speeds, there would
be four, and for larger speeds, there would be none). With the
origin at the top of the building, the intersection points are
defined by the following system of equations:
x2 + z2 + 2zR = 0, z =v2
0
2g−
gx2
2v20
.
Upon eliminating z, this becomes a biquadratic equation for x:
x4
(
g
2v20
)2
+ x2
(
1
2−
gR
v20
)
+
(
v20
4g+ R
)
v20
g= 0.
Hence the speed by which the real-valued solutions disappear
can be found from the condition that the discriminant vanishes:(
1
2−
gR
v20
)2
=1
4+
gR
v20
=⇒gR
v20
= 2.
Bearing in mind that due to the energy conservation law, at
the ground level the squared speed is increased by 4gR. Thus
we finally obtain
vmin =√
v20 + 4gR = 3
√
gR
2.
Alternative solution using the fact that if a ball
is thrown from a point A to a point B (possibly at
different heights) with the minimal required launch-
ing speed, the initial and terminal velocities are equal.
This fact may be known to some of the contestants, but it
can be also easily derived. Indeed, suppose that when start-
ing with velocity v0 at point A, the ball will hit after time τ
the point B, and |v0| is the minimal speed by which hitting is
possible. Now, let us rotate the vector v0 slightly by adding
to it a perpendicular small vector u ⊥ v0 (|u| ≪ |v0|). With
the launching velocity v1 = v0 + u, the trajectory of the ball
will still go almost through point B: near the pont B, the
displacement of the trajectory cannot change linearly with |u|.
Indeed, a linear in |u| displacement would mean that with es-
sentially the same speed |v1| ≈ |v0|, it would be possible to
throw over point B, which is in a contradiction with the op-
timality of the original trajectory. Hence, the displacement
vector r0(τ) − r1(τ) needs to be parallel to the trajectory, i.e.
to the velocity vB of the ball at point B. Here, r0(t) [r1(t)]
is the radius vector of the ball as a function of time when it
was launched with velocity v0 [v1]. In a free-falling system of
— page 1 of 5 —Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) When the stone is thrown vertically upwards, it
can reach the point x = 0, z = v20/2g (as it follows from the
energy conservation law). Comparing this with the inequality
z ≤ z0 − kx2 we conclude that
z0 = v20/2g. [0.3 pts]
Let us consider the asymptotics z → −∞; the trajectory of
the stone is a parabola, and at this limit, the horizontal dis-
placement (for the given z) is very sensitive with respect to the
curvature of the parabola: the flatter the parabola, the larger
the displacement. The parabola has the flattest shape when
the stone is thrown horizontally, x = v0t and z = −gt2/2, i.e.
its trajectory is given by z = −gx2/2v20. Now, let us recall
that z ≤ z0 − kx2, i.e. −gx2/2v20 ≤ z0 − kx2 ⇒ k ≤ g/2v2
0.
Note that k < g/2v20 would imply that there is a gap between
the parabolic region z ≤ z0 − kx2 and the given trajectory
z = −gx2/2v20 . This trajectory is supposed to be optimal for
hitting targets far below (z → −∞), so there should be no such
a gap, and hence, we can exclude the option k < g/2v20. This
leaves us with
k = g/2v20. [0.5 pts]
ii. (1.2 pts) Let us note that the
stone trajectory is reversible and due
to the energy conservation law, one
can equivalently ask, what is the min-
imal initial speed needed for a stone
to be thrown from the topmost point
of the spherical building down to the
ground without hitting the roof, and what is the respective tra-
jectory. It is easy to understand that the trajectory either needs
to touch the roof, or start horizontally from the topmost point
with the curvature radius equal to R. Indeed, if neither were
the case, it would be possible to keep the same throwing angle
and just reduce the speed a little bit — the stone would still
reach the ground without hitting the roof. Further, if it were
tangent at the topmost point, the trajectory wouldn’t touch
nor intersect the roof anywhere else, because the curvature of
the parabola has maximum at its topmost point. Then, it
would be possible to keep the initial speed constant, and in-
crease slightly the throwing angle (from horizontal to slightly
upwards): the new trajectory wouldn’t be neither tangent at
the top nor touch the roof at any other point; now we can re-
duce the initial speed as we argued previously. So we conclude
that the optimal trajectory needs to touch the roof somewhere,
as shown in Fig.
iii. (2.5 pts) The brute force approach would be writing down
the condition that the optimal trajectory intersects with the
building at two points and touches at one. This would be de-
scribed by a fourth order algebraic equation and therefore, it is
not realistic to accomplish such a solution within a reasonable
time frame.
Note that the interior of the building needs to lie inside the
region where the targets can be hit with a stone thrown from
the top with initial speed vmin. Indeed, if we can throw over
the building, we can hit anything inside by lowering the throw-
ing angle. On the other hand, the boundary of the targetable
region needs to touch the building. Indeed, if there were a
gap, it would be possible to hit a target just above the point
where the optimal trajectory touches the building; the traject-
ory through that target wouldn’t touch the building anywhere,
hence we arrive at a contradiction.
So, with v0 corresponding to the optimal trajectory, the tar-
getable region touches the building; due to symmetry, overall
there are two touching points (for smaller speeds, there would
be four, and for larger speeds, there would be none). With the
origin at the top of the building, the intersection points are
defined by the following system of equations:
x2 + z2 + 2zR = 0, z =v2
0
2g−
gx2
2v20
.
Upon eliminating z, this becomes a biquadratic equation for x:
x4
(
g
2v20
)2
+ x2
(
1
2−
gR
v20
)
+
(
v20
4g+ R
)
v20
g= 0.
Hence the speed by which the real-valued solutions disappear
can be found from the condition that the discriminant vanishes:(
1
2−
gR
v20
)2
=1
4+
gR
v20
=⇒gR
v20
= 2.
Bearing in mind that due to the energy conservation law, at
the ground level the squared speed is increased by 4gR. Thus
we finally obtain
vmin =√
v20 + 4gR = 3
√
gR
2.
Alternative solution using the fact that if a ball
is thrown from a point A to a point B (possibly at
different heights) with the minimal required launch-
ing speed, the initial and terminal velocities are equal.
This fact may be known to some of the contestants, but it
can be also easily derived. Indeed, suppose that when start-
ing with velocity v0 at point A, the ball will hit after time τ
the point B, and |v0| is the minimal speed by which hitting is
possible. Now, let us rotate the vector v0 slightly by adding
to it a perpendicular small vector u ⊥ v0 (|u| ≪ |v0|). With
the launching velocity v1 = v0 + u, the trajectory of the ball
will still go almost through point B: near the pont B, the
displacement of the trajectory cannot change linearly with |u|.
Indeed, a linear in |u| displacement would mean that with es-
sentially the same speed |v1| ≈ |v0|, it would be possible to
throw over point B, which is in a contradiction with the op-
timality of the original trajectory. Hence, the displacement
vector r0(τ) − r1(τ) needs to be parallel to the trajectory, i.e.
to the velocity vB of the ball at point B. Here, r0(t) [r1(t)]
is the radius vector of the ball as a function of time when it
was launched with velocity v0 [v1]. In a free-falling system of
— page 1 of 5 —
84
Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) When the stone is thrown vertically upwards, it
can reach the point x = 0, z = v20/2g (as it follows from the
energy conservation law). Comparing this with the inequality
z ≤ z0 − kx2 we conclude that
z0 = v20/2g. [0.3 pts]
Let us consider the asymptotics z → −∞; the trajectory of
the stone is a parabola, and at this limit, the horizontal dis-
placement (for the given z) is very sensitive with respect to the
curvature of the parabola: the flatter the parabola, the larger
the displacement. The parabola has the flattest shape when
the stone is thrown horizontally, x = v0t and z = −gt2/2, i.e.
its trajectory is given by z = −gx2/2v20. Now, let us recall
that z ≤ z0 − kx2, i.e. −gx2/2v20 ≤ z0 − kx2 ⇒ k ≤ g/2v2
0.
Note that k < g/2v20 would imply that there is a gap between
the parabolic region z ≤ z0 − kx2 and the given trajectory
z = −gx2/2v20 . This trajectory is supposed to be optimal for
hitting targets far below (z → −∞), so there should be no such
a gap, and hence, we can exclude the option k < g/2v20. This
leaves us with
k = g/2v20. [0.5 pts]
ii. (1.2 pts) Let us note that the
stone trajectory is reversible and due
to the energy conservation law, one
can equivalently ask, what is the min-
imal initial speed needed for a stone
to be thrown from the topmost point
of the spherical building down to the
ground without hitting the roof, and what is the respective tra-
jectory. It is easy to understand that the trajectory either needs
to touch the roof, or start horizontally from the topmost point
with the curvature radius equal to R. Indeed, if neither were
the case, it would be possible to keep the same throwing angle
and just reduce the speed a little bit — the stone would still
reach the ground without hitting the roof. Further, if it were
tangent at the topmost point, the trajectory wouldn’t touch
nor intersect the roof anywhere else, because the curvature of
the parabola has maximum at its topmost point. Then, it
would be possible to keep the initial speed constant, and in-
crease slightly the throwing angle (from horizontal to slightly
upwards): the new trajectory wouldn’t be neither tangent at
the top nor touch the roof at any other point; now we can re-
duce the initial speed as we argued previously. So we conclude
that the optimal trajectory needs to touch the roof somewhere,
as shown in Fig.
iii. (2.5 pts) The brute force approach would be writing down
the condition that the optimal trajectory intersects with the
building at two points and touches at one. This would be de-
scribed by a fourth order algebraic equation and therefore, it is
not realistic to accomplish such a solution within a reasonable
time frame.
Note that the interior of the building needs to lie inside the
region where the targets can be hit with a stone thrown from
the top with initial speed vmin. Indeed, if we can throw over
the building, we can hit anything inside by lowering the throw-
ing angle. On the other hand, the boundary of the targetable
region needs to touch the building. Indeed, if there were a
gap, it would be possible to hit a target just above the point
where the optimal trajectory touches the building; the traject-
ory through that target wouldn’t touch the building anywhere,
hence we arrive at a contradiction.
So, with v0 corresponding to the optimal trajectory, the tar-
getable region touches the building; due to symmetry, overall
there are two touching points (for smaller speeds, there would
be four, and for larger speeds, there would be none). With the
origin at the top of the building, the intersection points are
defined by the following system of equations:
x2 + z2 + 2zR = 0, z =v2
0
2g−
gx2
2v20
.
Upon eliminating z, this becomes a biquadratic equation for x:
x4
(
g
2v20
)2
+ x2
(
1
2−
gR
v20
)
+
(
v20
4g+ R
)
v20
g= 0.
Hence the speed by which the real-valued solutions disappear
can be found from the condition that the discriminant vanishes:(
1
2−
gR
v20
)2
=1
4+
gR
v20
=⇒gR
v20
= 2.
Bearing in mind that due to the energy conservation law, at
the ground level the squared speed is increased by 4gR. Thus
we finally obtain
vmin =√
v20 + 4gR = 3
√
gR
2.
Alternative solution using the fact that if a ball
is thrown from a point A to a point B (possibly at
different heights) with the minimal required launch-
ing speed, the initial and terminal velocities are equal.
This fact may be known to some of the contestants, but it
can be also easily derived. Indeed, suppose that when start-
ing with velocity v0 at point A, the ball will hit after time τ
the point B, and |v0| is the minimal speed by which hitting is
possible. Now, let us rotate the vector v0 slightly by adding
to it a perpendicular small vector u ⊥ v0 (|u| ≪ |v0|). With
the launching velocity v1 = v0 + u, the trajectory of the ball
will still go almost through point B: near the pont B, the
displacement of the trajectory cannot change linearly with |u|.
Indeed, a linear in |u| displacement would mean that with es-
sentially the same speed |v1| ≈ |v0|, it would be possible to
throw over point B, which is in a contradiction with the op-
timality of the original trajectory. Hence, the displacement
vector r0(τ) − r1(τ) needs to be parallel to the trajectory, i.e.
to the velocity vB of the ball at point B. Here, r0(t) [r1(t)]
is the radius vector of the ball as a function of time when it
was launched with velocity v0 [v1]. In a free-falling system of
— page 1 of 5 —reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,
(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.
We start again throwing a ball from the point O at the top
of the building, and notice that for the optimal trajctory, at
the point P , where it touches the building, the velocity is per-
pendicular both to the radius vector QP (where Q denotes the
building’s centre), and to the launching velocity. Hence, QP
and the launching velocity are parallel. Let O be the origin,
and let us denote ∠OQP = α. Then the trajectory is given by
z = x cot α −gx2
2v2 sin2 α.
Point P with coordinates x = R sin α and z = R(cos α − 1) be-
longs to this parabola, hence R = gR2
2v2 , from where we obtain
the previous result.
Part B. Mist (4 points)
i. (0.8 pts) In the plane’s reference frame, along the channel
between two streamlines the volume flux of air (volume flow
rate) is constant due to continuity. The volume flux is the
product of speed and channel’s cross-section area, which, due
to the two-dimensional geometry, is proportional to the channel
width and can be measured from the Fig. Due to the absence of
wind, the unperturbed air’s speed in the plane’s frame is just v0.
So, upon measuring the dimensions a = 10 mm and b = 13 mm
(see Fig), we can write v0a = ub and hence u = v0ab. Since at
point P , the streamlines are horizontal where all the velocities
are parallel, the vector addition is reduced to the scalar addi-
tion: the air’s ground speed vP = v0 − u = v0(1 −ab) = 23 m/s.
ii. (1.2 pts) Although the dynamic pressure 1
2ρv2 is relatively
small, it gives rise to some adiabatic expansion and compres-
sion. In expanding regions the temperature will drop and hence,
the pressure of saturated vapours will also drop. If the dew
point is reached, a stream of droplets will appear. This process
will start in a point where the adiabatic expansion is maximal,
i.e. where the hydrostatic pressure is minimal and consequently,
as it follows from the Bernoulli’s law p + 1
2ρv2 = const, the dy-
namic pressure is maximal: in the place where the air speed in
wing’s frame is maximal and the streamline distance minimal.
Such a point Q is marked in Fig.
iii. (2 pts) First we need to calculate the dew point for the air
of given water content (since the relative pressure change will
be small, we can ignore the dependence of the dew point on
pressure). The water vapour pressure is pw = psar = 2.08 kPa.
The relative change of the pressure of the saturated vapour is
small, so we can linearize its temperature dependence:
psa − pw
Ta − T=
psb − psa
Tb − Ta
=⇒ Ta − T = (Tb − Ta)(1 − r)psa
psb − psa
;
numerically T ≈ 291.5 K. Further we need to relate the air
speed to the temperature. To this end we need to use the en-
ergy conservation law. A convenient ready-to-use form of it is
provided by the Bernoulli’s law. Applying this law will give
a good approximation of the reality, but strictly speaking, it
needs to be modified to take into account the compressibility
of air and the associated expansion/contraction work. Con-
sider one mole of air, which has the mass µ and the volume
V = RT/p. Apparently the process is fast and the air par-
cels are large, so that heat transfer across the air parcels is
negligible. Additionally, the process is subsonic; all together
we can conclude that the process is adiabatic. Consider a seg-
ment of a tube formed by the streamlines. Let us denote the
physical quantities at its one end by index 1, and at the other
end — by index 2. Then, while one mole of gas flows into
the tube at one end, as much flows out at the other end. The
inflow carries in kinetic energy 1
2µv2
1 , and the outflow carries
out 1
2µv2
2. The inflowing gas receives work due to the pushing
gas equal to p1V1 = RT1, the outflowing gas performs work
p2V2 = RT2. Let’s define molar heat capacities CV = µcV and
Cp = µcp. The inflow carries in heat energy CV RT1, and the
outflow carries out CV RT2. All together, the energy balance
can be written as 1
2µv2 + CpT = const. From this we can
easily express ∆ v2
2= 1
Cv2
crit(a2
c2 − 1) = cp∆T , where c is the
streamline distance at the point Q, and further
vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s,
where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that
in reality, the required speed is probably somewhat higher, be-
cause for a fast condensation, a considerable over-saturation is
needed. However, within an order of magnitude, this estimate
remains valid.
Part C. Magnetic straws (4.5 points)
i. (0.8 pts) Due to the superconduct-
ing walls, the magnetic field lines cannot
cross the walls, so the flux is constant
along the tube. For a closed contour in-
side the tube, there should be no circu-
lation of the magnetic field, hence the
field lines cannot be curved, and the field
needs to be homogeneous. The field lines
close from outside the tube, similarly to a solenoid.
ii. (1.2 pts) Let us consider the change of the magnetic energy
when the tube is stretched (virtually) by a small amount ∆l.
Note that the magnetic flux trough the tube is conserved: any
change of flux would imply a non-zero electromotive force dΦ
dt,
and for a zero resistivity, an infinite current. So, the induc-
tion B = Φ
πr2 . The energy density of the magnetic field is B2
2µ0
.
— page 2 of 5 —
85
reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,
(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.
We start again throwing a ball from the point O at the top
of the building, and notice that for the optimal trajctory, at
the point P , where it touches the building, the velocity is per-
pendicular both to the radius vector QP (where Q denotes the
building’s centre), and to the launching velocity. Hence, QP
and the launching velocity are parallel. Let O be the origin,
and let us denote ∠OQP = α. Then the trajectory is given by
z = x cot α −gx2
2v2 sin2 α.
Point P with coordinates x = R sin α and z = R(cos α − 1) be-
longs to this parabola, hence R = gR2
2v2 , from where we obtain
the previous result.
Part B. Mist (4 points)
i. (0.8 pts) In the plane’s reference frame, along the channel
between two streamlines the volume flux of air (volume flow
rate) is constant due to continuity. The volume flux is the
product of speed and channel’s cross-section area, which, due
to the two-dimensional geometry, is proportional to the channel
width and can be measured from the Fig. Due to the absence of
wind, the unperturbed air’s speed in the plane’s frame is just v0.
So, upon measuring the dimensions a = 10 mm and b = 13 mm
(see Fig), we can write v0a = ub and hence u = v0ab. Since at
point P , the streamlines are horizontal where all the velocities
are parallel, the vector addition is reduced to the scalar addi-
tion: the air’s ground speed vP = v0 − u = v0(1 −ab) = 23 m/s.
ii. (1.2 pts) Although the dynamic pressure 1
2ρv2 is relatively
small, it gives rise to some adiabatic expansion and compres-
sion. In expanding regions the temperature will drop and hence,
the pressure of saturated vapours will also drop. If the dew
point is reached, a stream of droplets will appear. This process
will start in a point where the adiabatic expansion is maximal,
i.e. where the hydrostatic pressure is minimal and consequently,
as it follows from the Bernoulli’s law p + 1
2ρv2 = const, the dy-
namic pressure is maximal: in the place where the air speed in
wing’s frame is maximal and the streamline distance minimal.
Such a point Q is marked in Fig.
iii. (2 pts) First we need to calculate the dew point for the air
of given water content (since the relative pressure change will
be small, we can ignore the dependence of the dew point on
pressure). The water vapour pressure is pw = psar = 2.08 kPa.
The relative change of the pressure of the saturated vapour is
small, so we can linearize its temperature dependence:
psa − pw
Ta − T=
psb − psa
Tb − Ta
=⇒ Ta − T = (Tb − Ta)(1 − r)psa
psb − psa
;
numerically T ≈ 291.5 K. Further we need to relate the air
speed to the temperature. To this end we need to use the en-
ergy conservation law. A convenient ready-to-use form of it is
provided by the Bernoulli’s law. Applying this law will give
a good approximation of the reality, but strictly speaking, it
needs to be modified to take into account the compressibility
of air and the associated expansion/contraction work. Con-
sider one mole of air, which has the mass µ and the volume
V = RT/p. Apparently the process is fast and the air par-
cels are large, so that heat transfer across the air parcels is
negligible. Additionally, the process is subsonic; all together
we can conclude that the process is adiabatic. Consider a seg-
ment of a tube formed by the streamlines. Let us denote the
physical quantities at its one end by index 1, and at the other
end — by index 2. Then, while one mole of gas flows into
the tube at one end, as much flows out at the other end. The
inflow carries in kinetic energy 1
2µv2
1 , and the outflow carries
out 1
2µv2
2. The inflowing gas receives work due to the pushing
gas equal to p1V1 = RT1, the outflowing gas performs work
p2V2 = RT2. Let’s define molar heat capacities CV = µcV and
Cp = µcp. The inflow carries in heat energy CV RT1, and the
outflow carries out CV RT2. All together, the energy balance
can be written as 1
2µv2 + CpT = const. From this we can
easily express ∆ v2
2= 1
Cv2
crit(a2
c2 − 1) = cp∆T , where c is the
streamline distance at the point Q, and further
vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s,
where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that
in reality, the required speed is probably somewhat higher, be-
cause for a fast condensation, a considerable over-saturation is
needed. However, within an order of magnitude, this estimate
remains valid.
Part C. Magnetic straws (4.5 points)
i. (0.8 pts) Due to the superconduct-
ing walls, the magnetic field lines cannot
cross the walls, so the flux is constant
along the tube. For a closed contour in-
side the tube, there should be no circu-
lation of the magnetic field, hence the
field lines cannot be curved, and the field
needs to be homogeneous. The field lines
close from outside the tube, similarly to a solenoid.
ii. (1.2 pts) Let us consider the change of the magnetic energy
when the tube is stretched (virtually) by a small amount ∆l.
Note that the magnetic flux trough the tube is conserved: any
change of flux would imply a non-zero electromotive force dΦ
dt,
and for a zero resistivity, an infinite current. So, the induc-
tion B = Φ
πr2 . The energy density of the magnetic field is B2
2µ0
.
— page 2 of 5 —
reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,
(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.
We start again throwing a ball from the point O at the top
of the building, and notice that for the optimal trajctory, at
the point P , where it touches the building, the velocity is per-
pendicular both to the radius vector QP (where Q denotes the
building’s centre), and to the launching velocity. Hence, QP
and the launching velocity are parallel. Let O be the origin,
and let us denote ∠OQP = α. Then the trajectory is given by
z = x cot α −gx2
2v2 sin2 α.
Point P with coordinates x = R sin α and z = R(cos α − 1) be-
longs to this parabola, hence R = gR2
2v2 , from where we obtain
the previous result.
Part B. Mist (4 points)
i. (0.8 pts) In the plane’s reference frame, along the channel
between two streamlines the volume flux of air (volume flow
rate) is constant due to continuity. The volume flux is the
product of speed and channel’s cross-section area, which, due
to the two-dimensional geometry, is proportional to the channel
width and can be measured from the Fig. Due to the absence of
wind, the unperturbed air’s speed in the plane’s frame is just v0.
So, upon measuring the dimensions a = 10 mm and b = 13 mm
(see Fig), we can write v0a = ub and hence u = v0ab. Since at
point P , the streamlines are horizontal where all the velocities
are parallel, the vector addition is reduced to the scalar addi-
tion: the air’s ground speed vP = v0 − u = v0(1 −ab) = 23 m/s.
ii. (1.2 pts) Although the dynamic pressure 1
2ρv2 is relatively
small, it gives rise to some adiabatic expansion and compres-
sion. In expanding regions the temperature will drop and hence,
the pressure of saturated vapours will also drop. If the dew
point is reached, a stream of droplets will appear. This process
will start in a point where the adiabatic expansion is maximal,
i.e. where the hydrostatic pressure is minimal and consequently,
as it follows from the Bernoulli’s law p + 1
2ρv2 = const, the dy-
namic pressure is maximal: in the place where the air speed in
wing’s frame is maximal and the streamline distance minimal.
Such a point Q is marked in Fig.
iii. (2 pts) First we need to calculate the dew point for the air
of given water content (since the relative pressure change will
be small, we can ignore the dependence of the dew point on
pressure). The water vapour pressure is pw = psar = 2.08 kPa.
The relative change of the pressure of the saturated vapour is
small, so we can linearize its temperature dependence:
psa − pw
Ta − T=
psb − psa
Tb − Ta
=⇒ Ta − T = (Tb − Ta)(1 − r)psa
psb − psa
;
numerically T ≈ 291.5 K. Further we need to relate the air
speed to the temperature. To this end we need to use the en-
ergy conservation law. A convenient ready-to-use form of it is
provided by the Bernoulli’s law. Applying this law will give
a good approximation of the reality, but strictly speaking, it
needs to be modified to take into account the compressibility
of air and the associated expansion/contraction work. Con-
sider one mole of air, which has the mass µ and the volume
V = RT/p. Apparently the process is fast and the air par-
cels are large, so that heat transfer across the air parcels is
negligible. Additionally, the process is subsonic; all together
we can conclude that the process is adiabatic. Consider a seg-
ment of a tube formed by the streamlines. Let us denote the
physical quantities at its one end by index 1, and at the other
end — by index 2. Then, while one mole of gas flows into
the tube at one end, as much flows out at the other end. The
inflow carries in kinetic energy 1
2µv2
1 , and the outflow carries
out 1
2µv2
2. The inflowing gas receives work due to the pushing
gas equal to p1V1 = RT1, the outflowing gas performs work
p2V2 = RT2. Let’s define molar heat capacities CV = µcV and
Cp = µcp. The inflow carries in heat energy CV RT1, and the
outflow carries out CV RT2. All together, the energy balance
can be written as 1
2µv2 + CpT = const. From this we can
easily express ∆ v2
2= 1
Cv2
crit(a2
c2 − 1) = cp∆T , where c is the
streamline distance at the point Q, and further
vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s,
where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that
in reality, the required speed is probably somewhat higher, be-
cause for a fast condensation, a considerable over-saturation is
needed. However, within an order of magnitude, this estimate
remains valid.
Part C. Magnetic straws (4.5 points)
i. (0.8 pts) Due to the superconduct-
ing walls, the magnetic field lines cannot
cross the walls, so the flux is constant
along the tube. For a closed contour in-
side the tube, there should be no circu-
lation of the magnetic field, hence the
field lines cannot be curved, and the field
needs to be homogeneous. The field lines
close from outside the tube, similarly to a solenoid.
ii. (1.2 pts) Let us consider the change of the magnetic energy
when the tube is stretched (virtually) by a small amount ∆l.
Note that the magnetic flux trough the tube is conserved: any
change of flux would imply a non-zero electromotive force dΦ
dt,
and for a zero resistivity, an infinite current. So, the induc-
tion B = Φ
πr2 . The energy density of the magnetic field is B2
2µ0
.
— page 2 of 5 —reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,
(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.
We start again throwing a ball from the point O at the top
of the building, and notice that for the optimal trajctory, at
the point P , where it touches the building, the velocity is per-
pendicular both to the radius vector QP (where Q denotes the
building’s centre), and to the launching velocity. Hence, QP
and the launching velocity are parallel. Let O be the origin,
and let us denote ∠OQP = α. Then the trajectory is given by
z = x cot α −gx2
2v2 sin2 α.
Point P with coordinates x = R sin α and z = R(cos α − 1) be-
longs to this parabola, hence R = gR2
2v2 , from where we obtain
the previous result.
Part B. Mist (4 points)
i. (0.8 pts) In the plane’s reference frame, along the channel
between two streamlines the volume flux of air (volume flow
rate) is constant due to continuity. The volume flux is the
product of speed and channel’s cross-section area, which, due
to the two-dimensional geometry, is proportional to the channel
width and can be measured from the Fig. Due to the absence of
wind, the unperturbed air’s speed in the plane’s frame is just v0.
So, upon measuring the dimensions a = 10 mm and b = 13 mm
(see Fig), we can write v0a = ub and hence u = v0ab. Since at
point P , the streamlines are horizontal where all the velocities
are parallel, the vector addition is reduced to the scalar addi-
tion: the air’s ground speed vP = v0 − u = v0(1 −ab) = 23 m/s.
ii. (1.2 pts) Although the dynamic pressure 1
2ρv2 is relatively
small, it gives rise to some adiabatic expansion and compres-
sion. In expanding regions the temperature will drop and hence,
the pressure of saturated vapours will also drop. If the dew
point is reached, a stream of droplets will appear. This process
will start in a point where the adiabatic expansion is maximal,
i.e. where the hydrostatic pressure is minimal and consequently,
as it follows from the Bernoulli’s law p + 1
2ρv2 = const, the dy-
namic pressure is maximal: in the place where the air speed in
wing’s frame is maximal and the streamline distance minimal.
Such a point Q is marked in Fig.
iii. (2 pts) First we need to calculate the dew point for the air
of given water content (since the relative pressure change will
be small, we can ignore the dependence of the dew point on
pressure). The water vapour pressure is pw = psar = 2.08 kPa.
The relative change of the pressure of the saturated vapour is
small, so we can linearize its temperature dependence:
psa − pw
Ta − T=
psb − psa
Tb − Ta
=⇒ Ta − T = (Tb − Ta)(1 − r)psa
psb − psa
;
numerically T ≈ 291.5 K. Further we need to relate the air
speed to the temperature. To this end we need to use the en-
ergy conservation law. A convenient ready-to-use form of it is
provided by the Bernoulli’s law. Applying this law will give
a good approximation of the reality, but strictly speaking, it
needs to be modified to take into account the compressibility
of air and the associated expansion/contraction work. Con-
sider one mole of air, which has the mass µ and the volume
V = RT/p. Apparently the process is fast and the air par-
cels are large, so that heat transfer across the air parcels is
negligible. Additionally, the process is subsonic; all together
we can conclude that the process is adiabatic. Consider a seg-
ment of a tube formed by the streamlines. Let us denote the
physical quantities at its one end by index 1, and at the other
end — by index 2. Then, while one mole of gas flows into
the tube at one end, as much flows out at the other end. The
inflow carries in kinetic energy 1
2µv2
1 , and the outflow carries
out 1
2µv2
2. The inflowing gas receives work due to the pushing
gas equal to p1V1 = RT1, the outflowing gas performs work
p2V2 = RT2. Let’s define molar heat capacities CV = µcV and
Cp = µcp. The inflow carries in heat energy CV RT1, and the
outflow carries out CV RT2. All together, the energy balance
can be written as 1
2µv2 + CpT = const. From this we can
easily express ∆ v2
2= 1
Cv2
crit(a2
c2 − 1) = cp∆T , where c is the
streamline distance at the point Q, and further
vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s,
where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that
in reality, the required speed is probably somewhat higher, be-
cause for a fast condensation, a considerable over-saturation is
needed. However, within an order of magnitude, this estimate
remains valid.
Part C. Magnetic straws (4.5 points)
i. (0.8 pts) Due to the superconduct-
ing walls, the magnetic field lines cannot
cross the walls, so the flux is constant
along the tube. For a closed contour in-
side the tube, there should be no circu-
lation of the magnetic field, hence the
field lines cannot be curved, and the field
needs to be homogeneous. The field lines
close from outside the tube, similarly to a solenoid.
ii. (1.2 pts) Let us consider the change of the magnetic energy
when the tube is stretched (virtually) by a small amount ∆l.
Note that the magnetic flux trough the tube is conserved: any
change of flux would imply a non-zero electromotive force dΦ
dt,
and for a zero resistivity, an infinite current. So, the induc-
tion B = Φ
πr2 . The energy density of the magnetic field is B2
2µ0
.
— page 2 of 5 —
86
reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,
(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.
We start again throwing a ball from the point O at the top
of the building, and notice that for the optimal trajctory, at
the point P , where it touches the building, the velocity is per-
pendicular both to the radius vector QP (where Q denotes the
building’s centre), and to the launching velocity. Hence, QP
and the launching velocity are parallel. Let O be the origin,
and let us denote ∠OQP = α. Then the trajectory is given by
z = x cot α −gx2
2v2 sin2 α.
Point P with coordinates x = R sin α and z = R(cos α − 1) be-
longs to this parabola, hence R = gR2
2v2 , from where we obtain
the previous result.
Part B. Mist (4 points)
i. (0.8 pts) In the plane’s reference frame, along the channel
between two streamlines the volume flux of air (volume flow
rate) is constant due to continuity. The volume flux is the
product of speed and channel’s cross-section area, which, due
to the two-dimensional geometry, is proportional to the channel
width and can be measured from the Fig. Due to the absence of
wind, the unperturbed air’s speed in the plane’s frame is just v0.
So, upon measuring the dimensions a = 10 mm and b = 13 mm
(see Fig), we can write v0a = ub and hence u = v0ab. Since at
point P , the streamlines are horizontal where all the velocities
are parallel, the vector addition is reduced to the scalar addi-
tion: the air’s ground speed vP = v0 − u = v0(1 −ab) = 23 m/s.
ii. (1.2 pts) Although the dynamic pressure 1
2ρv2 is relatively
small, it gives rise to some adiabatic expansion and compres-
sion. In expanding regions the temperature will drop and hence,
the pressure of saturated vapours will also drop. If the dew
point is reached, a stream of droplets will appear. This process
will start in a point where the adiabatic expansion is maximal,
i.e. where the hydrostatic pressure is minimal and consequently,
as it follows from the Bernoulli’s law p + 1
2ρv2 = const, the dy-
namic pressure is maximal: in the place where the air speed in
wing’s frame is maximal and the streamline distance minimal.
Such a point Q is marked in Fig.
iii. (2 pts) First we need to calculate the dew point for the air
of given water content (since the relative pressure change will
be small, we can ignore the dependence of the dew point on
pressure). The water vapour pressure is pw = psar = 2.08 kPa.
The relative change of the pressure of the saturated vapour is
small, so we can linearize its temperature dependence:
psa − pw
Ta − T=
psb − psa
Tb − Ta
=⇒ Ta − T = (Tb − Ta)(1 − r)psa
psb − psa
;
numerically T ≈ 291.5 K. Further we need to relate the air
speed to the temperature. To this end we need to use the en-
ergy conservation law. A convenient ready-to-use form of it is
provided by the Bernoulli’s law. Applying this law will give
a good approximation of the reality, but strictly speaking, it
needs to be modified to take into account the compressibility
of air and the associated expansion/contraction work. Con-
sider one mole of air, which has the mass µ and the volume
V = RT/p. Apparently the process is fast and the air par-
cels are large, so that heat transfer across the air parcels is
negligible. Additionally, the process is subsonic; all together
we can conclude that the process is adiabatic. Consider a seg-
ment of a tube formed by the streamlines. Let us denote the
physical quantities at its one end by index 1, and at the other
end — by index 2. Then, while one mole of gas flows into
the tube at one end, as much flows out at the other end. The
inflow carries in kinetic energy 1
2µv2
1 , and the outflow carries
out 1
2µv2
2. The inflowing gas receives work due to the pushing
gas equal to p1V1 = RT1, the outflowing gas performs work
p2V2 = RT2. Let’s define molar heat capacities CV = µcV and
Cp = µcp. The inflow carries in heat energy CV RT1, and the
outflow carries out CV RT2. All together, the energy balance
can be written as 1
2µv2 + CpT = const. From this we can
easily express ∆ v2
2= 1
Cv2
crit(a2
c2 − 1) = cp∆T , where c is the
streamline distance at the point Q, and further
vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s,
where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that
in reality, the required speed is probably somewhat higher, be-
cause for a fast condensation, a considerable over-saturation is
needed. However, within an order of magnitude, this estimate
remains valid.
Part C. Magnetic straws (4.5 points)
i. (0.8 pts) Due to the superconduct-
ing walls, the magnetic field lines cannot
cross the walls, so the flux is constant
along the tube. For a closed contour in-
side the tube, there should be no circu-
lation of the magnetic field, hence the
field lines cannot be curved, and the field
needs to be homogeneous. The field lines
close from outside the tube, similarly to a solenoid.
ii. (1.2 pts) Let us consider the change of the magnetic energy
when the tube is stretched (virtually) by a small amount ∆l.
Note that the magnetic flux trough the tube is conserved: any
change of flux would imply a non-zero electromotive force dΦ
dt,
and for a zero resistivity, an infinite current. So, the induc-
tion B = Φ
πr2 . The energy density of the magnetic field is B2
2µ0
.
— page 2 of 5 —
87
reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,
(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.
We start again throwing a ball from the point O at the top
of the building, and notice that for the optimal trajctory, at
the point P , where it touches the building, the velocity is per-
pendicular both to the radius vector QP (where Q denotes the
building’s centre), and to the launching velocity. Hence, QP
and the launching velocity are parallel. Let O be the origin,
and let us denote ∠OQP = α. Then the trajectory is given by
z = x cot α −gx2
2v2 sin2 α.
Point P with coordinates x = R sin α and z = R(cos α − 1) be-
longs to this parabola, hence R = gR2
2v2 , from where we obtain
the previous result.
Part B. Mist (4 points)
i. (0.8 pts) In the plane’s reference frame, along the channel
between two streamlines the volume flux of air (volume flow
rate) is constant due to continuity. The volume flux is the
product of speed and channel’s cross-section area, which, due
to the two-dimensional geometry, is proportional to the channel
width and can be measured from the Fig. Due to the absence of
wind, the unperturbed air’s speed in the plane’s frame is just v0.
So, upon measuring the dimensions a = 10 mm and b = 13 mm
(see Fig), we can write v0a = ub and hence u = v0ab. Since at
point P , the streamlines are horizontal where all the velocities
are parallel, the vector addition is reduced to the scalar addi-
tion: the air’s ground speed vP = v0 − u = v0(1 −ab) = 23 m/s.
ii. (1.2 pts) Although the dynamic pressure 1
2ρv2 is relatively
small, it gives rise to some adiabatic expansion and compres-
sion. In expanding regions the temperature will drop and hence,
the pressure of saturated vapours will also drop. If the dew
point is reached, a stream of droplets will appear. This process
will start in a point where the adiabatic expansion is maximal,
i.e. where the hydrostatic pressure is minimal and consequently,
as it follows from the Bernoulli’s law p + 1
2ρv2 = const, the dy-
namic pressure is maximal: in the place where the air speed in
wing’s frame is maximal and the streamline distance minimal.
Such a point Q is marked in Fig.
iii. (2 pts) First we need to calculate the dew point for the air
of given water content (since the relative pressure change will
be small, we can ignore the dependence of the dew point on
pressure). The water vapour pressure is pw = psar = 2.08 kPa.
The relative change of the pressure of the saturated vapour is
small, so we can linearize its temperature dependence:
psa − pw
Ta − T=
psb − psa
Tb − Ta
=⇒ Ta − T = (Tb − Ta)(1 − r)psa
psb − psa
;
numerically T ≈ 291.5 K. Further we need to relate the air
speed to the temperature. To this end we need to use the en-
ergy conservation law. A convenient ready-to-use form of it is
provided by the Bernoulli’s law. Applying this law will give
a good approximation of the reality, but strictly speaking, it
needs to be modified to take into account the compressibility
of air and the associated expansion/contraction work. Con-
sider one mole of air, which has the mass µ and the volume
V = RT/p. Apparently the process is fast and the air par-
cels are large, so that heat transfer across the air parcels is
negligible. Additionally, the process is subsonic; all together
we can conclude that the process is adiabatic. Consider a seg-
ment of a tube formed by the streamlines. Let us denote the
physical quantities at its one end by index 1, and at the other
end — by index 2. Then, while one mole of gas flows into
the tube at one end, as much flows out at the other end. The
inflow carries in kinetic energy 1
2µv2
1 , and the outflow carries
out 1
2µv2
2. The inflowing gas receives work due to the pushing
gas equal to p1V1 = RT1, the outflowing gas performs work
p2V2 = RT2. Let’s define molar heat capacities CV = µcV and
Cp = µcp. The inflow carries in heat energy CV RT1, and the
outflow carries out CV RT2. All together, the energy balance
can be written as 1
2µv2 + CpT = const. From this we can
easily express ∆ v2
2= 1
Cv2
crit(a2
c2 − 1) = cp∆T , where c is the
streamline distance at the point Q, and further
vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s,
where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that
in reality, the required speed is probably somewhat higher, be-
cause for a fast condensation, a considerable over-saturation is
needed. However, within an order of magnitude, this estimate
remains valid.
Part C. Magnetic straws (4.5 points)
i. (0.8 pts) Due to the superconduct-
ing walls, the magnetic field lines cannot
cross the walls, so the flux is constant
along the tube. For a closed contour in-
side the tube, there should be no circu-
lation of the magnetic field, hence the
field lines cannot be curved, and the field
needs to be homogeneous. The field lines
close from outside the tube, similarly to a solenoid.
ii. (1.2 pts) Let us consider the change of the magnetic energy
when the tube is stretched (virtually) by a small amount ∆l.
Note that the magnetic flux trough the tube is conserved: any
change of flux would imply a non-zero electromotive force dΦ
dt,
and for a zero resistivity, an infinite current. So, the induc-
tion B = Φ
πr2 . The energy density of the magnetic field is B2
2µ0
.
— page 2 of 5 —
reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,
(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.
We start again throwing a ball from the point O at the top
of the building, and notice that for the optimal trajctory, at
the point P , where it touches the building, the velocity is per-
pendicular both to the radius vector QP (where Q denotes the
building’s centre), and to the launching velocity. Hence, QP
and the launching velocity are parallel. Let O be the origin,
and let us denote ∠OQP = α. Then the trajectory is given by
z = x cot α −gx2
2v2 sin2 α.
Point P with coordinates x = R sin α and z = R(cos α − 1) be-
longs to this parabola, hence R = gR2
2v2 , from where we obtain
the previous result.
Part B. Mist (4 points)
i. (0.8 pts) In the plane’s reference frame, along the channel
between two streamlines the volume flux of air (volume flow
rate) is constant due to continuity. The volume flux is the
product of speed and channel’s cross-section area, which, due
to the two-dimensional geometry, is proportional to the channel
width and can be measured from the Fig. Due to the absence of
wind, the unperturbed air’s speed in the plane’s frame is just v0.
So, upon measuring the dimensions a = 10 mm and b = 13 mm
(see Fig), we can write v0a = ub and hence u = v0ab. Since at
point P , the streamlines are horizontal where all the velocities
are parallel, the vector addition is reduced to the scalar addi-
tion: the air’s ground speed vP = v0 − u = v0(1 −ab) = 23 m/s.
ii. (1.2 pts) Although the dynamic pressure 1
2ρv2 is relatively
small, it gives rise to some adiabatic expansion and compres-
sion. In expanding regions the temperature will drop and hence,
the pressure of saturated vapours will also drop. If the dew
point is reached, a stream of droplets will appear. This process
will start in a point where the adiabatic expansion is maximal,
i.e. where the hydrostatic pressure is minimal and consequently,
as it follows from the Bernoulli’s law p + 1
2ρv2 = const, the dy-
namic pressure is maximal: in the place where the air speed in
wing’s frame is maximal and the streamline distance minimal.
Such a point Q is marked in Fig.
iii. (2 pts) First we need to calculate the dew point for the air
of given water content (since the relative pressure change will
be small, we can ignore the dependence of the dew point on
pressure). The water vapour pressure is pw = psar = 2.08 kPa.
The relative change of the pressure of the saturated vapour is
small, so we can linearize its temperature dependence:
psa − pw
Ta − T=
psb − psa
Tb − Ta
=⇒ Ta − T = (Tb − Ta)(1 − r)psa
psb − psa
;
numerically T ≈ 291.5 K. Further we need to relate the air
speed to the temperature. To this end we need to use the en-
ergy conservation law. A convenient ready-to-use form of it is
provided by the Bernoulli’s law. Applying this law will give
a good approximation of the reality, but strictly speaking, it
needs to be modified to take into account the compressibility
of air and the associated expansion/contraction work. Con-
sider one mole of air, which has the mass µ and the volume
V = RT/p. Apparently the process is fast and the air par-
cels are large, so that heat transfer across the air parcels is
negligible. Additionally, the process is subsonic; all together
we can conclude that the process is adiabatic. Consider a seg-
ment of a tube formed by the streamlines. Let us denote the
physical quantities at its one end by index 1, and at the other
end — by index 2. Then, while one mole of gas flows into
the tube at one end, as much flows out at the other end. The
inflow carries in kinetic energy 1
2µv2
1 , and the outflow carries
out 1
2µv2
2. The inflowing gas receives work due to the pushing
gas equal to p1V1 = RT1, the outflowing gas performs work
p2V2 = RT2. Let’s define molar heat capacities CV = µcV and
Cp = µcp. The inflow carries in heat energy CV RT1, and the
outflow carries out CV RT2. All together, the energy balance
can be written as 1
2µv2 + CpT = const. From this we can
easily express ∆ v2
2= 1
Cv2
crit(a2
c2 − 1) = cp∆T , where c is the
streamline distance at the point Q, and further
vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s,
where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that
in reality, the required speed is probably somewhat higher, be-
cause for a fast condensation, a considerable over-saturation is
needed. However, within an order of magnitude, this estimate
remains valid.
Part C. Magnetic straws (4.5 points)
i. (0.8 pts) Due to the superconduct-
ing walls, the magnetic field lines cannot
cross the walls, so the flux is constant
along the tube. For a closed contour in-
side the tube, there should be no circu-
lation of the magnetic field, hence the
field lines cannot be curved, and the field
needs to be homogeneous. The field lines
close from outside the tube, similarly to a solenoid.
ii. (1.2 pts) Let us consider the change of the magnetic energy
when the tube is stretched (virtually) by a small amount ∆l.
Note that the magnetic flux trough the tube is conserved: any
change of flux would imply a non-zero electromotive force dΦ
dt,
and for a zero resistivity, an infinite current. So, the induc-
tion B = Φ
πr2 . The energy density of the magnetic field is B2
2µ0
.
— page 2 of 5 —Thus, the change of the magnetic energy is calculated as
∆W =B2
2µ0
πr2∆l =Φ2
2µ0πr2∆l.
This energy increase is achieved owing to the work done by the
stretching force, ∆W = T ∆l. Hence, the force
T =Φ2
2µ0πr2.
iii. (2.5 pts) Let us analyse, what would be the change of
the magnetic energy when one of the straws is displaced to a
small distance. The magnetic field inside the tubes will remain
constant due to the conservation of magnetic flux, but outside,
the magnetic field will be changed. The magnetic field out-
side the straws is defined by the following condition: there is
no circulation of B (because there are no currents outside the
straws); there are no sources of the field lines, other than the
endpoints of the straws; each of the endpoints of the straws is
a source of streamlines with a fixed magnetic flux ±Φ. These
are exactly the same condition as those which define the elec-
tric field of four charges ±Q. We know that if the distance
between charges is much larger than the geometrical size of
a charge, the charges can be considered as point charges (the
electric field near the charges remains almost constant, so that
the respective contribution to the change of the overall electric
field energy is negligible). Therefore we can conclude that the
endpoints of the straws can be considered as magnetic point
charges. In order to calculate the force between two magnetic
charges (magnetic monopoles), we need to establish the corres-
pondence between magnetic and electric quantities.
For two electric charges Q separated by a distance a, the
force is F = 1
4πε0
Q2
a2 , and at the position of one charge, the elec-
tric field of the other charge has energy density w = 1
32π2ε0
Q2
a4 ;
hence we can write F = 8πwa2. This is a universal expression
for the force (for the case when the field lines have the same
shape as in the case of two opposite and equal by modulus elec-
tric charges) relying only on the energy density, and not related
to the nature of the field; so we can apply it to the magnetic
field. Indeed, the force can be calculated as a derivative of
the full field energy with respect to a virtual displacement of
a field line source (electric or magnetic charge); if the energy
densities of two fields are respectively equal at one point, they
are equal everywhere, and so are equal the full field energies.
As it follows from the Gauss law, for a point source of a fixed
magnetic flux Φ at a distance a, the induction B = 1
4πΦ
a2 . So,
the energy density w = B2
2µ0
= 1
32π2µ0
Φ2
a4 , hence
F =1
4πµ0
Φ2
a2.
For the two straws, we have four magnetic charges. The lon-
gitudinal (along a straw axis) forces cancel out (the diagonally
positioned pairs of same-sign-charges push in opposite direc-
tions). The normal force is a superposition of the attraction
due to the two pairs of opposite charges, F1 = 1
4πµ0
Φ2
l2 , and
the repulsive forces of diagonal pairs, F2 =√
2
8πµ0
Φ2
2l2 . The net
attractive force will be
F = 2(F1 − F2) =4 −
√2
8πµ0
Φ2
l2.
Alternative solution based on dipole energy calcu-
lation. It is known that the axial component of the magnetic
induction created by a solenoidal current of surface density j
is proportional to the solid angle Ω under which the current is
seen from the given point:
B = µ0jΩ/4π;
this can be easily derived from the Biot-Savart law. Let the
distance between the tubes be a (we’ll take derivative over a),
and let us consider a first tube’s point which has a coordinate
x (with 0 ≤ x ≤ l) from where the direction to the one end-
point of the other tube forms an angle α = arctan a/x with
the tube’s axis. From that point, the open circular face of the
other tube forms a solid angle Ω = πr2 sin2 α cos α/a2, so that
its contribution to the axial magnetic field at the point x
B =µ0jr2 sin2 α cos α
4a2=
Φ sin2 α cos α
4πa2.
The solenoidal current at that point forms a magnetic dipole
dm = πr2jdx = Φdx/µ0,
which has potential energy
dU = Bdm = sin2 α cos αΦ2dx/4πµ0a2.
When integrating over x, α varies from arctan a/l to 0, so that
U1 =
∫
dU =
∫
sin2 α cos αΦ2dx
4πµ0a2=
∫
cos αΦ2dα
4πµ0a.
Bearing in mind that the other end-circle of the other tube con-
tributes the same amount to the magnetic interaction energy,
we find
U = 2U1 =Φ2
2πµ0
(
1
a−
1√
a2 + l2
)
.
Upon taking derivative over a and using a = l we obtain the
same result for F as previously.
— page 3 of 5 —
88
Thus, the change of the magnetic energy is calculated as
∆W =B2
2µ0
πr2∆l =Φ2
2µ0πr2∆l.
This energy increase is achieved owing to the work done by the
stretching force, ∆W = T ∆l. Hence, the force
T =Φ2
2µ0πr2.
iii. (2.5 pts) Let us analyse, what would be the change of
the magnetic energy when one of the straws is displaced to a
small distance. The magnetic field inside the tubes will remain
constant due to the conservation of magnetic flux, but outside,
the magnetic field will be changed. The magnetic field out-
side the straws is defined by the following condition: there is
no circulation of B (because there are no currents outside the
straws); there are no sources of the field lines, other than the
endpoints of the straws; each of the endpoints of the straws is
a source of streamlines with a fixed magnetic flux ±Φ. These
are exactly the same condition as those which define the elec-
tric field of four charges ±Q. We know that if the distance
between charges is much larger than the geometrical size of
a charge, the charges can be considered as point charges (the
electric field near the charges remains almost constant, so that
the respective contribution to the change of the overall electric
field energy is negligible). Therefore we can conclude that the
endpoints of the straws can be considered as magnetic point
charges. In order to calculate the force between two magnetic
charges (magnetic monopoles), we need to establish the corres-
pondence between magnetic and electric quantities.
For two electric charges Q separated by a distance a, the
force is F = 1
4πε0
Q2
a2 , and at the position of one charge, the elec-
tric field of the other charge has energy density w = 1
32π2ε0
Q2
a4 ;
hence we can write F = 8πwa2. This is a universal expression
for the force (for the case when the field lines have the same
shape as in the case of two opposite and equal by modulus elec-
tric charges) relying only on the energy density, and not related
to the nature of the field; so we can apply it to the magnetic
field. Indeed, the force can be calculated as a derivative of
the full field energy with respect to a virtual displacement of
a field line source (electric or magnetic charge); if the energy
densities of two fields are respectively equal at one point, they
are equal everywhere, and so are equal the full field energies.
As it follows from the Gauss law, for a point source of a fixed
magnetic flux Φ at a distance a, the induction B = 1
4πΦ
a2 . So,
the energy density w = B2
2µ0
= 1
32π2µ0
Φ2
a4 , hence
F =1
4πµ0
Φ2
a2.
For the two straws, we have four magnetic charges. The lon-
gitudinal (along a straw axis) forces cancel out (the diagonally
positioned pairs of same-sign-charges push in opposite direc-
tions). The normal force is a superposition of the attraction
due to the two pairs of opposite charges, F1 = 1
4πµ0
Φ2
l2 , and
the repulsive forces of diagonal pairs, F2 =√
2
8πµ0
Φ2
2l2 . The net
attractive force will be
F = 2(F1 − F2) =4 −
√2
8πµ0
Φ2
l2.
Alternative solution based on dipole energy calcu-
lation. It is known that the axial component of the magnetic
induction created by a solenoidal current of surface density j
is proportional to the solid angle Ω under which the current is
seen from the given point:
B = µ0jΩ/4π;
this can be easily derived from the Biot-Savart law. Let the
distance between the tubes be a (we’ll take derivative over a),
and let us consider a first tube’s point which has a coordinate
x (with 0 ≤ x ≤ l) from where the direction to the one end-
point of the other tube forms an angle α = arctan a/x with
the tube’s axis. From that point, the open circular face of the
other tube forms a solid angle Ω = πr2 sin2 α cos α/a2, so that
its contribution to the axial magnetic field at the point x
B =µ0jr2 sin2 α cos α
4a2=
Φ sin2 α cos α
4πa2.
The solenoidal current at that point forms a magnetic dipole
dm = πr2jdx = Φdx/µ0,
which has potential energy
dU = Bdm = sin2 α cos αΦ2dx/4πµ0a2.
When integrating over x, α varies from arctan a/l to 0, so that
U1 =
∫
dU =
∫
sin2 α cos αΦ2dx
4πµ0a2=
∫
cos αΦ2dα
4πµ0a.
Bearing in mind that the other end-circle of the other tube con-
tributes the same amount to the magnetic interaction energy,
we find
U = 2U1 =Φ2
2πµ0
(
1
a−
1√
a2 + l2
)
.
Upon taking derivative over a and using a = l we obtain the
same result for F as previously.
— page 3 of 5 —
Thus, the change of the magnetic energy is calculated as
∆W =B2
2µ0
πr2∆l =Φ2
2µ0πr2∆l.
This energy increase is achieved owing to the work done by the
stretching force, ∆W = T ∆l. Hence, the force
T =Φ2
2µ0πr2.
iii. (2.5 pts) Let us analyse, what would be the change of
the magnetic energy when one of the straws is displaced to a
small distance. The magnetic field inside the tubes will remain
constant due to the conservation of magnetic flux, but outside,
the magnetic field will be changed. The magnetic field out-
side the straws is defined by the following condition: there is
no circulation of B (because there are no currents outside the
straws); there are no sources of the field lines, other than the
endpoints of the straws; each of the endpoints of the straws is
a source of streamlines with a fixed magnetic flux ±Φ. These
are exactly the same condition as those which define the elec-
tric field of four charges ±Q. We know that if the distance
between charges is much larger than the geometrical size of
a charge, the charges can be considered as point charges (the
electric field near the charges remains almost constant, so that
the respective contribution to the change of the overall electric
field energy is negligible). Therefore we can conclude that the
endpoints of the straws can be considered as magnetic point
charges. In order to calculate the force between two magnetic
charges (magnetic monopoles), we need to establish the corres-
pondence between magnetic and electric quantities.
For two electric charges Q separated by a distance a, the
force is F = 1
4πε0
Q2
a2 , and at the position of one charge, the elec-
tric field of the other charge has energy density w = 1
32π2ε0
Q2
a4 ;
hence we can write F = 8πwa2. This is a universal expression
for the force (for the case when the field lines have the same
shape as in the case of two opposite and equal by modulus elec-
tric charges) relying only on the energy density, and not related
to the nature of the field; so we can apply it to the magnetic
field. Indeed, the force can be calculated as a derivative of
the full field energy with respect to a virtual displacement of
a field line source (electric or magnetic charge); if the energy
densities of two fields are respectively equal at one point, they
are equal everywhere, and so are equal the full field energies.
As it follows from the Gauss law, for a point source of a fixed
magnetic flux Φ at a distance a, the induction B = 1
4πΦ
a2 . So,
the energy density w = B2
2µ0
= 1
32π2µ0
Φ2
a4 , hence
F =1
4πµ0
Φ2
a2.
For the two straws, we have four magnetic charges. The lon-
gitudinal (along a straw axis) forces cancel out (the diagonally
positioned pairs of same-sign-charges push in opposite direc-
tions). The normal force is a superposition of the attraction
due to the two pairs of opposite charges, F1 = 1
4πµ0
Φ2
l2 , and
the repulsive forces of diagonal pairs, F2 =√
2
8πµ0
Φ2
2l2 . The net
attractive force will be
F = 2(F1 − F2) =4 −
√2
8πµ0
Φ2
l2.
Alternative solution based on dipole energy calcu-
lation. It is known that the axial component of the magnetic
induction created by a solenoidal current of surface density j
is proportional to the solid angle Ω under which the current is
seen from the given point:
B = µ0jΩ/4π;
this can be easily derived from the Biot-Savart law. Let the
distance between the tubes be a (we’ll take derivative over a),
and let us consider a first tube’s point which has a coordinate
x (with 0 ≤ x ≤ l) from where the direction to the one end-
point of the other tube forms an angle α = arctan a/x with
the tube’s axis. From that point, the open circular face of the
other tube forms a solid angle Ω = πr2 sin2 α cos α/a2, so that
its contribution to the axial magnetic field at the point x
B =µ0jr2 sin2 α cos α
4a2=
Φ sin2 α cos α
4πa2.
The solenoidal current at that point forms a magnetic dipole
dm = πr2jdx = Φdx/µ0,
which has potential energy
dU = Bdm = sin2 α cos αΦ2dx/4πµ0a2.
When integrating over x, α varies from arctan a/l to 0, so that
U1 =
∫
dU =
∫
sin2 α cos αΦ2dx
4πµ0a2=
∫
cos αΦ2dα
4πµ0a.
Bearing in mind that the other end-circle of the other tube con-
tributes the same amount to the magnetic interaction energy,
we find
U = 2U1 =Φ2
2πµ0
(
1
a−
1√
a2 + l2
)
.
Upon taking derivative over a and using a = l we obtain the
same result for F as previously.
— page 3 of 5 —
89
PrObLEm T2. kELvIn waTEr drOPPEr (8 POInTs)
Thus, the change of the magnetic energy is calculated as
∆W =B2
2µ0
πr2∆l =Φ2
2µ0πr2∆l.
This energy increase is achieved owing to the work done by the
stretching force, ∆W = T ∆l. Hence, the force
T =Φ2
2µ0πr2.
iii. (2.5 pts) Let us analyse, what would be the change of
the magnetic energy when one of the straws is displaced to a
small distance. The magnetic field inside the tubes will remain
constant due to the conservation of magnetic flux, but outside,
the magnetic field will be changed. The magnetic field out-
side the straws is defined by the following condition: there is
no circulation of B (because there are no currents outside the
straws); there are no sources of the field lines, other than the
endpoints of the straws; each of the endpoints of the straws is
a source of streamlines with a fixed magnetic flux ±Φ. These
are exactly the same condition as those which define the elec-
tric field of four charges ±Q. We know that if the distance
between charges is much larger than the geometrical size of
a charge, the charges can be considered as point charges (the
electric field near the charges remains almost constant, so that
the respective contribution to the change of the overall electric
field energy is negligible). Therefore we can conclude that the
endpoints of the straws can be considered as magnetic point
charges. In order to calculate the force between two magnetic
charges (magnetic monopoles), we need to establish the corres-
pondence between magnetic and electric quantities.
For two electric charges Q separated by a distance a, the
force is F = 1
4πε0
Q2
a2 , and at the position of one charge, the elec-
tric field of the other charge has energy density w = 1
32π2ε0
Q2
a4 ;
hence we can write F = 8πwa2. This is a universal expression
for the force (for the case when the field lines have the same
shape as in the case of two opposite and equal by modulus elec-
tric charges) relying only on the energy density, and not related
to the nature of the field; so we can apply it to the magnetic
field. Indeed, the force can be calculated as a derivative of
the full field energy with respect to a virtual displacement of
a field line source (electric or magnetic charge); if the energy
densities of two fields are respectively equal at one point, they
are equal everywhere, and so are equal the full field energies.
As it follows from the Gauss law, for a point source of a fixed
magnetic flux Φ at a distance a, the induction B = 1
4πΦ
a2 . So,
the energy density w = B2
2µ0
= 1
32π2µ0
Φ2
a4 , hence
F =1
4πµ0
Φ2
a2.
For the two straws, we have four magnetic charges. The lon-
gitudinal (along a straw axis) forces cancel out (the diagonally
positioned pairs of same-sign-charges push in opposite direc-
tions). The normal force is a superposition of the attraction
due to the two pairs of opposite charges, F1 = 1
4πµ0
Φ2
l2 , and
the repulsive forces of diagonal pairs, F2 =√
2
8πµ0
Φ2
2l2 . The net
attractive force will be
F = 2(F1 − F2) =4 −
√2
8πµ0
Φ2
l2.
Alternative solution based on dipole energy calcu-
lation. It is known that the axial component of the magnetic
induction created by a solenoidal current of surface density j
is proportional to the solid angle Ω under which the current is
seen from the given point:
B = µ0jΩ/4π;
this can be easily derived from the Biot-Savart law. Let the
distance between the tubes be a (we’ll take derivative over a),
and let us consider a first tube’s point which has a coordinate
x (with 0 ≤ x ≤ l) from where the direction to the one end-
point of the other tube forms an angle α = arctan a/x with
the tube’s axis. From that point, the open circular face of the
other tube forms a solid angle Ω = πr2 sin2 α cos α/a2, so that
its contribution to the axial magnetic field at the point x
B =µ0jr2 sin2 α cos α
4a2=
Φ sin2 α cos α
4πa2.
The solenoidal current at that point forms a magnetic dipole
dm = πr2jdx = Φdx/µ0,
which has potential energy
dU = Bdm = sin2 α cos αΦ2dx/4πµ0a2.
When integrating over x, α varies from arctan a/l to 0, so that
U1 =
∫
dU =
∫
sin2 α cos αΦ2dx
4πµ0a2=
∫
cos αΦ2dα
4πµ0a.
Bearing in mind that the other end-circle of the other tube con-
tributes the same amount to the magnetic interaction energy,
we find
U = 2U1 =Φ2
2πµ0
(
1
a−
1√
a2 + l2
)
.
Upon taking derivative over a and using a = l we obtain the
same result for F as previously.
— page 3 of 5 —Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) Let us write the force balance for the droplet.
Since d ≪ r, we can neglect the force π4
∆pd2 due to the excess
pressure ∆p inside the tube. So, the gravity force 4
3πr3
maxρg
is balanced by the capillary force. When the droplet separates
from the tube, the water surface forms in the vicinity of the
nozzle a “neck”, which has vertical tangent. In the horizontal
cross-section of that “neck”, the capillary force is vertical and
can be calculated as πσd. So,
rmax = 3
√
3σd
4ρg.
ii. (1.2 pts) Since d ≪ r, we can neglect the change of the
droplet’s capacitance due to the tube. On the one hand, the
droplet’s potential is ϕ; on the other hand, it is 1
4πε0
Q
r. So,
Q = 4πε0ϕr.
iii. (1.6 pts) Excess pressure inside the droplet is caused by
the capillary pressure 2σ/r (increases the inside pressure), and
by the electrostatic pressure 1
2ε0E2 = 1
2ε0ϕ2/r2 (decreases the
pressure). So, the sign of the excess pressure will change, if1
2ε0ϕ2
max/r2 = 2σ/r, hence
ϕmax = 2√
σr/ε0.
The expression for the electrostatic pressure used above can
be derived as follows. The electrostatic force acting on a surface
charge of density σ and surface area S is given by F = σS · E,
where E is the field at the site without the field created by the
surface charge element itself. Note that this force is perpen-
dicular to the surface, so F/S can be interpreted as a pressure.
The surface charge gives rise to a field drop on the surface equal
to ∆E = σ/ε0 (which follows from the Gauss law); inside the
droplet, there is no field due to the conductivity of the droplet:
E − 1
2∆E = 0; outside the droplet, there is field E = E + 1
2∆E,
therefore E = 1
2E = 1
2∆E. Bringing everything together, we
obtain the expression used above.
Note that alternatively, this expression can be derived by
considering a virtual displacement of a capacitor’s surface and
comparing the pressure work p∆V with the change of the elec-
trostatic field energy 1
2ε0E2∆V .
Finally, the answer to the question can be also derived from
the requirement that the mechanical work dA done for an in-
finitesimal droplet inflation needs to be zero. From the en-
ergy conservation law, dW + dWel
= σ d(4πr2) + 1
2ϕ2
max dCd,
where the droplet’s capacitance Cd = 4πε0r; the electrical work
dWel
= ϕmaxdq = 4πε0ϕ2maxdr. Putting dW = 0 we obtain an
equation for ϕmax, which recovers the earlier result.
Part B. Two pipes (4 points)
i. (1.2 pts) This is basically the same as Part A-ii, except
that the surroundings’ potential is that of the surrounding
electrode, −U/2 (where U = q/C is the capacitor’s voltage)
and droplet has the ground potential (0). As it is not defined
which electrode is the positive one, opposite sign of the po-
tential may be chosen, if done consistently. Note that since
the cylindrical electrode is long, it shields effectively the en-
vironment’s (ground, wall, etc) potential. So, relative to its
surroundings, the droplet’s potential is U/2. Using the result
of Part A we obtain
Q = 2πε0Urmax = 2πε0qrmax/C.
ii. (1.5 pts) The sign of the droplet’s charge is the same as
that of the capacitor’s opposite plate (which is connected to
the farther electrode). So, when the droplet falls into the bowl,
it will increase the capacitor’s charge by Q:
dq = 2πε0UrmaxdN = 2πε0rmaxndtq
C,
where dN = ndt is the number of droplets which fall during
the time dt This is a simple linear differential equation which
is solved easily to obtain
q = q0eγt, γ =2πε0rmaxn
C=
πε0n
C3
√
6σd
ρg.
iii. (1.3 pts) The droplets can reach the bowls if their mech-
anical energy mgH (where m is the droplet’s mass) is large
enough to overcome the electrostatic push: The droplet starts
at the point where the electric potential is 0, which is the sum of
the potential U/2, due to the electrode, and of its self-generated
potential −U/2. Its motion is not affected by the self-generated
field, so it needs to fall from the potential U/2 down to the po-
tential −U/2, resulting in the change of the electrostatic energy
equal to UQ ≤ mgH , where Q = 2πε0Urmax (see above). So,
Umax =mgH
2πε0Umaxrmax
,
∴ Umax =
√
Hσd
2ε0rmax
= 6
√
H3gσ2ρd2
6ε30
.
— page 4 of 5 —
90
Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) Let us write the force balance for the droplet.
Since d ≪ r, we can neglect the force π4
∆pd2 due to the excess
pressure ∆p inside the tube. So, the gravity force 4
3πr3
maxρg
is balanced by the capillary force. When the droplet separates
from the tube, the water surface forms in the vicinity of the
nozzle a “neck”, which has vertical tangent. In the horizontal
cross-section of that “neck”, the capillary force is vertical and
can be calculated as πσd. So,
rmax = 3
√
3σd
4ρg.
ii. (1.2 pts) Since d ≪ r, we can neglect the change of the
droplet’s capacitance due to the tube. On the one hand, the
droplet’s potential is ϕ; on the other hand, it is 1
4πε0
Q
r. So,
Q = 4πε0ϕr.
iii. (1.6 pts) Excess pressure inside the droplet is caused by
the capillary pressure 2σ/r (increases the inside pressure), and
by the electrostatic pressure 1
2ε0E2 = 1
2ε0ϕ2/r2 (decreases the
pressure). So, the sign of the excess pressure will change, if1
2ε0ϕ2
max/r2 = 2σ/r, hence
ϕmax = 2√
σr/ε0.
The expression for the electrostatic pressure used above can
be derived as follows. The electrostatic force acting on a surface
charge of density σ and surface area S is given by F = σS · E,
where E is the field at the site without the field created by the
surface charge element itself. Note that this force is perpen-
dicular to the surface, so F/S can be interpreted as a pressure.
The surface charge gives rise to a field drop on the surface equal
to ∆E = σ/ε0 (which follows from the Gauss law); inside the
droplet, there is no field due to the conductivity of the droplet:
E − 1
2∆E = 0; outside the droplet, there is field E = E + 1
2∆E,
therefore E = 1
2E = 1
2∆E. Bringing everything together, we
obtain the expression used above.
Note that alternatively, this expression can be derived by
considering a virtual displacement of a capacitor’s surface and
comparing the pressure work p∆V with the change of the elec-
trostatic field energy 1
2ε0E2∆V .
Finally, the answer to the question can be also derived from
the requirement that the mechanical work dA done for an in-
finitesimal droplet inflation needs to be zero. From the en-
ergy conservation law, dW + dWel
= σ d(4πr2) + 1
2ϕ2
max dCd,
where the droplet’s capacitance Cd = 4πε0r; the electrical work
dWel
= ϕmaxdq = 4πε0ϕ2maxdr. Putting dW = 0 we obtain an
equation for ϕmax, which recovers the earlier result.
Part B. Two pipes (4 points)
i. (1.2 pts) This is basically the same as Part A-ii, except
that the surroundings’ potential is that of the surrounding
electrode, −U/2 (where U = q/C is the capacitor’s voltage)
and droplet has the ground potential (0). As it is not defined
which electrode is the positive one, opposite sign of the po-
tential may be chosen, if done consistently. Note that since
the cylindrical electrode is long, it shields effectively the en-
vironment’s (ground, wall, etc) potential. So, relative to its
surroundings, the droplet’s potential is U/2. Using the result
of Part A we obtain
Q = 2πε0Urmax = 2πε0qrmax/C.
ii. (1.5 pts) The sign of the droplet’s charge is the same as
that of the capacitor’s opposite plate (which is connected to
the farther electrode). So, when the droplet falls into the bowl,
it will increase the capacitor’s charge by Q:
dq = 2πε0UrmaxdN = 2πε0rmaxndtq
C,
where dN = ndt is the number of droplets which fall during
the time dt This is a simple linear differential equation which
is solved easily to obtain
q = q0eγt, γ =2πε0rmaxn
C=
πε0n
C3
√
6σd
ρg.
iii. (1.3 pts) The droplets can reach the bowls if their mech-
anical energy mgH (where m is the droplet’s mass) is large
enough to overcome the electrostatic push: The droplet starts
at the point where the electric potential is 0, which is the sum of
the potential U/2, due to the electrode, and of its self-generated
potential −U/2. Its motion is not affected by the self-generated
field, so it needs to fall from the potential U/2 down to the po-
tential −U/2, resulting in the change of the electrostatic energy
equal to UQ ≤ mgH , where Q = 2πε0Urmax (see above). So,
Umax =mgH
2πε0Umaxrmax
,
∴ Umax =
√
Hσd
2ε0rmax
= 6
√
H3gσ2ρd2
6ε30
.
— page 4 of 5 —
Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) Let us write the force balance for the droplet.
Since d ≪ r, we can neglect the force π4
∆pd2 due to the excess
pressure ∆p inside the tube. So, the gravity force 4
3πr3
maxρg
is balanced by the capillary force. When the droplet separates
from the tube, the water surface forms in the vicinity of the
nozzle a “neck”, which has vertical tangent. In the horizontal
cross-section of that “neck”, the capillary force is vertical and
can be calculated as πσd. So,
rmax = 3
√
3σd
4ρg.
ii. (1.2 pts) Since d ≪ r, we can neglect the change of the
droplet’s capacitance due to the tube. On the one hand, the
droplet’s potential is ϕ; on the other hand, it is 1
4πε0
Q
r. So,
Q = 4πε0ϕr.
iii. (1.6 pts) Excess pressure inside the droplet is caused by
the capillary pressure 2σ/r (increases the inside pressure), and
by the electrostatic pressure 1
2ε0E2 = 1
2ε0ϕ2/r2 (decreases the
pressure). So, the sign of the excess pressure will change, if1
2ε0ϕ2
max/r2 = 2σ/r, hence
ϕmax = 2√
σr/ε0.
The expression for the electrostatic pressure used above can
be derived as follows. The electrostatic force acting on a surface
charge of density σ and surface area S is given by F = σS · E,
where E is the field at the site without the field created by the
surface charge element itself. Note that this force is perpen-
dicular to the surface, so F/S can be interpreted as a pressure.
The surface charge gives rise to a field drop on the surface equal
to ∆E = σ/ε0 (which follows from the Gauss law); inside the
droplet, there is no field due to the conductivity of the droplet:
E − 1
2∆E = 0; outside the droplet, there is field E = E + 1
2∆E,
therefore E = 1
2E = 1
2∆E. Bringing everything together, we
obtain the expression used above.
Note that alternatively, this expression can be derived by
considering a virtual displacement of a capacitor’s surface and
comparing the pressure work p∆V with the change of the elec-
trostatic field energy 1
2ε0E2∆V .
Finally, the answer to the question can be also derived from
the requirement that the mechanical work dA done for an in-
finitesimal droplet inflation needs to be zero. From the en-
ergy conservation law, dW + dWel
= σ d(4πr2) + 1
2ϕ2
max dCd,
where the droplet’s capacitance Cd = 4πε0r; the electrical work
dWel
= ϕmaxdq = 4πε0ϕ2maxdr. Putting dW = 0 we obtain an
equation for ϕmax, which recovers the earlier result.
Part B. Two pipes (4 points)
i. (1.2 pts) This is basically the same as Part A-ii, except
that the surroundings’ potential is that of the surrounding
electrode, −U/2 (where U = q/C is the capacitor’s voltage)
and droplet has the ground potential (0). As it is not defined
which electrode is the positive one, opposite sign of the po-
tential may be chosen, if done consistently. Note that since
the cylindrical electrode is long, it shields effectively the en-
vironment’s (ground, wall, etc) potential. So, relative to its
surroundings, the droplet’s potential is U/2. Using the result
of Part A we obtain
Q = 2πε0Urmax = 2πε0qrmax/C.
ii. (1.5 pts) The sign of the droplet’s charge is the same as
that of the capacitor’s opposite plate (which is connected to
the farther electrode). So, when the droplet falls into the bowl,
it will increase the capacitor’s charge by Q:
dq = 2πε0UrmaxdN = 2πε0rmaxndtq
C,
where dN = ndt is the number of droplets which fall during
the time dt This is a simple linear differential equation which
is solved easily to obtain
q = q0eγt, γ =2πε0rmaxn
C=
πε0n
C3
√
6σd
ρg.
iii. (1.3 pts) The droplets can reach the bowls if their mech-
anical energy mgH (where m is the droplet’s mass) is large
enough to overcome the electrostatic push: The droplet starts
at the point where the electric potential is 0, which is the sum of
the potential U/2, due to the electrode, and of its self-generated
potential −U/2. Its motion is not affected by the self-generated
field, so it needs to fall from the potential U/2 down to the po-
tential −U/2, resulting in the change of the electrostatic energy
equal to UQ ≤ mgH , where Q = 2πε0Urmax (see above). So,
Umax =mgH
2πε0Umaxrmax
,
∴ Umax =
√
Hσd
2ε0rmax
= 6
√
H3gσ2ρd2
6ε30
.
— page 4 of 5 —
91
Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) Let us write the force balance for the droplet.
Since d ≪ r, we can neglect the force π4
∆pd2 due to the excess
pressure ∆p inside the tube. So, the gravity force 4
3πr3
maxρg
is balanced by the capillary force. When the droplet separates
from the tube, the water surface forms in the vicinity of the
nozzle a “neck”, which has vertical tangent. In the horizontal
cross-section of that “neck”, the capillary force is vertical and
can be calculated as πσd. So,
rmax = 3
√
3σd
4ρg.
ii. (1.2 pts) Since d ≪ r, we can neglect the change of the
droplet’s capacitance due to the tube. On the one hand, the
droplet’s potential is ϕ; on the other hand, it is 1
4πε0
Q
r. So,
Q = 4πε0ϕr.
iii. (1.6 pts) Excess pressure inside the droplet is caused by
the capillary pressure 2σ/r (increases the inside pressure), and
by the electrostatic pressure 1
2ε0E2 = 1
2ε0ϕ2/r2 (decreases the
pressure). So, the sign of the excess pressure will change, if1
2ε0ϕ2
max/r2 = 2σ/r, hence
ϕmax = 2√
σr/ε0.
The expression for the electrostatic pressure used above can
be derived as follows. The electrostatic force acting on a surface
charge of density σ and surface area S is given by F = σS · E,
where E is the field at the site without the field created by the
surface charge element itself. Note that this force is perpen-
dicular to the surface, so F/S can be interpreted as a pressure.
The surface charge gives rise to a field drop on the surface equal
to ∆E = σ/ε0 (which follows from the Gauss law); inside the
droplet, there is no field due to the conductivity of the droplet:
E − 1
2∆E = 0; outside the droplet, there is field E = E + 1
2∆E,
therefore E = 1
2E = 1
2∆E. Bringing everything together, we
obtain the expression used above.
Note that alternatively, this expression can be derived by
considering a virtual displacement of a capacitor’s surface and
comparing the pressure work p∆V with the change of the elec-
trostatic field energy 1
2ε0E2∆V .
Finally, the answer to the question can be also derived from
the requirement that the mechanical work dA done for an in-
finitesimal droplet inflation needs to be zero. From the en-
ergy conservation law, dW + dWel
= σ d(4πr2) + 1
2ϕ2
max dCd,
where the droplet’s capacitance Cd = 4πε0r; the electrical work
dWel
= ϕmaxdq = 4πε0ϕ2maxdr. Putting dW = 0 we obtain an
equation for ϕmax, which recovers the earlier result.
Part B. Two pipes (4 points)
i. (1.2 pts) This is basically the same as Part A-ii, except
that the surroundings’ potential is that of the surrounding
electrode, −U/2 (where U = q/C is the capacitor’s voltage)
and droplet has the ground potential (0). As it is not defined
which electrode is the positive one, opposite sign of the po-
tential may be chosen, if done consistently. Note that since
the cylindrical electrode is long, it shields effectively the en-
vironment’s (ground, wall, etc) potential. So, relative to its
surroundings, the droplet’s potential is U/2. Using the result
of Part A we obtain
Q = 2πε0Urmax = 2πε0qrmax/C.
ii. (1.5 pts) The sign of the droplet’s charge is the same as
that of the capacitor’s opposite plate (which is connected to
the farther electrode). So, when the droplet falls into the bowl,
it will increase the capacitor’s charge by Q:
dq = 2πε0UrmaxdN = 2πε0rmaxndtq
C,
where dN = ndt is the number of droplets which fall during
the time dt This is a simple linear differential equation which
is solved easily to obtain
q = q0eγt, γ =2πε0rmaxn
C=
πε0n
C3
√
6σd
ρg.
iii. (1.3 pts) The droplets can reach the bowls if their mech-
anical energy mgH (where m is the droplet’s mass) is large
enough to overcome the electrostatic push: The droplet starts
at the point where the electric potential is 0, which is the sum of
the potential U/2, due to the electrode, and of its self-generated
potential −U/2. Its motion is not affected by the self-generated
field, so it needs to fall from the potential U/2 down to the po-
tential −U/2, resulting in the change of the electrostatic energy
equal to UQ ≤ mgH , where Q = 2πε0Urmax (see above). So,
Umax =mgH
2πε0Umaxrmax
,
∴ Umax =
√
Hσd
2ε0rmax
= 6
√
H3gσ2ρd2
6ε30
.
— page 4 of 5 —
92
Problem T3. Protostar formation (9 points)i. (0.8 pts)
T = const =⇒ pV = const
V ∝ r3
∴ p ∝ r−3 =⇒p(r1)
p(r0)= 23 = 8.
ii. (1 pt) During the period considered the pressure is negli-
gible. Therefore the gas is in free fall. By Gauss’ theorem and
symmetry, the gravitational field at any point in the ball is
equivalent to the one generated when all the mass closer to the
center is compressed into the center. Moreover, while the ball
has not yet shrunk much, the field strength on its surface does
not change much either. The acceleration of the outermost
layer stays approximately constant. Thus,
t ≈
√
2(r0 − r2)
g
where
g ≈Gm
r20
,
∴ t ≈
√
2r20(r0 − r2)
Gm=
√
0.1r30
Gm.
iii. (2.5 pts) Gravitationally the outer layer of the ball is in-
fluenced by the rest just as the rest were compressed into a
point mass. Therefore we have Keplerian motion: the fall of
any part of the outer layer consists in a halfperiod of an ultra-
elliptical orbit. The ellipse is degenerate into a line; its foci are
at the ends of the line; one focus is at the center of the ball (by
Kepler’s 1st law) and the other one is at r0, see figure (instead
of a degenerate ellipse, a strongly elliptical ellipse is depicted).
The period of the orbit is determined by the longer semiaxis of
the ellipse (by Kepler’s 3rd law). The longer semiaxis is r0/2
and we are interested in half a period. Thus, the answer is
equal to the halfperiod of a circular orbit of radius r0/2:
(
2π
2tr→0
)2r0
2=
Gm
(r0/2)2=⇒ tr→0 = π
√
r30
8Gm.
Alternatively, one may write the energy conservation lawr2
2−
Gmr
= E (that in turn is obtainable from Newton’s
II law r = − Gmr2 ) with E = − Gm
r0
, separate the variables
(drdt
= −
√
2E + 2Gmr
) and write the integral t = −∫
dr√2E+ 2Gm
r
.
This integral is probably not calculable during the limitted
time given during the Olympiad, but a possible approach can
be sketched as follows. Substituting√
2E + 2Gmr
= ξ and√
2E = υ, one gets
t∞
4Gm=
∫
∞
0
dξ
(υ2 − ξ2)2
=1
4υ3
∫
∞
0
[
υ
(υ − ξ)2+
υ
(υ + ξ)2+
1
υ − ξ+
1
υ + ξ
]
dξ.
Here (after shifting the variable) one can use∫
dξξ
= ln ξ and∫
dξ
ξ2 = − 1
ξ, finally getting the same answer as by Kepler’s laws.
iv. (1.7 pts) By Clapeyron–Mendeleyev law,
p =mRT0
µV.
Work done by gravity to compress the ball is
W = −
∫
p dV = −mRT0
µ
∫ 4
3πr3
3
4
3πr3
0
dV
V=
3mRT0
µln
r0
r3
.
The temperature stays constant, so the internal energy does not
change; hence, according to the 1st law of thermodynamics, the
compression work W is the heat radiated.
v. (1 pt) The collapse continues adiabatically.
pV γ = const =⇒ T V γ−1 = const.
∴ T ∝ V 1−γ∝ r3−3γ
∴ T = T0
(r3
r
)3γ−3
.
vi. (2 pts) During the collapse, the gravitational energy is con-
verted into heat. Since r3 ≫ r4, The released gravitational en-
ergy can be estimated as ∆Π = −Gm2(r−14 −r−1
3 ) ≈ −Gm2/r4
(exact calculation by integration adds a prefactor 3
5); the ter-
minal heat energy is estimated as ∆Q = cVmµ
(T4 − T0) ≈
cVmµ
T4 (the approximation T4 ≫ T0 follows from the result
of the previous question, when combined with r3 ≫ r4). So,
∆Q = Rγ−1
mµ
T4 ≈mµ
RT4. For the temperature T4, we can use
the result of the previous question, T4 = T0
(
r3
r4
)3γ−3
. Since
initial full energy was approximately zero, ∆Q + ∆Π ≈ 0, we
obtain
Gm2
r4
≈m
µRT0
(
r3
r4
)3γ−3
=⇒ r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
.
Therefore,
T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
.
Alternatively, one can obtain the result by approximately
equating the hydrostatic pressure ρr4Gmr2
4
to the gas pressure
p4 = ρ
µRT4; the result will be exactly the same as given above.
— page 5 of 5 —
PrObLEm T3. PrOTOsTar FOrmaTIOn (9 POInTs)
93
Problem T3. Protostar formation (9 points)i. (0.8 pts)
T = const =⇒ pV = const
V ∝ r3
∴ p ∝ r−3 =⇒p(r1)
p(r0)= 23 = 8.
ii. (1 pt) During the period considered the pressure is negli-
gible. Therefore the gas is in free fall. By Gauss’ theorem and
symmetry, the gravitational field at any point in the ball is
equivalent to the one generated when all the mass closer to the
center is compressed into the center. Moreover, while the ball
has not yet shrunk much, the field strength on its surface does
not change much either. The acceleration of the outermost
layer stays approximately constant. Thus,
t ≈
√
2(r0 − r2)
g
where
g ≈Gm
r20
,
∴ t ≈
√
2r20(r0 − r2)
Gm=
√
0.1r30
Gm.
iii. (2.5 pts) Gravitationally the outer layer of the ball is in-
fluenced by the rest just as the rest were compressed into a
point mass. Therefore we have Keplerian motion: the fall of
any part of the outer layer consists in a halfperiod of an ultra-
elliptical orbit. The ellipse is degenerate into a line; its foci are
at the ends of the line; one focus is at the center of the ball (by
Kepler’s 1st law) and the other one is at r0, see figure (instead
of a degenerate ellipse, a strongly elliptical ellipse is depicted).
The period of the orbit is determined by the longer semiaxis of
the ellipse (by Kepler’s 3rd law). The longer semiaxis is r0/2
and we are interested in half a period. Thus, the answer is
equal to the halfperiod of a circular orbit of radius r0/2:
(
2π
2tr→0
)2r0
2=
Gm
(r0/2)2=⇒ tr→0 = π
√
r30
8Gm.
Alternatively, one may write the energy conservation lawr2
2−
Gmr
= E (that in turn is obtainable from Newton’s
II law r = − Gmr2 ) with E = − Gm
r0
, separate the variables
(drdt
= −
√
2E + 2Gmr
) and write the integral t = −∫
dr√2E+ 2Gm
r
.
This integral is probably not calculable during the limitted
time given during the Olympiad, but a possible approach can
be sketched as follows. Substituting√
2E + 2Gmr
= ξ and√
2E = υ, one gets
t∞
4Gm=
∫
∞
0
dξ
(υ2 − ξ2)2
=1
4υ3
∫
∞
0
[
υ
(υ − ξ)2+
υ
(υ + ξ)2+
1
υ − ξ+
1
υ + ξ
]
dξ.
Here (after shifting the variable) one can use∫
dξξ
= ln ξ and∫
dξ
ξ2 = − 1
ξ, finally getting the same answer as by Kepler’s laws.
iv. (1.7 pts) By Clapeyron–Mendeleyev law,
p =mRT0
µV.
Work done by gravity to compress the ball is
W = −
∫
p dV = −mRT0
µ
∫ 4
3πr3
3
4
3πr3
0
dV
V=
3mRT0
µln
r0
r3
.
The temperature stays constant, so the internal energy does not
change; hence, according to the 1st law of thermodynamics, the
compression work W is the heat radiated.
v. (1 pt) The collapse continues adiabatically.
pV γ = const =⇒ T V γ−1 = const.
∴ T ∝ V 1−γ∝ r3−3γ
∴ T = T0
(r3
r
)3γ−3
.
vi. (2 pts) During the collapse, the gravitational energy is con-
verted into heat. Since r3 ≫ r4, The released gravitational en-
ergy can be estimated as ∆Π = −Gm2(r−14 −r−1
3 ) ≈ −Gm2/r4
(exact calculation by integration adds a prefactor 3
5); the ter-
minal heat energy is estimated as ∆Q = cVmµ
(T4 − T0) ≈
cVmµ
T4 (the approximation T4 ≫ T0 follows from the result
of the previous question, when combined with r3 ≫ r4). So,
∆Q = Rγ−1
mµ
T4 ≈mµ
RT4. For the temperature T4, we can use
the result of the previous question, T4 = T0
(
r3
r4
)3γ−3
. Since
initial full energy was approximately zero, ∆Q + ∆Π ≈ 0, we
obtain
Gm2
r4
≈m
µRT0
(
r3
r4
)3γ−3
=⇒ r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
.
Therefore,
T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
.
Alternatively, one can obtain the result by approximately
equating the hydrostatic pressure ρr4Gmr2
4
to the gas pressure
p4 = ρ
µRT4; the result will be exactly the same as given above.
— page 5 of 5 —
Problem T3. Protostar formation (9 points)i. (0.8 pts)
T = const =⇒ pV = const
V ∝ r3
∴ p ∝ r−3 =⇒p(r1)
p(r0)= 23 = 8.
ii. (1 pt) During the period considered the pressure is negli-
gible. Therefore the gas is in free fall. By Gauss’ theorem and
symmetry, the gravitational field at any point in the ball is
equivalent to the one generated when all the mass closer to the
center is compressed into the center. Moreover, while the ball
has not yet shrunk much, the field strength on its surface does
not change much either. The acceleration of the outermost
layer stays approximately constant. Thus,
t ≈
√
2(r0 − r2)
g
where
g ≈Gm
r20
,
∴ t ≈
√
2r20(r0 − r2)
Gm=
√
0.1r30
Gm.
iii. (2.5 pts) Gravitationally the outer layer of the ball is in-
fluenced by the rest just as the rest were compressed into a
point mass. Therefore we have Keplerian motion: the fall of
any part of the outer layer consists in a halfperiod of an ultra-
elliptical orbit. The ellipse is degenerate into a line; its foci are
at the ends of the line; one focus is at the center of the ball (by
Kepler’s 1st law) and the other one is at r0, see figure (instead
of a degenerate ellipse, a strongly elliptical ellipse is depicted).
The period of the orbit is determined by the longer semiaxis of
the ellipse (by Kepler’s 3rd law). The longer semiaxis is r0/2
and we are interested in half a period. Thus, the answer is
equal to the halfperiod of a circular orbit of radius r0/2:
(
2π
2tr→0
)2r0
2=
Gm
(r0/2)2=⇒ tr→0 = π
√
r30
8Gm.
Alternatively, one may write the energy conservation lawr2
2−
Gmr
= E (that in turn is obtainable from Newton’s
II law r = − Gmr2 ) with E = − Gm
r0
, separate the variables
(drdt
= −
√
2E + 2Gmr
) and write the integral t = −∫
dr√2E+ 2Gm
r
.
This integral is probably not calculable during the limitted
time given during the Olympiad, but a possible approach can
be sketched as follows. Substituting√
2E + 2Gmr
= ξ and√
2E = υ, one gets
t∞
4Gm=
∫
∞
0
dξ
(υ2 − ξ2)2
=1
4υ3
∫
∞
0
[
υ
(υ − ξ)2+
υ
(υ + ξ)2+
1
υ − ξ+
1
υ + ξ
]
dξ.
Here (after shifting the variable) one can use∫
dξξ
= ln ξ and∫
dξ
ξ2 = − 1
ξ, finally getting the same answer as by Kepler’s laws.
iv. (1.7 pts) By Clapeyron–Mendeleyev law,
p =mRT0
µV.
Work done by gravity to compress the ball is
W = −
∫
p dV = −mRT0
µ
∫ 4
3πr3
3
4
3πr3
0
dV
V=
3mRT0
µln
r0
r3
.
The temperature stays constant, so the internal energy does not
change; hence, according to the 1st law of thermodynamics, the
compression work W is the heat radiated.
v. (1 pt) The collapse continues adiabatically.
pV γ = const =⇒ T V γ−1 = const.
∴ T ∝ V 1−γ∝ r3−3γ
∴ T = T0
(r3
r
)3γ−3
.
vi. (2 pts) During the collapse, the gravitational energy is con-
verted into heat. Since r3 ≫ r4, The released gravitational en-
ergy can be estimated as ∆Π = −Gm2(r−14 −r−1
3 ) ≈ −Gm2/r4
(exact calculation by integration adds a prefactor 3
5); the ter-
minal heat energy is estimated as ∆Q = cVmµ
(T4 − T0) ≈
cVmµ
T4 (the approximation T4 ≫ T0 follows from the result
of the previous question, when combined with r3 ≫ r4). So,
∆Q = Rγ−1
mµ
T4 ≈mµ
RT4. For the temperature T4, we can use
the result of the previous question, T4 = T0
(
r3
r4
)3γ−3
. Since
initial full energy was approximately zero, ∆Q + ∆Π ≈ 0, we
obtain
Gm2
r4
≈m
µRT0
(
r3
r4
)3γ−3
=⇒ r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
.
Therefore,
T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
.
Alternatively, one can obtain the result by approximately
equating the hydrostatic pressure ρr4Gmr2
4
to the gas pressure
p4 = ρ
µRT4; the result will be exactly the same as given above.
— page 5 of 5 —
94
Problem T3. Protostar formation (9 points)i. (0.8 pts)
T = const =⇒ pV = const
V ∝ r3
∴ p ∝ r−3 =⇒p(r1)
p(r0)= 23 = 8.
ii. (1 pt) During the period considered the pressure is negli-
gible. Therefore the gas is in free fall. By Gauss’ theorem and
symmetry, the gravitational field at any point in the ball is
equivalent to the one generated when all the mass closer to the
center is compressed into the center. Moreover, while the ball
has not yet shrunk much, the field strength on its surface does
not change much either. The acceleration of the outermost
layer stays approximately constant. Thus,
t ≈
√
2(r0 − r2)
g
where
g ≈Gm
r20
,
∴ t ≈
√
2r20(r0 − r2)
Gm=
√
0.1r30
Gm.
iii. (2.5 pts) Gravitationally the outer layer of the ball is in-
fluenced by the rest just as the rest were compressed into a
point mass. Therefore we have Keplerian motion: the fall of
any part of the outer layer consists in a halfperiod of an ultra-
elliptical orbit. The ellipse is degenerate into a line; its foci are
at the ends of the line; one focus is at the center of the ball (by
Kepler’s 1st law) and the other one is at r0, see figure (instead
of a degenerate ellipse, a strongly elliptical ellipse is depicted).
The period of the orbit is determined by the longer semiaxis of
the ellipse (by Kepler’s 3rd law). The longer semiaxis is r0/2
and we are interested in half a period. Thus, the answer is
equal to the halfperiod of a circular orbit of radius r0/2:
(
2π
2tr→0
)2r0
2=
Gm
(r0/2)2=⇒ tr→0 = π
√
r30
8Gm.
Alternatively, one may write the energy conservation lawr2
2−
Gmr
= E (that in turn is obtainable from Newton’s
II law r = − Gmr2 ) with E = − Gm
r0
, separate the variables
(drdt
= −
√
2E + 2Gmr
) and write the integral t = −∫
dr√2E+ 2Gm
r
.
This integral is probably not calculable during the limitted
time given during the Olympiad, but a possible approach can
be sketched as follows. Substituting√
2E + 2Gmr
= ξ and√
2E = υ, one gets
t∞
4Gm=
∫
∞
0
dξ
(υ2 − ξ2)2
=1
4υ3
∫
∞
0
[
υ
(υ − ξ)2+
υ
(υ + ξ)2+
1
υ − ξ+
1
υ + ξ
]
dξ.
Here (after shifting the variable) one can use∫
dξξ
= ln ξ and∫
dξ
ξ2 = − 1
ξ, finally getting the same answer as by Kepler’s laws.
iv. (1.7 pts) By Clapeyron–Mendeleyev law,
p =mRT0
µV.
Work done by gravity to compress the ball is
W = −
∫
p dV = −mRT0
µ
∫ 4
3πr3
3
4
3πr3
0
dV
V=
3mRT0
µln
r0
r3
.
The temperature stays constant, so the internal energy does not
change; hence, according to the 1st law of thermodynamics, the
compression work W is the heat radiated.
v. (1 pt) The collapse continues adiabatically.
pV γ = const =⇒ T V γ−1 = const.
∴ T ∝ V 1−γ∝ r3−3γ
∴ T = T0
(r3
r
)3γ−3
.
vi. (2 pts) During the collapse, the gravitational energy is con-
verted into heat. Since r3 ≫ r4, The released gravitational en-
ergy can be estimated as ∆Π = −Gm2(r−14 −r−1
3 ) ≈ −Gm2/r4
(exact calculation by integration adds a prefactor 3
5); the ter-
minal heat energy is estimated as ∆Q = cVmµ
(T4 − T0) ≈
cVmµ
T4 (the approximation T4 ≫ T0 follows from the result
of the previous question, when combined with r3 ≫ r4). So,
∆Q = Rγ−1
mµ
T4 ≈mµ
RT4. For the temperature T4, we can use
the result of the previous question, T4 = T0
(
r3
r4
)3γ−3
. Since
initial full energy was approximately zero, ∆Q + ∆Π ≈ 0, we
obtain
Gm2
r4
≈m
µRT0
(
r3
r4
)3γ−3
=⇒ r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
.
Therefore,
T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
.
Alternatively, one can obtain the result by approximately
equating the hydrostatic pressure ρr4Gmr2
4
to the gas pressure
p4 = ρ
µRT4; the result will be exactly the same as given above.
— page 5 of 5 —
95
Grading scheme: Theory The 43rd International Physics Olympiad — July 2012
Grading scheme: Theory
General rules This grading scheme describes the number of
points allotted for each term entering a useful formula. These
terms don’t need to be separately described: if a formula is
written correctly, full marks (the sum of the marks of all the
terms of that formula) are given. If a formula is not written
explicitly, but it is clear that individual terms are written bear-
ing the equation in mind (eg. indicated on a diagram), marks
for these terms will be given. Some points are allotted for
mathematical calculations.
If a certain term of a useful formula is written incor-
rectly, 0.1 is subtracted for a minor mistake (eg. missing non-
dimensional factor); no mark is given if the mistake is major
(with non-matching dimensionality). The same rule is applied
to the marking of mathematical calculations: each minor mis-
take leads to a subtraction of 0.1 pts (as long as the remaining
score for that particular calculation remains positive), and no
marks are given in the case of dimensional mistakes.
No penalty is applied in these cases when a mistake is clearly
just a rewriting typo (i.e. when there is no mistake in the draft).
If formula is written without deriving: if it is simple
enough to be derived in head, full marks, otherwise zero
marks.
If there two solutions on Solution sheets, one correct and
another incorrect: only the one wich corresponds to the An-
swer Sheets is taken into account. What is crossed out is
never considered.
No penalty is applied for propagating errors unless the cal-
culations are significantly simplified (in which case mathemat-
ical calculations are credited partially, according to the degree
of simplification, with marking granularity of 0.1 pts).
— page 1 of 5 —
The 43rd International Physics Olympiad — July 2012
Grading scheme: Theory
General rules This grading scheme describes the number of
points allotted for each term entering a useful formula. These
terms don’t need to be separately described: if a formula is
written correctly, full marks (the sum of the marks of all the
terms of that formula) are given. If a formula is not written
explicitly, but it is clear that individual terms are written bear-
ing the equation in mind (eg. indicated on a diagram), marks
for these terms will be given. Some points are allotted for
mathematical calculations.
If a certain term of a useful formula is written incor-
rectly, 0.1 is subtracted for a minor mistake (eg. missing non-
dimensional factor); no mark is given if the mistake is major
(with non-matching dimensionality). The same rule is applied
to the marking of mathematical calculations: each minor mis-
take leads to a subtraction of 0.1 pts (as long as the remaining
score for that particular calculation remains positive), and no
marks are given in the case of dimensional mistakes.
No penalty is applied in these cases when a mistake is clearly
just a rewriting typo (i.e. when there is no mistake in the draft).
If formula is written without deriving: if it is simple
enough to be derived in head, full marks, otherwise zero
marks.
If there two solutions on Solution sheets, one correct and
another incorrect: only the one wich corresponds to the An-
swer Sheets is taken into account. What is crossed out is
never considered.
No penalty is applied for propagating errors unless the cal-
culations are significantly simplified (in which case mathemat-
ical calculations are credited partially, according to the degree
of simplification, with marking granularity of 0.1 pts).
— page 1 of 5 —
96
Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) If the parameters are derived using the particular
cases of throwing up and throwing horizontally:
for throwing up, zg = v20/2 — 0.2 pts;
from where z0 = v20/2g — 0.1 pts;
noticing that for horizontally thrown ball,
the trajectory has the same shape as z = −kx2 — 0.2 pts;
finding this trajectory, z = −gx2/2v20 — 0.2 pts;
Concluding k = g/2v20 — 0.1 pts;
If, instead of studying the trajectory of a horizontally
thrown ball (for which 0.5 pts were allocated), a trajectory
of a ball thrown at 45 degrees is studied:
distance is max. when the angle is 45 — 0.2
Finding this maximal distance v2/g — 0.2
Obtaining k — 0.1
If the parameters are derived using condition that the quad-
ratic equation for the tangent of the throwing angle has exactly
one real solution:
Requiring x = v cos αt — 0.1 pts;
Requiring z = v sin αt − gt2/2 — 0.2 pts;
Eliminating t: z = x tan α − gx2/v2 cos2 α — 0.1 pts;
Obtaining z − x tan α + gx2/v2(1 + tan2 α) = 0 — 0.1 pts;
Requiring that the discriminant is 0 — 0.2 pts;
Obtaining z0 = v20/2g, k = gx2/2v2
0 — 0.1 pts;
(If one of the two is incorrect — 0 pts for the last line)
ii. (1.2 pts)
Trajectory hits the sphere when descending — 0.7 pts(if the top of the parabola is higher than 5
2R — 0.5 pts);
Trajectory touches the sphere when ascending — 0.5 pts.
Trajectory touches the sphere at its top oris clearly non-parabolic or starts inside the sphere
or intersects the sphere: total — 0 pts.
iii. (2.5 pts) If the analysis is based on a trajectory which is
wrong in principle, 0 pts.
For minimal speed, z = z0 − kx2 touches the sphere — 1.5 pts
Shifting the origin to the sphere’s top for simplicity — 0.1 pts
Then z =v2
0
2g−
gx2
2v20
— 0.2 pts
and x2 + z2 + 2zR = 0 — 0.2 pts
⇒ x4
(
g
2v2
0
)2
+ x2
(
1
2−
gR
v2
0
)
+(
v2
0
4g+ R
)
v2
0
g= 0 — 0.1 pts
(full 0.6 pts if equivalent eq. is obtained for non-shifted origin)
Discriminant equals 0 — 0.2 pts
from where v20 = 0.5gR — 0.1 pts
hence vmin =√
4.5gR — 0.1 pts
If alternative solution is followed:
Shifting the origin to the sphere’s top for simplicity — 0.1 pts
Noting that the touching point vel. is ⊥ to launching vel.— 0.6 pts
Angle from centre to touching pt P = launching angle — 0.6 pts
Cond. that P belongs to the trajectory — 0.6 pts
From this cond., final answer obtained — 0.6 pts
For a brute force approach:
Obtaining 4th order equationfor intersection points x (or y) — 0.5 pts
which is reduced to cubic(divided by x) — 0.1 pts
Mentioning that it has exactly one pos. root — 0.2 pts(equivalently, an extrememum coincides with a root or there
are exactly two distinct real roots.)
Obtaining quadratic eq. for x-coord. of extrema — 0.2 pts
Finding the roots of it — 0.2 pts
Selecting the larger root x∗ of it — 0.2 pts
Requiring that x∗ is the root of the cubic eq — 0.2 pts
Obtaining the speed as a function of α — 0.2 pts
Finding the minimum of that function — 0.7 pts
Part B. Air flow around a wing (4 points)
i. (0.8 pts)
Using the wing’s frame of reference — 0.1 pts
taking streamline distance measurment at P — 0.2 pts(if off by 1 mm or more, 0.1 pts)
measuring unperturbed streamline distance — 0.1 pts(if off by 1 mm or more, 0 pts)
at P , streamlines are horiz. ⇒ scalar adding — 0.1 pts
Writing continuity condition — 0.2 pts
Finding final answer — 0.1 pts
(if the speed in wing’s frame is given as final anser, points
are lost for 4th and 6th line)
ii. (1.2 pts)
Writing down continuity condition — 0.2 pts(or stating: smaller streamline distance ⇒ larger speed)
Writing Bernoulli’s law — 0.4 pts0 pts out of 0.4 if the Bernoulli law includes ρgh
(or stating that larger speed ⇒ smaller pressure)
Writing adiabatic law — 0.4 pts
— page 2 of 5 —
PrObLEm T1. FOcus On skETchEs (13 POInTs)
97
Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) If the parameters are derived using the particular
cases of throwing up and throwing horizontally:
for throwing up, zg = v20/2 — 0.2 pts;
from where z0 = v20/2g — 0.1 pts;
noticing that for horizontally thrown ball,
the trajectory has the same shape as z = −kx2 — 0.2 pts;
finding this trajectory, z = −gx2/2v20 — 0.2 pts;
Concluding k = g/2v20 — 0.1 pts;
If, instead of studying the trajectory of a horizontally
thrown ball (for which 0.5 pts were allocated), a trajectory
of a ball thrown at 45 degrees is studied:
distance is max. when the angle is 45 — 0.2
Finding this maximal distance v2/g — 0.2
Obtaining k — 0.1
If the parameters are derived using condition that the quad-
ratic equation for the tangent of the throwing angle has exactly
one real solution:
Requiring x = v cos αt — 0.1 pts;
Requiring z = v sin αt − gt2/2 — 0.2 pts;
Eliminating t: z = x tan α − gx2/v2 cos2 α — 0.1 pts;
Obtaining z − x tan α + gx2/v2(1 + tan2 α) = 0 — 0.1 pts;
Requiring that the discriminant is 0 — 0.2 pts;
Obtaining z0 = v20/2g, k = gx2/2v2
0 — 0.1 pts;
(If one of the two is incorrect — 0 pts for the last line)
ii. (1.2 pts)
Trajectory hits the sphere when descending — 0.7 pts(if the top of the parabola is higher than 5
2R — 0.5 pts);
Trajectory touches the sphere when ascending — 0.5 pts.
Trajectory touches the sphere at its top oris clearly non-parabolic or starts inside the sphere
or intersects the sphere: total — 0 pts.
iii. (2.5 pts) If the analysis is based on a trajectory which is
wrong in principle, 0 pts.
For minimal speed, z = z0 − kx2 touches the sphere — 1.5 pts
Shifting the origin to the sphere’s top for simplicity — 0.1 pts
Then z =v2
0
2g−
gx2
2v20
— 0.2 pts
and x2 + z2 + 2zR = 0 — 0.2 pts
⇒ x4
(
g
2v2
0
)2
+ x2
(
1
2−
gR
v2
0
)
+(
v2
0
4g+ R
)
v2
0
g= 0 — 0.1 pts
(full 0.6 pts if equivalent eq. is obtained for non-shifted origin)
Discriminant equals 0 — 0.2 pts
from where v20 = 0.5gR — 0.1 pts
hence vmin =√
4.5gR — 0.1 pts
If alternative solution is followed:
Shifting the origin to the sphere’s top for simplicity — 0.1 pts
Noting that the touching point vel. is ⊥ to launching vel.— 0.6 pts
Angle from centre to touching pt P = launching angle — 0.6 pts
Cond. that P belongs to the trajectory — 0.6 pts
From this cond., final answer obtained — 0.6 pts
For a brute force approach:
Obtaining 4th order equationfor intersection points x (or y) — 0.5 pts
which is reduced to cubic(divided by x) — 0.1 pts
Mentioning that it has exactly one pos. root — 0.2 pts(equivalently, an extrememum coincides with a root or there
are exactly two distinct real roots.)
Obtaining quadratic eq. for x-coord. of extrema — 0.2 pts
Finding the roots of it — 0.2 pts
Selecting the larger root x∗ of it — 0.2 pts
Requiring that x∗ is the root of the cubic eq — 0.2 pts
Obtaining the speed as a function of α — 0.2 pts
Finding the minimum of that function — 0.7 pts
Part B. Air flow around a wing (4 points)
i. (0.8 pts)
Using the wing’s frame of reference — 0.1 pts
taking streamline distance measurment at P — 0.2 pts(if off by 1 mm or more, 0.1 pts)
measuring unperturbed streamline distance — 0.1 pts(if off by 1 mm or more, 0 pts)
at P , streamlines are horiz. ⇒ scalar adding — 0.1 pts
Writing continuity condition — 0.2 pts
Finding final answer — 0.1 pts
(if the speed in wing’s frame is given as final anser, points
are lost for 4th and 6th line)
ii. (1.2 pts)
Writing down continuity condition — 0.2 pts(or stating: smaller streamline distance ⇒ larger speed)
Writing Bernoulli’s law — 0.4 pts0 pts out of 0.4 if the Bernoulli law includes ρgh
(or stating that larger speed ⇒ smaller pressure)
Writing adiabatic law — 0.4 pts
— page 2 of 5 —Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) If the parameters are derived using the particular
cases of throwing up and throwing horizontally:
for throwing up, zg = v20/2 — 0.2 pts;
from where z0 = v20/2g — 0.1 pts;
noticing that for horizontally thrown ball,
the trajectory has the same shape as z = −kx2 — 0.2 pts;
finding this trajectory, z = −gx2/2v20 — 0.2 pts;
Concluding k = g/2v20 — 0.1 pts;
If, instead of studying the trajectory of a horizontally
thrown ball (for which 0.5 pts were allocated), a trajectory
of a ball thrown at 45 degrees is studied:
distance is max. when the angle is 45 — 0.2
Finding this maximal distance v2/g — 0.2
Obtaining k — 0.1
If the parameters are derived using condition that the quad-
ratic equation for the tangent of the throwing angle has exactly
one real solution:
Requiring x = v cos αt — 0.1 pts;
Requiring z = v sin αt − gt2/2 — 0.2 pts;
Eliminating t: z = x tan α − gx2/v2 cos2 α — 0.1 pts;
Obtaining z − x tan α + gx2/v2(1 + tan2 α) = 0 — 0.1 pts;
Requiring that the discriminant is 0 — 0.2 pts;
Obtaining z0 = v20/2g, k = gx2/2v2
0 — 0.1 pts;
(If one of the two is incorrect — 0 pts for the last line)
ii. (1.2 pts)
Trajectory hits the sphere when descending — 0.7 pts(if the top of the parabola is higher than 5
2R — 0.5 pts);
Trajectory touches the sphere when ascending — 0.5 pts.
Trajectory touches the sphere at its top oris clearly non-parabolic or starts inside the sphere
or intersects the sphere: total — 0 pts.
iii. (2.5 pts) If the analysis is based on a trajectory which is
wrong in principle, 0 pts.
For minimal speed, z = z0 − kx2 touches the sphere — 1.5 pts
Shifting the origin to the sphere’s top for simplicity — 0.1 pts
Then z =v2
0
2g−
gx2
2v20
— 0.2 pts
and x2 + z2 + 2zR = 0 — 0.2 pts
⇒ x4
(
g
2v2
0
)2
+ x2
(
1
2−
gR
v2
0
)
+(
v2
0
4g+ R
)
v2
0
g= 0 — 0.1 pts
(full 0.6 pts if equivalent eq. is obtained for non-shifted origin)
Discriminant equals 0 — 0.2 pts
from where v20 = 0.5gR — 0.1 pts
hence vmin =√
4.5gR — 0.1 pts
If alternative solution is followed:
Shifting the origin to the sphere’s top for simplicity — 0.1 pts
Noting that the touching point vel. is ⊥ to launching vel.— 0.6 pts
Angle from centre to touching pt P = launching angle — 0.6 pts
Cond. that P belongs to the trajectory — 0.6 pts
From this cond., final answer obtained — 0.6 pts
For a brute force approach:
Obtaining 4th order equationfor intersection points x (or y) — 0.5 pts
which is reduced to cubic(divided by x) — 0.1 pts
Mentioning that it has exactly one pos. root — 0.2 pts(equivalently, an extrememum coincides with a root or there
are exactly two distinct real roots.)
Obtaining quadratic eq. for x-coord. of extrema — 0.2 pts
Finding the roots of it — 0.2 pts
Selecting the larger root x∗ of it — 0.2 pts
Requiring that x∗ is the root of the cubic eq — 0.2 pts
Obtaining the speed as a function of α — 0.2 pts
Finding the minimum of that function — 0.7 pts
Part B. Air flow around a wing (4 points)
i. (0.8 pts)
Using the wing’s frame of reference — 0.1 pts
taking streamline distance measurment at P — 0.2 pts(if off by 1 mm or more, 0.1 pts)
measuring unperturbed streamline distance — 0.1 pts(if off by 1 mm or more, 0 pts)
at P , streamlines are horiz. ⇒ scalar adding — 0.1 pts
Writing continuity condition — 0.2 pts
Finding final answer — 0.1 pts
(if the speed in wing’s frame is given as final anser, points
are lost for 4th and 6th line)
ii. (1.2 pts)
Writing down continuity condition — 0.2 pts(or stating: smaller streamline distance ⇒ larger speed)
Writing Bernoulli’s law — 0.4 pts0 pts out of 0.4 if the Bernoulli law includes ρgh
(or stating that larger speed ⇒ smaller pressure)
Writing adiabatic law — 0.4 pts
— page 2 of 5 —
98
Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)
i. (0.8 pts) If the parameters are derived using the particular
cases of throwing up and throwing horizontally:
for throwing up, zg = v20/2 — 0.2 pts;
from where z0 = v20/2g — 0.1 pts;
noticing that for horizontally thrown ball,
the trajectory has the same shape as z = −kx2 — 0.2 pts;
finding this trajectory, z = −gx2/2v20 — 0.2 pts;
Concluding k = g/2v20 — 0.1 pts;
If, instead of studying the trajectory of a horizontally
thrown ball (for which 0.5 pts were allocated), a trajectory
of a ball thrown at 45 degrees is studied:
distance is max. when the angle is 45 — 0.2
Finding this maximal distance v2/g — 0.2
Obtaining k — 0.1
If the parameters are derived using condition that the quad-
ratic equation for the tangent of the throwing angle has exactly
one real solution:
Requiring x = v cos αt — 0.1 pts;
Requiring z = v sin αt − gt2/2 — 0.2 pts;
Eliminating t: z = x tan α − gx2/v2 cos2 α — 0.1 pts;
Obtaining z − x tan α + gx2/v2(1 + tan2 α) = 0 — 0.1 pts;
Requiring that the discriminant is 0 — 0.2 pts;
Obtaining z0 = v20/2g, k = gx2/2v2
0 — 0.1 pts;
(If one of the two is incorrect — 0 pts for the last line)
ii. (1.2 pts)
Trajectory hits the sphere when descending — 0.7 pts(if the top of the parabola is higher than 5
2R — 0.5 pts);
Trajectory touches the sphere when ascending — 0.5 pts.
Trajectory touches the sphere at its top oris clearly non-parabolic or starts inside the sphere
or intersects the sphere: total — 0 pts.
iii. (2.5 pts) If the analysis is based on a trajectory which is
wrong in principle, 0 pts.
For minimal speed, z = z0 − kx2 touches the sphere — 1.5 pts
Shifting the origin to the sphere’s top for simplicity — 0.1 pts
Then z =v2
0
2g−
gx2
2v20
— 0.2 pts
and x2 + z2 + 2zR = 0 — 0.2 pts
⇒ x4
(
g
2v2
0
)2
+ x2
(
1
2−
gR
v2
0
)
+(
v2
0
4g+ R
)
v2
0
g= 0 — 0.1 pts
(full 0.6 pts if equivalent eq. is obtained for non-shifted origin)
Discriminant equals 0 — 0.2 pts
from where v20 = 0.5gR — 0.1 pts
hence vmin =√
4.5gR — 0.1 pts
If alternative solution is followed:
Shifting the origin to the sphere’s top for simplicity — 0.1 pts
Noting that the touching point vel. is ⊥ to launching vel.— 0.6 pts
Angle from centre to touching pt P = launching angle — 0.6 pts
Cond. that P belongs to the trajectory — 0.6 pts
From this cond., final answer obtained — 0.6 pts
For a brute force approach:
Obtaining 4th order equationfor intersection points x (or y) — 0.5 pts
which is reduced to cubic(divided by x) — 0.1 pts
Mentioning that it has exactly one pos. root — 0.2 pts(equivalently, an extrememum coincides with a root or there
are exactly two distinct real roots.)
Obtaining quadratic eq. for x-coord. of extrema — 0.2 pts
Finding the roots of it — 0.2 pts
Selecting the larger root x∗ of it — 0.2 pts
Requiring that x∗ is the root of the cubic eq — 0.2 pts
Obtaining the speed as a function of α — 0.2 pts
Finding the minimum of that function — 0.7 pts
Part B. Air flow around a wing (4 points)
i. (0.8 pts)
Using the wing’s frame of reference — 0.1 pts
taking streamline distance measurment at P — 0.2 pts(if off by 1 mm or more, 0.1 pts)
measuring unperturbed streamline distance — 0.1 pts(if off by 1 mm or more, 0 pts)
at P , streamlines are horiz. ⇒ scalar adding — 0.1 pts
Writing continuity condition — 0.2 pts
Finding final answer — 0.1 pts
(if the speed in wing’s frame is given as final anser, points
are lost for 4th and 6th line)
ii. (1.2 pts)
Writing down continuity condition — 0.2 pts(or stating: smaller streamline distance ⇒ larger speed)
Writing Bernoulli’s law — 0.4 pts0 pts out of 0.4 if the Bernoulli law includes ρgh
(or stating that larger speed ⇒ smaller pressure)
Writing adiabatic law — 0.4 pts
— page 2 of 5 —(or stating that smaller pressure ⇒ lower temperature)
Finding Q as in the figure in the solutions — 0.2 pts(if Q marked below the wing’s tip — 0.1 pts)
iii. (2 pts)
Finding the dew point: idea of linearization — 0.2 pts
Expression and/or numerical value for dew point — 0.2 pts
Deriving 1
2µv2 + cpT = const:
1 mole of gas carries kin. en. 1
2µv2 — 0.2 pts
1 mole of gas carries heat en. CV T — 0.2 pts
work done on 1 mole of gas: p1V1 − p2V2 — 0.4 pts
en.: 1
2µ(v2
2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts
using ideal gas law obtaining 1
2v2 + cpT = const — 0.2 pts
Alternative approximate approach:
Bernoulli’s law 1
2ρv2 + p = const — 0.3 pts
0 pts out of 0.3 if the Bernoulli law includes ρgh
Adiabatic law pV γ = const — 0.3 pts
⇒ p1−γT γ = const — 0.2 pts
Approximation ∆p = γγ−1
pT
∆T — 0.2 pts
Leading to 1
2v2 + R
µ
cp
cp−cVT = const — 0.1 pts
Leading to 1
2v2 + cpT = const — 0.1 pts
And further (for either approach):
Bringing it to ∆v2
2= 1
2v2
crit(a2
c2 − 1) = cp∆T — 0.1 pts
Measuring a and c — 0.2 pts
Obtaining vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s — 0.1 pts
Part C. Magnetic straws (4.5 points)
i. (0.8 pts)
Lines straight and parallel inside — 0.6 pts
(if not drawn over the entire length, subtract 0.2)
(can be slightly curved close the tube’s end)
Lines curve outwards slightly after the exit — 0.2 pts
(if so curved that more than one closed loop on both sides is
depicted in Fig, 0 pts out of 0.2)
ii. (1.2 pts)
Expressing induction as B = Φ/πr2 — 0.2 pts
Stating that w = B2
2µ0
— 0.2 pts
Idea of using virtual lengthening — 0.2 pts
(Introducing a lengthening ∆l is enough)
Expressing ∆W = B2
2µ0
πr2∆l — 0.2 pts
(Same marks if W expressed for entire tube)
Equating ∆W = T ∆l — 0.2 pts
Expressing T = Φ2
2µ0πr2 . — 0.2 pts
iii. (2.5 pts)
Idea of using analogy with el. charges — 1 pt
(Sketch with a quadrupole conf. of el. charges is enough)
Finding the force between two magn. charges via matching el.
and magn. quantities is worth 1 pts in total, split down as
follows:
Expressing el. stat. force via en. density — 0.5 pts
Using the obtained Eq. to obtain F = 1
4πµ0
Φ2
a2 — 0.5 pts
Any other matching scheme is graded analogously; e.g. find-
ing such a Q which has the same en. density as Φ — 0.5 pts;
expressing the force between magn. charges (Φ) as the force
between the matching el. charges (Q) — 0.5 pts. Declaring
matching pairs Q ↔ Φ and 1
4πε0
↔1
4πµ0
without energy-based-
motivation gives only 0.5 pts.
Noting that to tubes comp. of force = 0 — 0.2 pts
When expressing F = 2(F1 − F2): for factor “2” — 0.1 pts
and for finding F2 — 0.2 pts
(if wrong sign for F2, subtract 0.1)
If alternative solution is followed
Plan to express inter. energy as a function of a
intending to find F as a derivative — 0.2 pts
Calculating B(x) — 0.8 pts
If estimated without dependance on x, 0.4
considering a tube as an array of dipoles — 0.3 pts
expressing dm = Sjdx — 0.3 pts
relating j to Φ — 0.2 pts
Expressing U =∫
B(x)dm — 0.4 pts
If estimated as BSjl, 0.2
Finding F = dU/da — 0.2 pts
Taking into account the factor “ 2”— 0.1 pts
— page 3 of 5 —
99
(or stating that smaller pressure ⇒ lower temperature)
Finding Q as in the figure in the solutions — 0.2 pts(if Q marked below the wing’s tip — 0.1 pts)
iii. (2 pts)
Finding the dew point: idea of linearization — 0.2 pts
Expression and/or numerical value for dew point — 0.2 pts
Deriving 1
2µv2 + cpT = const:
1 mole of gas carries kin. en. 1
2µv2 — 0.2 pts
1 mole of gas carries heat en. CV T — 0.2 pts
work done on 1 mole of gas: p1V1 − p2V2 — 0.4 pts
en.: 1
2µ(v2
2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts
using ideal gas law obtaining 1
2v2 + cpT = const — 0.2 pts
Alternative approximate approach:
Bernoulli’s law 1
2ρv2 + p = const — 0.3 pts
0 pts out of 0.3 if the Bernoulli law includes ρgh
Adiabatic law pV γ = const — 0.3 pts
⇒ p1−γT γ = const — 0.2 pts
Approximation ∆p = γγ−1
pT
∆T — 0.2 pts
Leading to 1
2v2 + R
µ
cp
cp−cVT = const — 0.1 pts
Leading to 1
2v2 + cpT = const — 0.1 pts
And further (for either approach):
Bringing it to ∆v2
2= 1
2v2
crit(a2
c2 − 1) = cp∆T — 0.1 pts
Measuring a and c — 0.2 pts
Obtaining vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s — 0.1 pts
Part C. Magnetic straws (4.5 points)
i. (0.8 pts)
Lines straight and parallel inside — 0.6 pts
(if not drawn over the entire length, subtract 0.2)
(can be slightly curved close the tube’s end)
Lines curve outwards slightly after the exit — 0.2 pts
(if so curved that more than one closed loop on both sides is
depicted in Fig, 0 pts out of 0.2)
ii. (1.2 pts)
Expressing induction as B = Φ/πr2 — 0.2 pts
Stating that w = B2
2µ0
— 0.2 pts
Idea of using virtual lengthening — 0.2 pts
(Introducing a lengthening ∆l is enough)
Expressing ∆W = B2
2µ0
πr2∆l — 0.2 pts
(Same marks if W expressed for entire tube)
Equating ∆W = T ∆l — 0.2 pts
Expressing T = Φ2
2µ0πr2 . — 0.2 pts
iii. (2.5 pts)
Idea of using analogy with el. charges — 1 pt
(Sketch with a quadrupole conf. of el. charges is enough)
Finding the force between two magn. charges via matching el.
and magn. quantities is worth 1 pts in total, split down as
follows:
Expressing el. stat. force via en. density — 0.5 pts
Using the obtained Eq. to obtain F = 1
4πµ0
Φ2
a2 — 0.5 pts
Any other matching scheme is graded analogously; e.g. find-
ing such a Q which has the same en. density as Φ — 0.5 pts;
expressing the force between magn. charges (Φ) as the force
between the matching el. charges (Q) — 0.5 pts. Declaring
matching pairs Q ↔ Φ and 1
4πε0
↔1
4πµ0
without energy-based-
motivation gives only 0.5 pts.
Noting that to tubes comp. of force = 0 — 0.2 pts
When expressing F = 2(F1 − F2): for factor “2” — 0.1 pts
and for finding F2 — 0.2 pts
(if wrong sign for F2, subtract 0.1)
If alternative solution is followed
Plan to express inter. energy as a function of a
intending to find F as a derivative — 0.2 pts
Calculating B(x) — 0.8 pts
If estimated without dependance on x, 0.4
considering a tube as an array of dipoles — 0.3 pts
expressing dm = Sjdx — 0.3 pts
relating j to Φ — 0.2 pts
Expressing U =∫
B(x)dm — 0.4 pts
If estimated as BSjl, 0.2
Finding F = dU/da — 0.2 pts
Taking into account the factor “ 2”— 0.1 pts
— page 3 of 5 —
100
(or stating that smaller pressure ⇒ lower temperature)
Finding Q as in the figure in the solutions — 0.2 pts(if Q marked below the wing’s tip — 0.1 pts)
iii. (2 pts)
Finding the dew point: idea of linearization — 0.2 pts
Expression and/or numerical value for dew point — 0.2 pts
Deriving 1
2µv2 + cpT = const:
1 mole of gas carries kin. en. 1
2µv2 — 0.2 pts
1 mole of gas carries heat en. CV T — 0.2 pts
work done on 1 mole of gas: p1V1 − p2V2 — 0.4 pts
en.: 1
2µ(v2
2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts
using ideal gas law obtaining 1
2v2 + cpT = const — 0.2 pts
Alternative approximate approach:
Bernoulli’s law 1
2ρv2 + p = const — 0.3 pts
0 pts out of 0.3 if the Bernoulli law includes ρgh
Adiabatic law pV γ = const — 0.3 pts
⇒ p1−γT γ = const — 0.2 pts
Approximation ∆p = γγ−1
pT
∆T — 0.2 pts
Leading to 1
2v2 + R
µ
cp
cp−cVT = const — 0.1 pts
Leading to 1
2v2 + cpT = const — 0.1 pts
And further (for either approach):
Bringing it to ∆v2
2= 1
2v2
crit(a2
c2 − 1) = cp∆T — 0.1 pts
Measuring a and c — 0.2 pts
Obtaining vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s — 0.1 pts
Part C. Magnetic straws (4.5 points)
i. (0.8 pts)
Lines straight and parallel inside — 0.6 pts
(if not drawn over the entire length, subtract 0.2)
(can be slightly curved close the tube’s end)
Lines curve outwards slightly after the exit — 0.2 pts
(if so curved that more than one closed loop on both sides is
depicted in Fig, 0 pts out of 0.2)
ii. (1.2 pts)
Expressing induction as B = Φ/πr2 — 0.2 pts
Stating that w = B2
2µ0
— 0.2 pts
Idea of using virtual lengthening — 0.2 pts
(Introducing a lengthening ∆l is enough)
Expressing ∆W = B2
2µ0
πr2∆l — 0.2 pts
(Same marks if W expressed for entire tube)
Equating ∆W = T ∆l — 0.2 pts
Expressing T = Φ2
2µ0πr2 . — 0.2 pts
iii. (2.5 pts)
Idea of using analogy with el. charges — 1 pt
(Sketch with a quadrupole conf. of el. charges is enough)
Finding the force between two magn. charges via matching el.
and magn. quantities is worth 1 pts in total, split down as
follows:
Expressing el. stat. force via en. density — 0.5 pts
Using the obtained Eq. to obtain F = 1
4πµ0
Φ2
a2 — 0.5 pts
Any other matching scheme is graded analogously; e.g. find-
ing such a Q which has the same en. density as Φ — 0.5 pts;
expressing the force between magn. charges (Φ) as the force
between the matching el. charges (Q) — 0.5 pts. Declaring
matching pairs Q ↔ Φ and 1
4πε0
↔1
4πµ0
without energy-based-
motivation gives only 0.5 pts.
Noting that to tubes comp. of force = 0 — 0.2 pts
When expressing F = 2(F1 − F2): for factor “2” — 0.1 pts
and for finding F2 — 0.2 pts
(if wrong sign for F2, subtract 0.1)
If alternative solution is followed
Plan to express inter. energy as a function of a
intending to find F as a derivative — 0.2 pts
Calculating B(x) — 0.8 pts
If estimated without dependance on x, 0.4
considering a tube as an array of dipoles — 0.3 pts
expressing dm = Sjdx — 0.3 pts
relating j to Φ — 0.2 pts
Expressing U =∫
B(x)dm — 0.4 pts
If estimated as BSjl, 0.2
Finding F = dU/da — 0.2 pts
Taking into account the factor “ 2”— 0.1 pts
— page 3 of 5 —
(or stating that smaller pressure ⇒ lower temperature)
Finding Q as in the figure in the solutions — 0.2 pts(if Q marked below the wing’s tip — 0.1 pts)
iii. (2 pts)
Finding the dew point: idea of linearization — 0.2 pts
Expression and/or numerical value for dew point — 0.2 pts
Deriving 1
2µv2 + cpT = const:
1 mole of gas carries kin. en. 1
2µv2 — 0.2 pts
1 mole of gas carries heat en. CV T — 0.2 pts
work done on 1 mole of gas: p1V1 − p2V2 — 0.4 pts
en.: 1
2µ(v2
2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts
using ideal gas law obtaining 1
2v2 + cpT = const — 0.2 pts
Alternative approximate approach:
Bernoulli’s law 1
2ρv2 + p = const — 0.3 pts
0 pts out of 0.3 if the Bernoulli law includes ρgh
Adiabatic law pV γ = const — 0.3 pts
⇒ p1−γT γ = const — 0.2 pts
Approximation ∆p = γγ−1
pT
∆T — 0.2 pts
Leading to 1
2v2 + R
µ
cp
cp−cVT = const — 0.1 pts
Leading to 1
2v2 + cpT = const — 0.1 pts
And further (for either approach):
Bringing it to ∆v2
2= 1
2v2
crit(a2
c2 − 1) = cp∆T — 0.1 pts
Measuring a and c — 0.2 pts
Obtaining vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s — 0.1 pts
Part C. Magnetic straws (4.5 points)
i. (0.8 pts)
Lines straight and parallel inside — 0.6 pts
(if not drawn over the entire length, subtract 0.2)
(can be slightly curved close the tube’s end)
Lines curve outwards slightly after the exit — 0.2 pts
(if so curved that more than one closed loop on both sides is
depicted in Fig, 0 pts out of 0.2)
ii. (1.2 pts)
Expressing induction as B = Φ/πr2 — 0.2 pts
Stating that w = B2
2µ0
— 0.2 pts
Idea of using virtual lengthening — 0.2 pts
(Introducing a lengthening ∆l is enough)
Expressing ∆W = B2
2µ0
πr2∆l — 0.2 pts
(Same marks if W expressed for entire tube)
Equating ∆W = T ∆l — 0.2 pts
Expressing T = Φ2
2µ0πr2 . — 0.2 pts
iii. (2.5 pts)
Idea of using analogy with el. charges — 1 pt
(Sketch with a quadrupole conf. of el. charges is enough)
Finding the force between two magn. charges via matching el.
and magn. quantities is worth 1 pts in total, split down as
follows:
Expressing el. stat. force via en. density — 0.5 pts
Using the obtained Eq. to obtain F = 1
4πµ0
Φ2
a2 — 0.5 pts
Any other matching scheme is graded analogously; e.g. find-
ing such a Q which has the same en. density as Φ — 0.5 pts;
expressing the force between magn. charges (Φ) as the force
between the matching el. charges (Q) — 0.5 pts. Declaring
matching pairs Q ↔ Φ and 1
4πε0
↔1
4πµ0
without energy-based-
motivation gives only 0.5 pts.
Noting that to tubes comp. of force = 0 — 0.2 pts
When expressing F = 2(F1 − F2): for factor “2” — 0.1 pts
and for finding F2 — 0.2 pts
(if wrong sign for F2, subtract 0.1)
If alternative solution is followed
Plan to express inter. energy as a function of a
intending to find F as a derivative — 0.2 pts
Calculating B(x) — 0.8 pts
If estimated without dependance on x, 0.4
considering a tube as an array of dipoles — 0.3 pts
expressing dm = Sjdx — 0.3 pts
relating j to Φ — 0.2 pts
Expressing U =∫
B(x)dm — 0.4 pts
If estimated as BSjl, 0.2
Finding F = dU/da — 0.2 pts
Taking into account the factor “ 2”— 0.1 pts
— page 3 of 5 —
101
(or stating that smaller pressure ⇒ lower temperature)
Finding Q as in the figure in the solutions — 0.2 pts(if Q marked below the wing’s tip — 0.1 pts)
iii. (2 pts)
Finding the dew point: idea of linearization — 0.2 pts
Expression and/or numerical value for dew point — 0.2 pts
Deriving 1
2µv2 + cpT = const:
1 mole of gas carries kin. en. 1
2µv2 — 0.2 pts
1 mole of gas carries heat en. CV T — 0.2 pts
work done on 1 mole of gas: p1V1 − p2V2 — 0.4 pts
en.: 1
2µ(v2
2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts
using ideal gas law obtaining 1
2v2 + cpT = const — 0.2 pts
Alternative approximate approach:
Bernoulli’s law 1
2ρv2 + p = const — 0.3 pts
0 pts out of 0.3 if the Bernoulli law includes ρgh
Adiabatic law pV γ = const — 0.3 pts
⇒ p1−γT γ = const — 0.2 pts
Approximation ∆p = γγ−1
pT
∆T — 0.2 pts
Leading to 1
2v2 + R
µ
cp
cp−cVT = const — 0.1 pts
Leading to 1
2v2 + cpT = const — 0.1 pts
And further (for either approach):
Bringing it to ∆v2
2= 1
2v2
crit(a2
c2 − 1) = cp∆T — 0.1 pts
Measuring a and c — 0.2 pts
Obtaining vcrit = c
√
2cp∆T
a2 − c2≈ 23 m/s — 0.1 pts
Part C. Magnetic straws (4.5 points)
i. (0.8 pts)
Lines straight and parallel inside — 0.6 pts
(if not drawn over the entire length, subtract 0.2)
(can be slightly curved close the tube’s end)
Lines curve outwards slightly after the exit — 0.2 pts
(if so curved that more than one closed loop on both sides is
depicted in Fig, 0 pts out of 0.2)
ii. (1.2 pts)
Expressing induction as B = Φ/πr2 — 0.2 pts
Stating that w = B2
2µ0
— 0.2 pts
Idea of using virtual lengthening — 0.2 pts
(Introducing a lengthening ∆l is enough)
Expressing ∆W = B2
2µ0
πr2∆l — 0.2 pts
(Same marks if W expressed for entire tube)
Equating ∆W = T ∆l — 0.2 pts
Expressing T = Φ2
2µ0πr2 . — 0.2 pts
iii. (2.5 pts)
Idea of using analogy with el. charges — 1 pt
(Sketch with a quadrupole conf. of el. charges is enough)
Finding the force between two magn. charges via matching el.
and magn. quantities is worth 1 pts in total, split down as
follows:
Expressing el. stat. force via en. density — 0.5 pts
Using the obtained Eq. to obtain F = 1
4πµ0
Φ2
a2 — 0.5 pts
Any other matching scheme is graded analogously; e.g. find-
ing such a Q which has the same en. density as Φ — 0.5 pts;
expressing the force between magn. charges (Φ) as the force
between the matching el. charges (Q) — 0.5 pts. Declaring
matching pairs Q ↔ Φ and 1
4πε0
↔1
4πµ0
without energy-based-
motivation gives only 0.5 pts.
Noting that to tubes comp. of force = 0 — 0.2 pts
When expressing F = 2(F1 − F2): for factor “2” — 0.1 pts
and for finding F2 — 0.2 pts
(if wrong sign for F2, subtract 0.1)
If alternative solution is followed
Plan to express inter. energy as a function of a
intending to find F as a derivative — 0.2 pts
Calculating B(x) — 0.8 pts
If estimated without dependance on x, 0.4
considering a tube as an array of dipoles — 0.3 pts
expressing dm = Sjdx — 0.3 pts
relating j to Φ — 0.2 pts
Expressing U =∫
B(x)dm — 0.4 pts
If estimated as BSjl, 0.2
Finding F = dU/da — 0.2 pts
Taking into account the factor “ 2”— 0.1 pts
— page 3 of 5 —
Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) For the terms entering the force balance of a
droplet immediately before separation from the nozzle, the
points are given as follows:
mg — 0.2 pts;
m = ρV — 0.1 pts;
V =4
3πr3
max — 0.1 pts;
πσd — 0.4 pts.
(if geometrically obtained cos α is included, 0.2 pts)
Force balance equation including these terms — 0.2 pts;
For expressing rmax from the equation — 0.2 pts.
ii. (1.2 pts)
Stating that the droplet’s potential is ϕ — 0.2 pts
(if used correctly, this does not need to be explicitly stated).
Expressing the droplet’s potential as1
4πε0
Q
r— 0.8 pts
(without correct sign — 0.6 pts).
Expressing Q from the obtained equation — 0.2 pts.
iii. (1.6 pts) The components of the excess pressure are graded
as follows.
2σ/r — 0.5 pts;
1
2ε0E2 — 0.4 pts;
bringing it to the form1
2ε0ϕ2/r2 — 0.2 pts.
Noticing that the two effects have opposite sign — 0.2 pts
(if used correctly, this does not need to be stated separately).
Equation stating that the excess pressure is 0 — 0.1 pts;
Expressing ϕmax from the obtained equation — 0.2 pts.
In the case of energy-balance-based solution, the distribu-
tion of marks is as follows.
surface energy change as 4πσd(r2) = 8πσrdr — 0.5 pts;
electrostatic energy change as 2πε0ϕ2maxdr — 0.5 pts;
electrostatic work as dAel
= 4πε0ϕ2maxdr — 0.3 pts;
equation stating that en. change equals to work — 0.1 pts;
expressing ϕmax from the obtained equation — 0.2 pts.
0 pts if energy are equated (without virtual displacement).
If factor 1
2missing before the electrostat. force (but other-
wise correct), -0.2 pts).
Part B. Two pipes (4 points)
i. (1.2 pts)
Stating that the surroundings’ potential is − U/2 — 0.4 pts;
(if not stated but used correctly — full marks; opposite signs
are allowed to be chosen consistently)
stating that the droplet’s potential is 0 — 0.4 pts
(if not stated but used correctly — full marks)
Applying formula U = q/C — 0.2 pts
Using the result of Part A with
ϕ = U/2 to obtain the final result — 0.2 pts.
the solutions with U instead of U/2 will qualify for 0.4 pts
for the first two lines (i.e. in total up to 0.8)
ii. (1.5 pts)
Stating that a droplet will increase the capacitor’s
charge by its own charge Q = 2πε0qrmax/C — 0.4 pts;
(0 pts if wrong sign)
(0.2 pts if redundant factor “2”)
Expressing dq = Q dN — 0.2 pts;
Substituting dN = n dt — 0.2 pts;
solving the obtained differential equation — 0.5 pts;
determining the integration constant from
the initial condition — 0.1 pts;
substituting rmax from above — 0.1 pts.
iii. (1.3 pts) Equation for the energy balance of a droplet:
expressing the droplet’s electrostatic energy change during
the fall as UQ — 0.6 pts
(0.3 for UQ/2)
expressing the droplet’s gravitational energy change as mgH
— 0.2 pts
noticing that at the limit voltage, droplet’s terminal kinetic
energy is zero — 0.3 pts
expressing energy conservation law equation — 0.1 pts
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the mgh line
and the kin. en. lines are applicable)
If instead of the energy balance, the force balance is used
(which is incorrect), partial credit for the terms entering the
equation is given as follows.
gravity force gm — 0.2 pts
electric field estimated as E ≈ U/(H − L/2) — 0.2 pts
(0.1, if not realized that this is an approximation)
electric force F = EQ — 0.1 pts
writing down force balance equation — 0.1 pts.
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the first line and
the force balance line are applicable)
— page 4 of 5 —
PrObLEm T2. kELvIn waTEr drOPPEr (8 POInTs)
102
Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) For the terms entering the force balance of a
droplet immediately before separation from the nozzle, the
points are given as follows:
mg — 0.2 pts;
m = ρV — 0.1 pts;
V =4
3πr3
max — 0.1 pts;
πσd — 0.4 pts.
(if geometrically obtained cos α is included, 0.2 pts)
Force balance equation including these terms — 0.2 pts;
For expressing rmax from the equation — 0.2 pts.
ii. (1.2 pts)
Stating that the droplet’s potential is ϕ — 0.2 pts
(if used correctly, this does not need to be explicitly stated).
Expressing the droplet’s potential as1
4πε0
Q
r— 0.8 pts
(without correct sign — 0.6 pts).
Expressing Q from the obtained equation — 0.2 pts.
iii. (1.6 pts) The components of the excess pressure are graded
as follows.
2σ/r — 0.5 pts;
1
2ε0E2 — 0.4 pts;
bringing it to the form1
2ε0ϕ2/r2 — 0.2 pts.
Noticing that the two effects have opposite sign — 0.2 pts
(if used correctly, this does not need to be stated separately).
Equation stating that the excess pressure is 0 — 0.1 pts;
Expressing ϕmax from the obtained equation — 0.2 pts.
In the case of energy-balance-based solution, the distribu-
tion of marks is as follows.
surface energy change as 4πσd(r2) = 8πσrdr — 0.5 pts;
electrostatic energy change as 2πε0ϕ2maxdr — 0.5 pts;
electrostatic work as dAel
= 4πε0ϕ2maxdr — 0.3 pts;
equation stating that en. change equals to work — 0.1 pts;
expressing ϕmax from the obtained equation — 0.2 pts.
0 pts if energy are equated (without virtual displacement).
If factor 1
2missing before the electrostat. force (but other-
wise correct), -0.2 pts).
Part B. Two pipes (4 points)
i. (1.2 pts)
Stating that the surroundings’ potential is − U/2 — 0.4 pts;
(if not stated but used correctly — full marks; opposite signs
are allowed to be chosen consistently)
stating that the droplet’s potential is 0 — 0.4 pts
(if not stated but used correctly — full marks)
Applying formula U = q/C — 0.2 pts
Using the result of Part A with
ϕ = U/2 to obtain the final result — 0.2 pts.
the solutions with U instead of U/2 will qualify for 0.4 pts
for the first two lines (i.e. in total up to 0.8)
ii. (1.5 pts)
Stating that a droplet will increase the capacitor’s
charge by its own charge Q = 2πε0qrmax/C — 0.4 pts;
(0 pts if wrong sign)
(0.2 pts if redundant factor “2”)
Expressing dq = Q dN — 0.2 pts;
Substituting dN = n dt — 0.2 pts;
solving the obtained differential equation — 0.5 pts;
determining the integration constant from
the initial condition — 0.1 pts;
substituting rmax from above — 0.1 pts.
iii. (1.3 pts) Equation for the energy balance of a droplet:
expressing the droplet’s electrostatic energy change during
the fall as UQ — 0.6 pts
(0.3 for UQ/2)
expressing the droplet’s gravitational energy change as mgH
— 0.2 pts
noticing that at the limit voltage, droplet’s terminal kinetic
energy is zero — 0.3 pts
expressing energy conservation law equation — 0.1 pts
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the mgh line
and the kin. en. lines are applicable)
If instead of the energy balance, the force balance is used
(which is incorrect), partial credit for the terms entering the
equation is given as follows.
gravity force gm — 0.2 pts
electric field estimated as E ≈ U/(H − L/2) — 0.2 pts
(0.1, if not realized that this is an approximation)
electric force F = EQ — 0.1 pts
writing down force balance equation — 0.1 pts.
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the first line and
the force balance line are applicable)
— page 4 of 5 —
Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) For the terms entering the force balance of a
droplet immediately before separation from the nozzle, the
points are given as follows:
mg — 0.2 pts;
m = ρV — 0.1 pts;
V =4
3πr3
max — 0.1 pts;
πσd — 0.4 pts.
(if geometrically obtained cos α is included, 0.2 pts)
Force balance equation including these terms — 0.2 pts;
For expressing rmax from the equation — 0.2 pts.
ii. (1.2 pts)
Stating that the droplet’s potential is ϕ — 0.2 pts
(if used correctly, this does not need to be explicitly stated).
Expressing the droplet’s potential as1
4πε0
Q
r— 0.8 pts
(without correct sign — 0.6 pts).
Expressing Q from the obtained equation — 0.2 pts.
iii. (1.6 pts) The components of the excess pressure are graded
as follows.
2σ/r — 0.5 pts;
1
2ε0E2 — 0.4 pts;
bringing it to the form1
2ε0ϕ2/r2 — 0.2 pts.
Noticing that the two effects have opposite sign — 0.2 pts
(if used correctly, this does not need to be stated separately).
Equation stating that the excess pressure is 0 — 0.1 pts;
Expressing ϕmax from the obtained equation — 0.2 pts.
In the case of energy-balance-based solution, the distribu-
tion of marks is as follows.
surface energy change as 4πσd(r2) = 8πσrdr — 0.5 pts;
electrostatic energy change as 2πε0ϕ2maxdr — 0.5 pts;
electrostatic work as dAel
= 4πε0ϕ2maxdr — 0.3 pts;
equation stating that en. change equals to work — 0.1 pts;
expressing ϕmax from the obtained equation — 0.2 pts.
0 pts if energy are equated (without virtual displacement).
If factor 1
2missing before the electrostat. force (but other-
wise correct), -0.2 pts).
Part B. Two pipes (4 points)
i. (1.2 pts)
Stating that the surroundings’ potential is − U/2 — 0.4 pts;
(if not stated but used correctly — full marks; opposite signs
are allowed to be chosen consistently)
stating that the droplet’s potential is 0 — 0.4 pts
(if not stated but used correctly — full marks)
Applying formula U = q/C — 0.2 pts
Using the result of Part A with
ϕ = U/2 to obtain the final result — 0.2 pts.
the solutions with U instead of U/2 will qualify for 0.4 pts
for the first two lines (i.e. in total up to 0.8)
ii. (1.5 pts)
Stating that a droplet will increase the capacitor’s
charge by its own charge Q = 2πε0qrmax/C — 0.4 pts;
(0 pts if wrong sign)
(0.2 pts if redundant factor “2”)
Expressing dq = Q dN — 0.2 pts;
Substituting dN = n dt — 0.2 pts;
solving the obtained differential equation — 0.5 pts;
determining the integration constant from
the initial condition — 0.1 pts;
substituting rmax from above — 0.1 pts.
iii. (1.3 pts) Equation for the energy balance of a droplet:
expressing the droplet’s electrostatic energy change during
the fall as UQ — 0.6 pts
(0.3 for UQ/2)
expressing the droplet’s gravitational energy change as mgH
— 0.2 pts
noticing that at the limit voltage, droplet’s terminal kinetic
energy is zero — 0.3 pts
expressing energy conservation law equation — 0.1 pts
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the mgh line
and the kin. en. lines are applicable)
If instead of the energy balance, the force balance is used
(which is incorrect), partial credit for the terms entering the
equation is given as follows.
gravity force gm — 0.2 pts
electric field estimated as E ≈ U/(H − L/2) — 0.2 pts
(0.1, if not realized that this is an approximation)
electric force F = EQ — 0.1 pts
writing down force balance equation — 0.1 pts.
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the first line and
the force balance line are applicable)
— page 4 of 5 —
Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) For the terms entering the force balance of a
droplet immediately before separation from the nozzle, the
points are given as follows:
mg — 0.2 pts;
m = ρV — 0.1 pts;
V =4
3πr3
max — 0.1 pts;
πσd — 0.4 pts.
(if geometrically obtained cos α is included, 0.2 pts)
Force balance equation including these terms — 0.2 pts;
For expressing rmax from the equation — 0.2 pts.
ii. (1.2 pts)
Stating that the droplet’s potential is ϕ — 0.2 pts
(if used correctly, this does not need to be explicitly stated).
Expressing the droplet’s potential as1
4πε0
Q
r— 0.8 pts
(without correct sign — 0.6 pts).
Expressing Q from the obtained equation — 0.2 pts.
iii. (1.6 pts) The components of the excess pressure are graded
as follows.
2σ/r — 0.5 pts;
1
2ε0E2 — 0.4 pts;
bringing it to the form1
2ε0ϕ2/r2 — 0.2 pts.
Noticing that the two effects have opposite sign — 0.2 pts
(if used correctly, this does not need to be stated separately).
Equation stating that the excess pressure is 0 — 0.1 pts;
Expressing ϕmax from the obtained equation — 0.2 pts.
In the case of energy-balance-based solution, the distribu-
tion of marks is as follows.
surface energy change as 4πσd(r2) = 8πσrdr — 0.5 pts;
electrostatic energy change as 2πε0ϕ2maxdr — 0.5 pts;
electrostatic work as dAel
= 4πε0ϕ2maxdr — 0.3 pts;
equation stating that en. change equals to work — 0.1 pts;
expressing ϕmax from the obtained equation — 0.2 pts.
0 pts if energy are equated (without virtual displacement).
If factor 1
2missing before the electrostat. force (but other-
wise correct), -0.2 pts).
Part B. Two pipes (4 points)
i. (1.2 pts)
Stating that the surroundings’ potential is − U/2 — 0.4 pts;
(if not stated but used correctly — full marks; opposite signs
are allowed to be chosen consistently)
stating that the droplet’s potential is 0 — 0.4 pts
(if not stated but used correctly — full marks)
Applying formula U = q/C — 0.2 pts
Using the result of Part A with
ϕ = U/2 to obtain the final result — 0.2 pts.
the solutions with U instead of U/2 will qualify for 0.4 pts
for the first two lines (i.e. in total up to 0.8)
ii. (1.5 pts)
Stating that a droplet will increase the capacitor’s
charge by its own charge Q = 2πε0qrmax/C — 0.4 pts;
(0 pts if wrong sign)
(0.2 pts if redundant factor “2”)
Expressing dq = Q dN — 0.2 pts;
Substituting dN = n dt — 0.2 pts;
solving the obtained differential equation — 0.5 pts;
determining the integration constant from
the initial condition — 0.1 pts;
substituting rmax from above — 0.1 pts.
iii. (1.3 pts) Equation for the energy balance of a droplet:
expressing the droplet’s electrostatic energy change during
the fall as UQ — 0.6 pts
(0.3 for UQ/2)
expressing the droplet’s gravitational energy change as mgH
— 0.2 pts
noticing that at the limit voltage, droplet’s terminal kinetic
energy is zero — 0.3 pts
expressing energy conservation law equation — 0.1 pts
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the mgh line
and the kin. en. lines are applicable)
If instead of the energy balance, the force balance is used
(which is incorrect), partial credit for the terms entering the
equation is given as follows.
gravity force gm — 0.2 pts
electric field estimated as E ≈ U/(H − L/2) — 0.2 pts
(0.1, if not realized that this is an approximation)
electric force F = EQ — 0.1 pts
writing down force balance equation — 0.1 pts.
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the first line and
the force balance line are applicable)
— page 4 of 5 —
103
Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)
i. (1.2 pts) For the terms entering the force balance of a
droplet immediately before separation from the nozzle, the
points are given as follows:
mg — 0.2 pts;
m = ρV — 0.1 pts;
V =4
3πr3
max — 0.1 pts;
πσd — 0.4 pts.
(if geometrically obtained cos α is included, 0.2 pts)
Force balance equation including these terms — 0.2 pts;
For expressing rmax from the equation — 0.2 pts.
ii. (1.2 pts)
Stating that the droplet’s potential is ϕ — 0.2 pts
(if used correctly, this does not need to be explicitly stated).
Expressing the droplet’s potential as1
4πε0
Q
r— 0.8 pts
(without correct sign — 0.6 pts).
Expressing Q from the obtained equation — 0.2 pts.
iii. (1.6 pts) The components of the excess pressure are graded
as follows.
2σ/r — 0.5 pts;
1
2ε0E2 — 0.4 pts;
bringing it to the form1
2ε0ϕ2/r2 — 0.2 pts.
Noticing that the two effects have opposite sign — 0.2 pts
(if used correctly, this does not need to be stated separately).
Equation stating that the excess pressure is 0 — 0.1 pts;
Expressing ϕmax from the obtained equation — 0.2 pts.
In the case of energy-balance-based solution, the distribu-
tion of marks is as follows.
surface energy change as 4πσd(r2) = 8πσrdr — 0.5 pts;
electrostatic energy change as 2πε0ϕ2maxdr — 0.5 pts;
electrostatic work as dAel
= 4πε0ϕ2maxdr — 0.3 pts;
equation stating that en. change equals to work — 0.1 pts;
expressing ϕmax from the obtained equation — 0.2 pts.
0 pts if energy are equated (without virtual displacement).
If factor 1
2missing before the electrostat. force (but other-
wise correct), -0.2 pts).
Part B. Two pipes (4 points)
i. (1.2 pts)
Stating that the surroundings’ potential is − U/2 — 0.4 pts;
(if not stated but used correctly — full marks; opposite signs
are allowed to be chosen consistently)
stating that the droplet’s potential is 0 — 0.4 pts
(if not stated but used correctly — full marks)
Applying formula U = q/C — 0.2 pts
Using the result of Part A with
ϕ = U/2 to obtain the final result — 0.2 pts.
the solutions with U instead of U/2 will qualify for 0.4 pts
for the first two lines (i.e. in total up to 0.8)
ii. (1.5 pts)
Stating that a droplet will increase the capacitor’s
charge by its own charge Q = 2πε0qrmax/C — 0.4 pts;
(0 pts if wrong sign)
(0.2 pts if redundant factor “2”)
Expressing dq = Q dN — 0.2 pts;
Substituting dN = n dt — 0.2 pts;
solving the obtained differential equation — 0.5 pts;
determining the integration constant from
the initial condition — 0.1 pts;
substituting rmax from above — 0.1 pts.
iii. (1.3 pts) Equation for the energy balance of a droplet:
expressing the droplet’s electrostatic energy change during
the fall as UQ — 0.6 pts
(0.3 for UQ/2)
expressing the droplet’s gravitational energy change as mgH
— 0.2 pts
noticing that at the limit voltage, droplet’s terminal kinetic
energy is zero — 0.3 pts
expressing energy conservation law equation — 0.1 pts
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the mgh line
and the kin. en. lines are applicable)
If instead of the energy balance, the force balance is used
(which is incorrect), partial credit for the terms entering the
equation is given as follows.
gravity force gm — 0.2 pts
electric field estimated as E ≈ U/(H − L/2) — 0.2 pts
(0.1, if not realized that this is an approximation)
electric force F = EQ — 0.1 pts
writing down force balance equation — 0.1 pts.
expressing the final answer — 0.1 pts.
(if bowl considered as a point charge, only the first line and
the force balance line are applicable)
— page 4 of 5 —
104
Problem T3. Protostar formation (9 points)i. (0.8 pts)
In thermodynamic equilibrium T = const — 0.2 pts.(if not stated but used correctly — full marks)
pV = const — 0.3 pts.
V ∝ r3 — 0.1 pts.
p ∝ r−3 — 0.1 pts.
p(r1)
p(r0)= 8 — 0.1 pts.
ii. (1 pt)
t ≈
√
2(r0 − r2)
g— 0.4 pts.
g ≈Gm
r20
— 0.4 pts.
t ≈
√
0.1r30
Gm— 0.2 pts.
iii. (2.5 pts)
First solution:
Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.
(equivalently one may mention the ∝ r−2 force coming from
Gauss’ law; if not stated, but used correctly — full marks)
Idea of the motion as an ultraelliptical orbit — 1 pt.
Period of the elliptical orbit is equal to the periodof the circular orbit of the same longer semiaxis — 0.4 pts.
(if not stated but used correctly — full marks)
The longer semiaxis is r0/2 — 0.1 pts.
Equation(s) for the period — 0.3 pts.
We need half a period — 0.1 pts.
Final answer tr→0 = π
√
r30
8Gm— 0.1 pts.
Alternative solution:
Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.
(equivalently one may mention the ∝ r−2 force coming from
Gauss’ law; if not stated, but used correctly — full marks)
Energy conservation as a differential equation — 0.3 pts.
(if only expressed through v (like mv2
2− GMm
r= E) — 0.1 pts.;
if the differential equation of Newton’s 2nd law (r = −Gmr2 ) is
given instead — 0.2 pts.)
t =
∫
dr√
2E + 2Gmr
(whichever the sign is) — 0.4 pts.
Integration and final answer — 1.3 pts.
iv. (1.7 pts)
Radiated heat equals the compression work — 1 pt.
[for mentioning or using the 1st law of thermodynamics (pos-
sibly in a wrong way) — 0.5 pts.]
W = −
∫
p dV or W = −∑
p ∆V — 0.3 pts.
(with either “+” or “−” — give full marks)
p =mRT0
µV— 0.2 pts.
Calculating the integral, W =3mRT0
µln
r0
r3
— 0.2 pts.
v. (1 pt)
The collapse continues adiabatically.(If used, but not written down, give full marks.) — 0.3 pts.
pV γ = const — 0.3 pts.
T ∝ V 1−γ — 0.2 pts.
T = T0
(r3
r
)3γ−3
— 0.2 pts.
vi. (2 pts)
In case of using energies (marks must not be subtracted for
fewer approximations, even in the final answer):
T4 = T0
(
r3
r4
)3γ−3
— 0.1 pts.
∆Q + ∆Π ≈ 0 — 0.4 pts.
∆Π ≈ −Gm2/r4 — 0.6 pts.
∆Q = mcV T4 — 0.4 pts.
cV ≈R
µ— 0.3 pts.
Final answer r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
— 0.1 pts.
Final answer T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
— 0.1 pts.
In case of using pressures (again, fewer approximations are per-
mitted):
T4 = T0
(
r3
r4
)3γ−3
— 0.1 pts.
p4 = phydrostatic — 0.4 pts.
p4 =ρ
µRT4 — 0.5 pts.
phydrostatic ≈ ρgr4 — 0.4 pts.
g ≈Gm
r24
— 0.4 pts.
Final answer r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
— 0.1 pts.
Final answer T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
— 0.1 pts.
— page 5 of 5 —
PrObLEm T3. PrOTOsTar FOrmaTIOn (9 POInTs)
105
Problem T3. Protostar formation (9 points)i. (0.8 pts)
In thermodynamic equilibrium T = const — 0.2 pts.(if not stated but used correctly — full marks)
pV = const — 0.3 pts.
V ∝ r3 — 0.1 pts.
p ∝ r−3 — 0.1 pts.
p(r1)
p(r0)= 8 — 0.1 pts.
ii. (1 pt)
t ≈
√
2(r0 − r2)
g— 0.4 pts.
g ≈Gm
r20
— 0.4 pts.
t ≈
√
0.1r30
Gm— 0.2 pts.
iii. (2.5 pts)
First solution:
Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.
(equivalently one may mention the ∝ r−2 force coming from
Gauss’ law; if not stated, but used correctly — full marks)
Idea of the motion as an ultraelliptical orbit — 1 pt.
Period of the elliptical orbit is equal to the periodof the circular orbit of the same longer semiaxis — 0.4 pts.
(if not stated but used correctly — full marks)
The longer semiaxis is r0/2 — 0.1 pts.
Equation(s) for the period — 0.3 pts.
We need half a period — 0.1 pts.
Final answer tr→0 = π
√
r30
8Gm— 0.1 pts.
Alternative solution:
Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.
(equivalently one may mention the ∝ r−2 force coming from
Gauss’ law; if not stated, but used correctly — full marks)
Energy conservation as a differential equation — 0.3 pts.
(if only expressed through v (like mv2
2− GMm
r= E) — 0.1 pts.;
if the differential equation of Newton’s 2nd law (r = −Gmr2 ) is
given instead — 0.2 pts.)
t =
∫
dr√
2E + 2Gmr
(whichever the sign is) — 0.4 pts.
Integration and final answer — 1.3 pts.
iv. (1.7 pts)
Radiated heat equals the compression work — 1 pt.
[for mentioning or using the 1st law of thermodynamics (pos-
sibly in a wrong way) — 0.5 pts.]
W = −
∫
p dV or W = −∑
p ∆V — 0.3 pts.
(with either “+” or “−” — give full marks)
p =mRT0
µV— 0.2 pts.
Calculating the integral, W =3mRT0
µln
r0
r3
— 0.2 pts.
v. (1 pt)
The collapse continues adiabatically.(If used, but not written down, give full marks.) — 0.3 pts.
pV γ = const — 0.3 pts.
T ∝ V 1−γ — 0.2 pts.
T = T0
(r3
r
)3γ−3
— 0.2 pts.
vi. (2 pts)
In case of using energies (marks must not be subtracted for
fewer approximations, even in the final answer):
T4 = T0
(
r3
r4
)3γ−3
— 0.1 pts.
∆Q + ∆Π ≈ 0 — 0.4 pts.
∆Π ≈ −Gm2/r4 — 0.6 pts.
∆Q = mcV T4 — 0.4 pts.
cV ≈R
µ— 0.3 pts.
Final answer r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
— 0.1 pts.
Final answer T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
— 0.1 pts.
In case of using pressures (again, fewer approximations are per-
mitted):
T4 = T0
(
r3
r4
)3γ−3
— 0.1 pts.
p4 = phydrostatic — 0.4 pts.
p4 =ρ
µRT4 — 0.5 pts.
phydrostatic ≈ ρgr4 — 0.4 pts.
g ≈Gm
r24
— 0.4 pts.
Final answer r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
— 0.1 pts.
Final answer T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
— 0.1 pts.
— page 5 of 5 —Problem T3. Protostar formation (9 points)i. (0.8 pts)
In thermodynamic equilibrium T = const — 0.2 pts.(if not stated but used correctly — full marks)
pV = const — 0.3 pts.
V ∝ r3 — 0.1 pts.
p ∝ r−3 — 0.1 pts.
p(r1)
p(r0)= 8 — 0.1 pts.
ii. (1 pt)
t ≈
√
2(r0 − r2)
g— 0.4 pts.
g ≈Gm
r20
— 0.4 pts.
t ≈
√
0.1r30
Gm— 0.2 pts.
iii. (2.5 pts)
First solution:
Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.
(equivalently one may mention the ∝ r−2 force coming from
Gauss’ law; if not stated, but used correctly — full marks)
Idea of the motion as an ultraelliptical orbit — 1 pt.
Period of the elliptical orbit is equal to the periodof the circular orbit of the same longer semiaxis — 0.4 pts.
(if not stated but used correctly — full marks)
The longer semiaxis is r0/2 — 0.1 pts.
Equation(s) for the period — 0.3 pts.
We need half a period — 0.1 pts.
Final answer tr→0 = π
√
r30
8Gm— 0.1 pts.
Alternative solution:
Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.
(equivalently one may mention the ∝ r−2 force coming from
Gauss’ law; if not stated, but used correctly — full marks)
Energy conservation as a differential equation — 0.3 pts.
(if only expressed through v (like mv2
2− GMm
r= E) — 0.1 pts.;
if the differential equation of Newton’s 2nd law (r = −Gmr2 ) is
given instead — 0.2 pts.)
t =
∫
dr√
2E + 2Gmr
(whichever the sign is) — 0.4 pts.
Integration and final answer — 1.3 pts.
iv. (1.7 pts)
Radiated heat equals the compression work — 1 pt.
[for mentioning or using the 1st law of thermodynamics (pos-
sibly in a wrong way) — 0.5 pts.]
W = −
∫
p dV or W = −∑
p ∆V — 0.3 pts.
(with either “+” or “−” — give full marks)
p =mRT0
µV— 0.2 pts.
Calculating the integral, W =3mRT0
µln
r0
r3
— 0.2 pts.
v. (1 pt)
The collapse continues adiabatically.(If used, but not written down, give full marks.) — 0.3 pts.
pV γ = const — 0.3 pts.
T ∝ V 1−γ — 0.2 pts.
T = T0
(r3
r
)3γ−3
— 0.2 pts.
vi. (2 pts)
In case of using energies (marks must not be subtracted for
fewer approximations, even in the final answer):
T4 = T0
(
r3
r4
)3γ−3
— 0.1 pts.
∆Q + ∆Π ≈ 0 — 0.4 pts.
∆Π ≈ −Gm2/r4 — 0.6 pts.
∆Q = mcV T4 — 0.4 pts.
cV ≈R
µ— 0.3 pts.
Final answer r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
— 0.1 pts.
Final answer T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
— 0.1 pts.
In case of using pressures (again, fewer approximations are per-
mitted):
T4 = T0
(
r3
r4
)3γ−3
— 0.1 pts.
p4 = phydrostatic — 0.4 pts.
p4 =ρ
µRT4 — 0.5 pts.
phydrostatic ≈ ρgr4 — 0.4 pts.
g ≈Gm
r24
— 0.4 pts.
Final answer r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
— 0.1 pts.
Final answer T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
— 0.1 pts.
— page 5 of 5 —
106
Problem T3. Protostar formation (9 points)i. (0.8 pts)
In thermodynamic equilibrium T = const — 0.2 pts.(if not stated but used correctly — full marks)
pV = const — 0.3 pts.
V ∝ r3 — 0.1 pts.
p ∝ r−3 — 0.1 pts.
p(r1)
p(r0)= 8 — 0.1 pts.
ii. (1 pt)
t ≈
√
2(r0 − r2)
g— 0.4 pts.
g ≈Gm
r20
— 0.4 pts.
t ≈
√
0.1r30
Gm— 0.2 pts.
iii. (2.5 pts)
First solution:
Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.
(equivalently one may mention the ∝ r−2 force coming from
Gauss’ law; if not stated, but used correctly — full marks)
Idea of the motion as an ultraelliptical orbit — 1 pt.
Period of the elliptical orbit is equal to the periodof the circular orbit of the same longer semiaxis — 0.4 pts.
(if not stated but used correctly — full marks)
The longer semiaxis is r0/2 — 0.1 pts.
Equation(s) for the period — 0.3 pts.
We need half a period — 0.1 pts.
Final answer tr→0 = π
√
r30
8Gm— 0.1 pts.
Alternative solution:
Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.
(equivalently one may mention the ∝ r−2 force coming from
Gauss’ law; if not stated, but used correctly — full marks)
Energy conservation as a differential equation — 0.3 pts.
(if only expressed through v (like mv2
2− GMm
r= E) — 0.1 pts.;
if the differential equation of Newton’s 2nd law (r = −Gmr2 ) is
given instead — 0.2 pts.)
t =
∫
dr√
2E + 2Gmr
(whichever the sign is) — 0.4 pts.
Integration and final answer — 1.3 pts.
iv. (1.7 pts)
Radiated heat equals the compression work — 1 pt.
[for mentioning or using the 1st law of thermodynamics (pos-
sibly in a wrong way) — 0.5 pts.]
W = −
∫
p dV or W = −∑
p ∆V — 0.3 pts.
(with either “+” or “−” — give full marks)
p =mRT0
µV— 0.2 pts.
Calculating the integral, W =3mRT0
µln
r0
r3
— 0.2 pts.
v. (1 pt)
The collapse continues adiabatically.(If used, but not written down, give full marks.) — 0.3 pts.
pV γ = const — 0.3 pts.
T ∝ V 1−γ — 0.2 pts.
T = T0
(r3
r
)3γ−3
— 0.2 pts.
vi. (2 pts)
In case of using energies (marks must not be subtracted for
fewer approximations, even in the final answer):
T4 = T0
(
r3
r4
)3γ−3
— 0.1 pts.
∆Q + ∆Π ≈ 0 — 0.4 pts.
∆Π ≈ −Gm2/r4 — 0.6 pts.
∆Q = mcV T4 — 0.4 pts.
cV ≈R
µ— 0.3 pts.
Final answer r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
— 0.1 pts.
Final answer T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
— 0.1 pts.
In case of using pressures (again, fewer approximations are per-
mitted):
T4 = T0
(
r3
r4
)3γ−3
— 0.1 pts.
p4 = phydrostatic — 0.4 pts.
p4 =ρ
µRT4 — 0.5 pts.
phydrostatic ≈ ρgr4 — 0.4 pts.
g ≈Gm
r24
— 0.4 pts.
Final answer r4 ≈ r3
(
RT0r3
µmG
)1
3γ−4
— 0.1 pts.
Final answer T4 ≈ T0
(
RT0r3
µmG
)
3γ−3
4−3γ
— 0.1 pts.
— page 5 of 5 —
107
The 43rd International Physics Olympiad — Experimental Competition
Tartu, Estonia — Thursday, July 19th 2012
• The examination lasts for 5 hours. There are 2 problems
worth in total 20 points. There are two tables in your
disposal (in two neighbouring cubicles), the apparatus of
Problem E1 is on one table and the apparatus of Prob-
lem E2 is on the other table; you can move freely between
these tables. However, you are not allowed to move
any piece of experimental setup from one table to
the other.
• Initially the experimental equipment on one table is
covered and on the other table is boxed. You must
neither remove the cover nor open the box nor
open the envelope with the problems before the
sound signal of the beginning of competition
(three short signals).
• You are not allowed to leave your working place
without permission. If you need any assistance (mal-
functioning equipment, broken calculator, need to
visit a restroom, etc), please raise the corresponding flag
(“help” or “toilet” with a long handle at your seat)
above your seat box walls and keep it raised until an or-
ganizer arrives.
• Use only the front side of the sheets of paper.
• For each problem, there are dedicated Solution Sheets
(see header for the number and pictogramme). Write
your solutions onto the appropriate Solution Sheets. For
each Problem, the Solution Sheets are numbered; use
the sheets according to the enumeration. Always mark
which Problem Part and Question you are deal-
ing with. Copy the final answers into the appropriate
boxes of the Answer Sheets. There are also Draft pa-
pers; use these for writing things which you don’t want
to be graded. If you have written something what you
don’t want to be graded onto the Solution Sheets (such
as initial and incorrect solutions), cross these out.
• If you need more paper for a certain problem, please raise
the flag “help” and tell an organizer the problem num-
ber; you are given two Solution sheets (you can do this
more than once).
• You should use as little text as possible: try to
explain your solution mainly with equations, numbers,
tables, symbols and diagrams. When textual explana-
tion is unavoidable, you are encouraged to provide Eng-
lish translation alongside with the text in your native
language (if you mistranslate, or don’t translate at all,
your native language text will be used during the Moder-
ation).
• Avoid unnecessary movements during the experimental
examination and do not shake the walls of your box: the
laser experiment requires stability.
• Do not look into the laser beam or its reflections! It may
permanently damage your eyes.
• The first single sound signal tells you that there are 30
min of solving time left; the second double sound signal
means that 5 min is left; the third triple sound signal
marks the end of solving time. After the third sound
signal you must stop writing immediately. Put all
the papers into the envelope at your desk. You are not
allowed to take any sheet of paper out of the room.
If you have finished solving before the final sound signal,
please raise your flag.
— page 1 of 4 —
The 43rd International Physics Olympiad — Experimental CompetitionTartu, Estonia — Thursday, July 19th 2012
108
The 43rd International Physics Olympiad — Experimental Competition
Tartu, Estonia — Thursday, July 19th 2012
• The examination lasts for 5 hours. There are 2 problems
worth in total 20 points. There are two tables in your
disposal (in two neighbouring cubicles), the apparatus of
Problem E1 is on one table and the apparatus of Prob-
lem E2 is on the other table; you can move freely between
these tables. However, you are not allowed to move
any piece of experimental setup from one table to
the other.
• Initially the experimental equipment on one table is
covered and on the other table is boxed. You must
neither remove the cover nor open the box nor
open the envelope with the problems before the
sound signal of the beginning of competition
(three short signals).
• You are not allowed to leave your working place
without permission. If you need any assistance (mal-
functioning equipment, broken calculator, need to
visit a restroom, etc), please raise the corresponding flag
(“help” or “toilet” with a long handle at your seat)
above your seat box walls and keep it raised until an or-
ganizer arrives.
• Use only the front side of the sheets of paper.
• For each problem, there are dedicated Solution Sheets
(see header for the number and pictogramme). Write
your solutions onto the appropriate Solution Sheets. For
each Problem, the Solution Sheets are numbered; use
the sheets according to the enumeration. Always mark
which Problem Part and Question you are deal-
ing with. Copy the final answers into the appropriate
boxes of the Answer Sheets. There are also Draft pa-
pers; use these for writing things which you don’t want
to be graded. If you have written something what you
don’t want to be graded onto the Solution Sheets (such
as initial and incorrect solutions), cross these out.
• If you need more paper for a certain problem, please raise
the flag “help” and tell an organizer the problem num-
ber; you are given two Solution sheets (you can do this
more than once).
• You should use as little text as possible: try to
explain your solution mainly with equations, numbers,
tables, symbols and diagrams. When textual explana-
tion is unavoidable, you are encouraged to provide Eng-
lish translation alongside with the text in your native
language (if you mistranslate, or don’t translate at all,
your native language text will be used during the Moder-
ation).
• Avoid unnecessary movements during the experimental
examination and do not shake the walls of your box: the
laser experiment requires stability.
• Do not look into the laser beam or its reflections! It may
permanently damage your eyes.
• The first single sound signal tells you that there are 30
min of solving time left; the second double sound signal
means that 5 min is left; the third triple sound signal
marks the end of solving time. After the third sound
signal you must stop writing immediately. Put all
the papers into the envelope at your desk. You are not
allowed to take any sheet of paper out of the room.
If you have finished solving before the final sound signal,
please raise your flag.
— page 1 of 4 —
The 43rd International Physics Olympiad — Experimental Competition
Tartu, Estonia — Thursday, July 19th 2012
• The examination lasts for 5 hours. There are 2 problems
worth in total 20 points. There are two tables in your
disposal (in two neighbouring cubicles), the apparatus of
Problem E1 is on one table and the apparatus of Prob-
lem E2 is on the other table; you can move freely between
these tables. However, you are not allowed to move
any piece of experimental setup from one table to
the other.
• Initially the experimental equipment on one table is
covered and on the other table is boxed. You must
neither remove the cover nor open the box nor
open the envelope with the problems before the
sound signal of the beginning of competition
(three short signals).
• You are not allowed to leave your working place
without permission. If you need any assistance (mal-
functioning equipment, broken calculator, need to
visit a restroom, etc), please raise the corresponding flag
(“help” or “toilet” with a long handle at your seat)
above your seat box walls and keep it raised until an or-
ganizer arrives.
• Use only the front side of the sheets of paper.
• For each problem, there are dedicated Solution Sheets
(see header for the number and pictogramme). Write
your solutions onto the appropriate Solution Sheets. For
each Problem, the Solution Sheets are numbered; use
the sheets according to the enumeration. Always mark
which Problem Part and Question you are deal-
ing with. Copy the final answers into the appropriate
boxes of the Answer Sheets. There are also Draft pa-
pers; use these for writing things which you don’t want
to be graded. If you have written something what you
don’t want to be graded onto the Solution Sheets (such
as initial and incorrect solutions), cross these out.
• If you need more paper for a certain problem, please raise
the flag “help” and tell an organizer the problem num-
ber; you are given two Solution sheets (you can do this
more than once).
• You should use as little text as possible: try to
explain your solution mainly with equations, numbers,
tables, symbols and diagrams. When textual explana-
tion is unavoidable, you are encouraged to provide Eng-
lish translation alongside with the text in your native
language (if you mistranslate, or don’t translate at all,
your native language text will be used during the Moder-
ation).
• Avoid unnecessary movements during the experimental
examination and do not shake the walls of your box: the
laser experiment requires stability.
• Do not look into the laser beam or its reflections! It may
permanently damage your eyes.
• The first single sound signal tells you that there are 30
min of solving time left; the second double sound signal
means that 5 min is left; the third triple sound signal
marks the end of solving time. After the third sound
signal you must stop writing immediately. Put all
the papers into the envelope at your desk. You are not
allowed to take any sheet of paper out of the room.
If you have finished solving before the final sound signal,
please raise your flag.
— page 1 of 4 —
The 43rd International Physics Olympiad — Experimental Competition
Tartu, Estonia — Thursday, July 19th 2012
• The examination lasts for 5 hours. There are 2 problems
worth in total 20 points. There are two tables in your
disposal (in two neighbouring cubicles), the apparatus of
Problem E1 is on one table and the apparatus of Prob-
lem E2 is on the other table; you can move freely between
these tables. However, you are not allowed to move
any piece of experimental setup from one table to
the other.
• Initially the experimental equipment on one table is
covered and on the other table is boxed. You must
neither remove the cover nor open the box nor
open the envelope with the problems before the
sound signal of the beginning of competition
(three short signals).
• You are not allowed to leave your working place
without permission. If you need any assistance (mal-
functioning equipment, broken calculator, need to
visit a restroom, etc), please raise the corresponding flag
(“help” or “toilet” with a long handle at your seat)
above your seat box walls and keep it raised until an or-
ganizer arrives.
• Use only the front side of the sheets of paper.
• For each problem, there are dedicated Solution Sheets
(see header for the number and pictogramme). Write
your solutions onto the appropriate Solution Sheets. For
each Problem, the Solution Sheets are numbered; use
the sheets according to the enumeration. Always mark
which Problem Part and Question you are deal-
ing with. Copy the final answers into the appropriate
boxes of the Answer Sheets. There are also Draft pa-
pers; use these for writing things which you don’t want
to be graded. If you have written something what you
don’t want to be graded onto the Solution Sheets (such
as initial and incorrect solutions), cross these out.
• If you need more paper for a certain problem, please raise
the flag “help” and tell an organizer the problem num-
ber; you are given two Solution sheets (you can do this
more than once).
• You should use as little text as possible: try to
explain your solution mainly with equations, numbers,
tables, symbols and diagrams. When textual explana-
tion is unavoidable, you are encouraged to provide Eng-
lish translation alongside with the text in your native
language (if you mistranslate, or don’t translate at all,
your native language text will be used during the Moder-
ation).
• Avoid unnecessary movements during the experimental
examination and do not shake the walls of your box: the
laser experiment requires stability.
• Do not look into the laser beam or its reflections! It may
permanently damage your eyes.
• The first single sound signal tells you that there are 30
min of solving time left; the second double sound signal
means that 5 min is left; the third triple sound signal
marks the end of solving time. After the third sound
signal you must stop writing immediately. Put all
the papers into the envelope at your desk. You are not
allowed to take any sheet of paper out of the room.
If you have finished solving before the final sound signal,
please raise your flag.
— page 1 of 4 —
109
Problem E1. The magnetic permeability of water(10 points)
The effect of a magnetic field on most of the substancesbut ferromagnetics is rather weak. This is because the energydensity of the magetic field in substances of relative magneticpermeability µ is given by formula w = 1
2µµ0
B2, and typically,
µ is very close to 1. Still, with suitable experimental techniquessuch effects are firmly observable. In this problem we study theeffect of a magnetic field, created by a permanent neodymiummagnet, on water and use the results to calculate the magneticpermeability of water. You are not asked to estimate anyuncertainties throughout this problem and you do notneed to take into account the effects of the surface ten-sion.The setup comprises of 1 a stand (the highlighted numbers
correspond to the numbers in the fig.), 3 a digital caliper, 4
a laser pointer, 5 a dish with water and 7 a cylindricalpermanent magnet in it (the magnet is axially magnetised).The dish is fixed to the base of the stand by the magnet’s pull.The laser is fixed to the caliper, the base of which is fastenedto the stand; the caliper allows horizontal displacement of thelaser. The on-off button of the laser can be kept down withthe help of 13 the white conical tube. The depth of the wa-ter above the magnet should be reasonably close to 1 mm (ifshallower, the water surface becomes so curved that it will bedifficult to take readings from the screen). 15 A cup of water
and 16 a syringe can be used for the water level adjustment
(to raise the level by 1 mm, add 13 ml of water). 2 A sheet ofgraph paper (the “screen”) is to be fixed to the vertical plate
with 14 small magnetic tablets. If the laser spot on the screenbecomes smeared, check for a dust on the water surface (andblow away).
The remaining legend for the figure is as follows: 6 the
point where the laser beam hits the screen; 11 the LCD screen
of the caliper, 10 the button which switches the caliper units
between millimeters and inches; 8 on-off switch; 9 buttonfor setting the origin of the caliper reading. Beneath the laserpointer, there is one more button of the caliper, which tempor-arily re-sets the origin (if you pushed it inadvertently, push itonce again to return to the normal measuring mode).
Numerical values for your calculations: horizontal distancebetween the magnet’s centre and the screen L0 = 490 mm.Check (and adjust, if needed) the alignment of the centre ofthe magnet in two perpendicular directions. The vertical axisof the magnet must intersect with the laser beam, and it mustalso intersect with 12 the black line on the support plate.Magnetic induction at the magnet’s axis, at the height of1 mm from the flat surface B0 = 0.50 T; Density of waterρw = 1000 kg/m3; free-fall acceleration g = 9.8 m/s2; va-cuum permeability µ0 = 4π × 10−7 H/m.
WARNINGS:
⋄ The laser orientation is pre-adjusted, do not move it!
⋄ Do not look into the laser beam or its reflections!
⋄ Do not try to remove the strong neodymium magnet!
⋄ Do not put magnetic materials close to the magnet!
⋄ Turn off the laser when not used, batteries drain in 1 h!
Part A. Qualitative shape of the water surface (1 points)When a cylindrical magnet is placed below water surface, thelatter becomes curved. Observe, what is the shape of the watersurface above the magnet; according to this observation, decideif the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).
Write the letter corresponding to the correct option into theAnswer Sheet, together with an inequality µ > 1 or µ < 1. Forthis part, you don’t need to motivate your answer.
Part B. Exact shape of the water surface (7 points)
Curving of the water surface can be checked with high sensitiv-ity by measuring the reflection of the laser beam by the surface.We use this effect to calculate the dependence of the depth ofthe water on the horizontal position above the magnet.
i. ( 1.6 pts) Measure the dependence of the height y of thelaser spot on the screen on the caliper reading x (c.f. figure).You should cover all the usable range of caliper displacements.Fill the results into the table in the Answer Sheet.
ii. (0.7 pts) Draw the graph of the obtained dependence.
iii. ( 0.7 pts) Using the obtained graph, determine the angleα0 between the beam and the horizontal area of the watersurface.
iv. ( 1.4 pts) Please note that the slope (tan β) of the watersurface can be expressed as follows:
tan β ≈ β ≈cos2 α0
2·
y − y0 − (x − x0) tan α0
L0 + x − x0
,
where y0 is the height of the laser spot on the screen whenthe beam is reflected from the water surface at the axis of themagnet, and x0 — the respective position of the caliper.
Calculate the values of the water surface slope and fill theminto the Table on the Answer Sheet. Please note that it maybe possible to simplify your calculations if you substitute somecombination of terms in the expression from the previous taskwith a reading from the last graph.
v. ( 1.6 pts) Calculate the height of the water surface relativeto the surface far from the magnet as a function of x and fill itinto the Table on the Answer Sheet.
vi. (1 pt) Draw the graph of the latter dependence. Indicateon it the region where the beam hits the water surface directlyabove the magnet.
Part C. Magnetic permeability (2 points)
Using the results of Part B, calculate the value of µ − 1 (theso-called magnetic susceptibility), where µ is the relative mag-netic permeability of the water. Write your final formula andthe numerical result into the Answer Sheet.
— page 2 of 4 —
Problem E1. The magnetic permeability of water(10 points)
The effect of a magnetic field on most of the substancesbut ferromagnetics is rather weak. This is because the energydensity of the magetic field in substances of relative magneticpermeability µ is given by formula w = 1
2µµ0
B2, and typically,
µ is very close to 1. Still, with suitable experimental techniquessuch effects are firmly observable. In this problem we study theeffect of a magnetic field, created by a permanent neodymiummagnet, on water and use the results to calculate the magneticpermeability of water. You are not asked to estimate anyuncertainties throughout this problem and you do notneed to take into account the effects of the surface ten-sion.The setup comprises of 1 a stand (the highlighted numbers
correspond to the numbers in the fig.), 3 a digital caliper, 4
a laser pointer, 5 a dish with water and 7 a cylindricalpermanent magnet in it (the magnet is axially magnetised).The dish is fixed to the base of the stand by the magnet’s pull.The laser is fixed to the caliper, the base of which is fastenedto the stand; the caliper allows horizontal displacement of thelaser. The on-off button of the laser can be kept down withthe help of 13 the white conical tube. The depth of the wa-ter above the magnet should be reasonably close to 1 mm (ifshallower, the water surface becomes so curved that it will bedifficult to take readings from the screen). 15 A cup of water
and 16 a syringe can be used for the water level adjustment
(to raise the level by 1 mm, add 13 ml of water). 2 A sheet ofgraph paper (the “screen”) is to be fixed to the vertical plate
with 14 small magnetic tablets. If the laser spot on the screenbecomes smeared, check for a dust on the water surface (andblow away).
The remaining legend for the figure is as follows: 6 the
point where the laser beam hits the screen; 11 the LCD screen
of the caliper, 10 the button which switches the caliper units
between millimeters and inches; 8 on-off switch; 9 buttonfor setting the origin of the caliper reading. Beneath the laserpointer, there is one more button of the caliper, which tempor-arily re-sets the origin (if you pushed it inadvertently, push itonce again to return to the normal measuring mode).
Numerical values for your calculations: horizontal distancebetween the magnet’s centre and the screen L0 = 490 mm.Check (and adjust, if needed) the alignment of the centre ofthe magnet in two perpendicular directions. The vertical axisof the magnet must intersect with the laser beam, and it mustalso intersect with 12 the black line on the support plate.Magnetic induction at the magnet’s axis, at the height of1 mm from the flat surface B0 = 0.50 T; Density of waterρw = 1000 kg/m3; free-fall acceleration g = 9.8 m/s2; va-cuum permeability µ0 = 4π × 10−7 H/m.
WARNINGS:
⋄ The laser orientation is pre-adjusted, do not move it!
⋄ Do not look into the laser beam or its reflections!
⋄ Do not try to remove the strong neodymium magnet!
⋄ Do not put magnetic materials close to the magnet!
⋄ Turn off the laser when not used, batteries drain in 1 h!
Part A. Qualitative shape of the water surface (1 points)When a cylindrical magnet is placed below water surface, thelatter becomes curved. Observe, what is the shape of the watersurface above the magnet; according to this observation, decideif the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).
Write the letter corresponding to the correct option into theAnswer Sheet, together with an inequality µ > 1 or µ < 1. Forthis part, you don’t need to motivate your answer.
Part B. Exact shape of the water surface (7 points)
Curving of the water surface can be checked with high sensitiv-ity by measuring the reflection of the laser beam by the surface.We use this effect to calculate the dependence of the depth ofthe water on the horizontal position above the magnet.
i. ( 1.6 pts) Measure the dependence of the height y of thelaser spot on the screen on the caliper reading x (c.f. figure).You should cover all the usable range of caliper displacements.Fill the results into the table in the Answer Sheet.
ii. (0.7 pts) Draw the graph of the obtained dependence.
iii. ( 0.7 pts) Using the obtained graph, determine the angleα0 between the beam and the horizontal area of the watersurface.
iv. ( 1.4 pts) Please note that the slope (tan β) of the watersurface can be expressed as follows:
tan β ≈ β ≈cos2 α0
2·
y − y0 − (x − x0) tan α0
L0 + x − x0
,
where y0 is the height of the laser spot on the screen whenthe beam is reflected from the water surface at the axis of themagnet, and x0 — the respective position of the caliper.
Calculate the values of the water surface slope and fill theminto the Table on the Answer Sheet. Please note that it maybe possible to simplify your calculations if you substitute somecombination of terms in the expression from the previous taskwith a reading from the last graph.
v. ( 1.6 pts) Calculate the height of the water surface relativeto the surface far from the magnet as a function of x and fill itinto the Table on the Answer Sheet.
vi. (1 pt) Draw the graph of the latter dependence. Indicateon it the region where the beam hits the water surface directlyabove the magnet.
Part C. Magnetic permeability (2 points)
Using the results of Part B, calculate the value of µ − 1 (theso-called magnetic susceptibility), where µ is the relative mag-netic permeability of the water. Write your final formula andthe numerical result into the Answer Sheet.
— page 2 of 4 —
PrObLEm E1. ThE magnETIc PErmEabILITy OF waTEr (10 POInTs)
110
Problem E1. The magnetic permeability of water(10 points)
The effect of a magnetic field on most of the substancesbut ferromagnetics is rather weak. This is because the energydensity of the magetic field in substances of relative magneticpermeability µ is given by formula w = 1
2µµ0
B2, and typically,
µ is very close to 1. Still, with suitable experimental techniquessuch effects are firmly observable. In this problem we study theeffect of a magnetic field, created by a permanent neodymiummagnet, on water and use the results to calculate the magneticpermeability of water. You are not asked to estimate anyuncertainties throughout this problem and you do notneed to take into account the effects of the surface ten-sion.The setup comprises of 1 a stand (the highlighted numbers
correspond to the numbers in the fig.), 3 a digital caliper, 4
a laser pointer, 5 a dish with water and 7 a cylindricalpermanent magnet in it (the magnet is axially magnetised).The dish is fixed to the base of the stand by the magnet’s pull.The laser is fixed to the caliper, the base of which is fastenedto the stand; the caliper allows horizontal displacement of thelaser. The on-off button of the laser can be kept down withthe help of 13 the white conical tube. The depth of the wa-ter above the magnet should be reasonably close to 1 mm (ifshallower, the water surface becomes so curved that it will bedifficult to take readings from the screen). 15 A cup of water
and 16 a syringe can be used for the water level adjustment
(to raise the level by 1 mm, add 13 ml of water). 2 A sheet ofgraph paper (the “screen”) is to be fixed to the vertical plate
with 14 small magnetic tablets. If the laser spot on the screenbecomes smeared, check for a dust on the water surface (andblow away).
The remaining legend for the figure is as follows: 6 the
point where the laser beam hits the screen; 11 the LCD screen
of the caliper, 10 the button which switches the caliper units
between millimeters and inches; 8 on-off switch; 9 buttonfor setting the origin of the caliper reading. Beneath the laserpointer, there is one more button of the caliper, which tempor-arily re-sets the origin (if you pushed it inadvertently, push itonce again to return to the normal measuring mode).
Numerical values for your calculations: horizontal distancebetween the magnet’s centre and the screen L0 = 490 mm.Check (and adjust, if needed) the alignment of the centre ofthe magnet in two perpendicular directions. The vertical axisof the magnet must intersect with the laser beam, and it mustalso intersect with 12 the black line on the support plate.Magnetic induction at the magnet’s axis, at the height of1 mm from the flat surface B0 = 0.50 T; Density of waterρw = 1000 kg/m3; free-fall acceleration g = 9.8 m/s2; va-cuum permeability µ0 = 4π × 10−7 H/m.
WARNINGS:
⋄ The laser orientation is pre-adjusted, do not move it!
⋄ Do not look into the laser beam or its reflections!
⋄ Do not try to remove the strong neodymium magnet!
⋄ Do not put magnetic materials close to the magnet!
⋄ Turn off the laser when not used, batteries drain in 1 h!
Part A. Qualitative shape of the water surface (1 points)When a cylindrical magnet is placed below water surface, thelatter becomes curved. Observe, what is the shape of the watersurface above the magnet; according to this observation, decideif the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).
Write the letter corresponding to the correct option into theAnswer Sheet, together with an inequality µ > 1 or µ < 1. Forthis part, you don’t need to motivate your answer.
Part B. Exact shape of the water surface (7 points)
Curving of the water surface can be checked with high sensitiv-ity by measuring the reflection of the laser beam by the surface.We use this effect to calculate the dependence of the depth ofthe water on the horizontal position above the magnet.
i. ( 1.6 pts) Measure the dependence of the height y of thelaser spot on the screen on the caliper reading x (c.f. figure).You should cover all the usable range of caliper displacements.Fill the results into the table in the Answer Sheet.
ii. (0.7 pts) Draw the graph of the obtained dependence.
iii. ( 0.7 pts) Using the obtained graph, determine the angleα0 between the beam and the horizontal area of the watersurface.
iv. ( 1.4 pts) Please note that the slope (tan β) of the watersurface can be expressed as follows:
tan β ≈ β ≈cos2 α0
2·
y − y0 − (x − x0) tan α0
L0 + x − x0
,
where y0 is the height of the laser spot on the screen whenthe beam is reflected from the water surface at the axis of themagnet, and x0 — the respective position of the caliper.
Calculate the values of the water surface slope and fill theminto the Table on the Answer Sheet. Please note that it maybe possible to simplify your calculations if you substitute somecombination of terms in the expression from the previous taskwith a reading from the last graph.
v. ( 1.6 pts) Calculate the height of the water surface relativeto the surface far from the magnet as a function of x and fill itinto the Table on the Answer Sheet.
vi. (1 pt) Draw the graph of the latter dependence. Indicateon it the region where the beam hits the water surface directlyabove the magnet.
Part C. Magnetic permeability (2 points)
Using the results of Part B, calculate the value of µ − 1 (theso-called magnetic susceptibility), where µ is the relative mag-netic permeability of the water. Write your final formula andthe numerical result into the Answer Sheet.
— page 2 of 4 —
Problem E1. The magnetic permeability of water(10 points)
The effect of a magnetic field on most of the substancesbut ferromagnetics is rather weak. This is because the energydensity of the magetic field in substances of relative magneticpermeability µ is given by formula w = 1
2µµ0
B2, and typically,
µ is very close to 1. Still, with suitable experimental techniquessuch effects are firmly observable. In this problem we study theeffect of a magnetic field, created by a permanent neodymiummagnet, on water and use the results to calculate the magneticpermeability of water. You are not asked to estimate anyuncertainties throughout this problem and you do notneed to take into account the effects of the surface ten-sion.The setup comprises of 1 a stand (the highlighted numbers
correspond to the numbers in the fig.), 3 a digital caliper, 4
a laser pointer, 5 a dish with water and 7 a cylindricalpermanent magnet in it (the magnet is axially magnetised).The dish is fixed to the base of the stand by the magnet’s pull.The laser is fixed to the caliper, the base of which is fastenedto the stand; the caliper allows horizontal displacement of thelaser. The on-off button of the laser can be kept down withthe help of 13 the white conical tube. The depth of the wa-ter above the magnet should be reasonably close to 1 mm (ifshallower, the water surface becomes so curved that it will bedifficult to take readings from the screen). 15 A cup of water
and 16 a syringe can be used for the water level adjustment
(to raise the level by 1 mm, add 13 ml of water). 2 A sheet ofgraph paper (the “screen”) is to be fixed to the vertical plate
with 14 small magnetic tablets. If the laser spot on the screenbecomes smeared, check for a dust on the water surface (andblow away).
The remaining legend for the figure is as follows: 6 the
point where the laser beam hits the screen; 11 the LCD screen
of the caliper, 10 the button which switches the caliper units
between millimeters and inches; 8 on-off switch; 9 buttonfor setting the origin of the caliper reading. Beneath the laserpointer, there is one more button of the caliper, which tempor-arily re-sets the origin (if you pushed it inadvertently, push itonce again to return to the normal measuring mode).
Numerical values for your calculations: horizontal distancebetween the magnet’s centre and the screen L0 = 490 mm.Check (and adjust, if needed) the alignment of the centre ofthe magnet in two perpendicular directions. The vertical axisof the magnet must intersect with the laser beam, and it mustalso intersect with 12 the black line on the support plate.Magnetic induction at the magnet’s axis, at the height of1 mm from the flat surface B0 = 0.50 T; Density of waterρw = 1000 kg/m3; free-fall acceleration g = 9.8 m/s2; va-cuum permeability µ0 = 4π × 10−7 H/m.
WARNINGS:
⋄ The laser orientation is pre-adjusted, do not move it!
⋄ Do not look into the laser beam or its reflections!
⋄ Do not try to remove the strong neodymium magnet!
⋄ Do not put magnetic materials close to the magnet!
⋄ Turn off the laser when not used, batteries drain in 1 h!
Part A. Qualitative shape of the water surface (1 points)When a cylindrical magnet is placed below water surface, thelatter becomes curved. Observe, what is the shape of the watersurface above the magnet; according to this observation, decideif the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).
Write the letter corresponding to the correct option into theAnswer Sheet, together with an inequality µ > 1 or µ < 1. Forthis part, you don’t need to motivate your answer.
Part B. Exact shape of the water surface (7 points)
Curving of the water surface can be checked with high sensitiv-ity by measuring the reflection of the laser beam by the surface.We use this effect to calculate the dependence of the depth ofthe water on the horizontal position above the magnet.
i. ( 1.6 pts) Measure the dependence of the height y of thelaser spot on the screen on the caliper reading x (c.f. figure).You should cover all the usable range of caliper displacements.Fill the results into the table in the Answer Sheet.
ii. (0.7 pts) Draw the graph of the obtained dependence.
iii. ( 0.7 pts) Using the obtained graph, determine the angleα0 between the beam and the horizontal area of the watersurface.
iv. ( 1.4 pts) Please note that the slope (tan β) of the watersurface can be expressed as follows:
tan β ≈ β ≈cos2 α0
2·
y − y0 − (x − x0) tan α0
L0 + x − x0
,
where y0 is the height of the laser spot on the screen whenthe beam is reflected from the water surface at the axis of themagnet, and x0 — the respective position of the caliper.
Calculate the values of the water surface slope and fill theminto the Table on the Answer Sheet. Please note that it maybe possible to simplify your calculations if you substitute somecombination of terms in the expression from the previous taskwith a reading from the last graph.
v. ( 1.6 pts) Calculate the height of the water surface relativeto the surface far from the magnet as a function of x and fill itinto the Table on the Answer Sheet.
vi. (1 pt) Draw the graph of the latter dependence. Indicateon it the region where the beam hits the water surface directlyabove the magnet.
Part C. Magnetic permeability (2 points)
Using the results of Part B, calculate the value of µ − 1 (theso-called magnetic susceptibility), where µ is the relative mag-netic permeability of the water. Write your final formula andthe numerical result into the Answer Sheet.
— page 2 of 4 —
Problem E1. The magnetic permeability of water(10 points)
The effect of a magnetic field on most of the substancesbut ferromagnetics is rather weak. This is because the energydensity of the magetic field in substances of relative magneticpermeability µ is given by formula w = 1
2µµ0
B2, and typically,
µ is very close to 1. Still, with suitable experimental techniquessuch effects are firmly observable. In this problem we study theeffect of a magnetic field, created by a permanent neodymiummagnet, on water and use the results to calculate the magneticpermeability of water. You are not asked to estimate anyuncertainties throughout this problem and you do notneed to take into account the effects of the surface ten-sion.The setup comprises of 1 a stand (the highlighted numbers
correspond to the numbers in the fig.), 3 a digital caliper, 4
a laser pointer, 5 a dish with water and 7 a cylindricalpermanent magnet in it (the magnet is axially magnetised).The dish is fixed to the base of the stand by the magnet’s pull.The laser is fixed to the caliper, the base of which is fastenedto the stand; the caliper allows horizontal displacement of thelaser. The on-off button of the laser can be kept down withthe help of 13 the white conical tube. The depth of the wa-ter above the magnet should be reasonably close to 1 mm (ifshallower, the water surface becomes so curved that it will bedifficult to take readings from the screen). 15 A cup of water
and 16 a syringe can be used for the water level adjustment
(to raise the level by 1 mm, add 13 ml of water). 2 A sheet ofgraph paper (the “screen”) is to be fixed to the vertical plate
with 14 small magnetic tablets. If the laser spot on the screenbecomes smeared, check for a dust on the water surface (andblow away).
The remaining legend for the figure is as follows: 6 the
point where the laser beam hits the screen; 11 the LCD screen
of the caliper, 10 the button which switches the caliper units
between millimeters and inches; 8 on-off switch; 9 buttonfor setting the origin of the caliper reading. Beneath the laserpointer, there is one more button of the caliper, which tempor-arily re-sets the origin (if you pushed it inadvertently, push itonce again to return to the normal measuring mode).
Numerical values for your calculations: horizontal distancebetween the magnet’s centre and the screen L0 = 490 mm.Check (and adjust, if needed) the alignment of the centre ofthe magnet in two perpendicular directions. The vertical axisof the magnet must intersect with the laser beam, and it mustalso intersect with 12 the black line on the support plate.Magnetic induction at the magnet’s axis, at the height of1 mm from the flat surface B0 = 0.50 T; Density of waterρw = 1000 kg/m3; free-fall acceleration g = 9.8 m/s2; va-cuum permeability µ0 = 4π × 10−7 H/m.
WARNINGS:
⋄ The laser orientation is pre-adjusted, do not move it!
⋄ Do not look into the laser beam or its reflections!
⋄ Do not try to remove the strong neodymium magnet!
⋄ Do not put magnetic materials close to the magnet!
⋄ Turn off the laser when not used, batteries drain in 1 h!
Part A. Qualitative shape of the water surface (1 points)When a cylindrical magnet is placed below water surface, thelatter becomes curved. Observe, what is the shape of the watersurface above the magnet; according to this observation, decideif the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).
Write the letter corresponding to the correct option into theAnswer Sheet, together with an inequality µ > 1 or µ < 1. Forthis part, you don’t need to motivate your answer.
Part B. Exact shape of the water surface (7 points)
Curving of the water surface can be checked with high sensitiv-ity by measuring the reflection of the laser beam by the surface.We use this effect to calculate the dependence of the depth ofthe water on the horizontal position above the magnet.
i. ( 1.6 pts) Measure the dependence of the height y of thelaser spot on the screen on the caliper reading x (c.f. figure).You should cover all the usable range of caliper displacements.Fill the results into the table in the Answer Sheet.
ii. (0.7 pts) Draw the graph of the obtained dependence.
iii. ( 0.7 pts) Using the obtained graph, determine the angleα0 between the beam and the horizontal area of the watersurface.
iv. ( 1.4 pts) Please note that the slope (tan β) of the watersurface can be expressed as follows:
tan β ≈ β ≈cos2 α0
2·
y − y0 − (x − x0) tan α0
L0 + x − x0
,
where y0 is the height of the laser spot on the screen whenthe beam is reflected from the water surface at the axis of themagnet, and x0 — the respective position of the caliper.
Calculate the values of the water surface slope and fill theminto the Table on the Answer Sheet. Please note that it maybe possible to simplify your calculations if you substitute somecombination of terms in the expression from the previous taskwith a reading from the last graph.
v. ( 1.6 pts) Calculate the height of the water surface relativeto the surface far from the magnet as a function of x and fill itinto the Table on the Answer Sheet.
vi. (1 pt) Draw the graph of the latter dependence. Indicateon it the region where the beam hits the water surface directlyabove the magnet.
Part C. Magnetic permeability (2 points)
Using the results of Part B, calculate the value of µ − 1 (theso-called magnetic susceptibility), where µ is the relative mag-netic permeability of the water. Write your final formula andthe numerical result into the Answer Sheet.
— page 2 of 4 —
Problem E1. The magnetic permeability of water(10 points)
The effect of a magnetic field on most of the substancesbut ferromagnetics is rather weak. This is because the energydensity of the magetic field in substances of relative magneticpermeability µ is given by formula w = 1
2µµ0
B2, and typically,
µ is very close to 1. Still, with suitable experimental techniquessuch effects are firmly observable. In this problem we study theeffect of a magnetic field, created by a permanent neodymiummagnet, on water and use the results to calculate the magneticpermeability of water. You are not asked to estimate anyuncertainties throughout this problem and you do notneed to take into account the effects of the surface ten-sion.The setup comprises of 1 a stand (the highlighted numbers
correspond to the numbers in the fig.), 3 a digital caliper, 4
a laser pointer, 5 a dish with water and 7 a cylindricalpermanent magnet in it (the magnet is axially magnetised).The dish is fixed to the base of the stand by the magnet’s pull.The laser is fixed to the caliper, the base of which is fastenedto the stand; the caliper allows horizontal displacement of thelaser. The on-off button of the laser can be kept down withthe help of 13 the white conical tube. The depth of the wa-ter above the magnet should be reasonably close to 1 mm (ifshallower, the water surface becomes so curved that it will bedifficult to take readings from the screen). 15 A cup of water
and 16 a syringe can be used for the water level adjustment
(to raise the level by 1 mm, add 13 ml of water). 2 A sheet ofgraph paper (the “screen”) is to be fixed to the vertical plate
with 14 small magnetic tablets. If the laser spot on the screenbecomes smeared, check for a dust on the water surface (andblow away).
The remaining legend for the figure is as follows: 6 the
point where the laser beam hits the screen; 11 the LCD screen
of the caliper, 10 the button which switches the caliper units
between millimeters and inches; 8 on-off switch; 9 buttonfor setting the origin of the caliper reading. Beneath the laserpointer, there is one more button of the caliper, which tempor-arily re-sets the origin (if you pushed it inadvertently, push itonce again to return to the normal measuring mode).
Numerical values for your calculations: horizontal distancebetween the magnet’s centre and the screen L0 = 490 mm.Check (and adjust, if needed) the alignment of the centre ofthe magnet in two perpendicular directions. The vertical axisof the magnet must intersect with the laser beam, and it mustalso intersect with 12 the black line on the support plate.Magnetic induction at the magnet’s axis, at the height of1 mm from the flat surface B0 = 0.50 T; Density of waterρw = 1000 kg/m3; free-fall acceleration g = 9.8 m/s2; va-cuum permeability µ0 = 4π × 10−7 H/m.
WARNINGS:
⋄ The laser orientation is pre-adjusted, do not move it!
⋄ Do not look into the laser beam or its reflections!
⋄ Do not try to remove the strong neodymium magnet!
⋄ Do not put magnetic materials close to the magnet!
⋄ Turn off the laser when not used, batteries drain in 1 h!
Part A. Qualitative shape of the water surface (1 points)When a cylindrical magnet is placed below water surface, thelatter becomes curved. Observe, what is the shape of the watersurface above the magnet; according to this observation, decideif the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).
Write the letter corresponding to the correct option into theAnswer Sheet, together with an inequality µ > 1 or µ < 1. Forthis part, you don’t need to motivate your answer.
Part B. Exact shape of the water surface (7 points)
Curving of the water surface can be checked with high sensitiv-ity by measuring the reflection of the laser beam by the surface.We use this effect to calculate the dependence of the depth ofthe water on the horizontal position above the magnet.
i. ( 1.6 pts) Measure the dependence of the height y of thelaser spot on the screen on the caliper reading x (c.f. figure).You should cover all the usable range of caliper displacements.Fill the results into the table in the Answer Sheet.
ii. (0.7 pts) Draw the graph of the obtained dependence.
iii. ( 0.7 pts) Using the obtained graph, determine the angleα0 between the beam and the horizontal area of the watersurface.
iv. ( 1.4 pts) Please note that the slope (tan β) of the watersurface can be expressed as follows:
tan β ≈ β ≈cos2 α0
2·
y − y0 − (x − x0) tan α0
L0 + x − x0
,
where y0 is the height of the laser spot on the screen whenthe beam is reflected from the water surface at the axis of themagnet, and x0 — the respective position of the caliper.
Calculate the values of the water surface slope and fill theminto the Table on the Answer Sheet. Please note that it maybe possible to simplify your calculations if you substitute somecombination of terms in the expression from the previous taskwith a reading from the last graph.
v. ( 1.6 pts) Calculate the height of the water surface relativeto the surface far from the magnet as a function of x and fill itinto the Table on the Answer Sheet.
vi. (1 pt) Draw the graph of the latter dependence. Indicateon it the region where the beam hits the water surface directlyabove the magnet.
Part C. Magnetic permeability (2 points)
Using the results of Part B, calculate the value of µ − 1 (theso-called magnetic susceptibility), where µ is the relative mag-netic permeability of the water. Write your final formula andthe numerical result into the Answer Sheet.
— page 2 of 4 —
111
Problem E1. The magnetic permeability of water(10 points)
The effect of a magnetic field on most of the substancesbut ferromagnetics is rather weak. This is because the energydensity of the magetic field in substances of relative magneticpermeability µ is given by formula w = 1
2µµ0
B2, and typically,
µ is very close to 1. Still, with suitable experimental techniquessuch effects are firmly observable. In this problem we study theeffect of a magnetic field, created by a permanent neodymiummagnet, on water and use the results to calculate the magneticpermeability of water. You are not asked to estimate anyuncertainties throughout this problem and you do notneed to take into account the effects of the surface ten-sion.The setup comprises of 1 a stand (the highlighted numbers
correspond to the numbers in the fig.), 3 a digital caliper, 4
a laser pointer, 5 a dish with water and 7 a cylindricalpermanent magnet in it (the magnet is axially magnetised).The dish is fixed to the base of the stand by the magnet’s pull.The laser is fixed to the caliper, the base of which is fastenedto the stand; the caliper allows horizontal displacement of thelaser. The on-off button of the laser can be kept down withthe help of 13 the white conical tube. The depth of the wa-ter above the magnet should be reasonably close to 1 mm (ifshallower, the water surface becomes so curved that it will bedifficult to take readings from the screen). 15 A cup of water
and 16 a syringe can be used for the water level adjustment
(to raise the level by 1 mm, add 13 ml of water). 2 A sheet ofgraph paper (the “screen”) is to be fixed to the vertical plate
with 14 small magnetic tablets. If the laser spot on the screenbecomes smeared, check for a dust on the water surface (andblow away).
The remaining legend for the figure is as follows: 6 the
point where the laser beam hits the screen; 11 the LCD screen
of the caliper, 10 the button which switches the caliper units
between millimeters and inches; 8 on-off switch; 9 buttonfor setting the origin of the caliper reading. Beneath the laserpointer, there is one more button of the caliper, which tempor-arily re-sets the origin (if you pushed it inadvertently, push itonce again to return to the normal measuring mode).
Numerical values for your calculations: horizontal distancebetween the magnet’s centre and the screen L0 = 490 mm.Check (and adjust, if needed) the alignment of the centre ofthe magnet in two perpendicular directions. The vertical axisof the magnet must intersect with the laser beam, and it mustalso intersect with 12 the black line on the support plate.Magnetic induction at the magnet’s axis, at the height of1 mm from the flat surface B0 = 0.50 T; Density of waterρw = 1000 kg/m3; free-fall acceleration g = 9.8 m/s2; va-cuum permeability µ0 = 4π × 10−7 H/m.
WARNINGS:
⋄ The laser orientation is pre-adjusted, do not move it!
⋄ Do not look into the laser beam or its reflections!
⋄ Do not try to remove the strong neodymium magnet!
⋄ Do not put magnetic materials close to the magnet!
⋄ Turn off the laser when not used, batteries drain in 1 h!
Part A. Qualitative shape of the water surface (1 points)When a cylindrical magnet is placed below water surface, thelatter becomes curved. Observe, what is the shape of the watersurface above the magnet; according to this observation, decideif the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).
Write the letter corresponding to the correct option into theAnswer Sheet, together with an inequality µ > 1 or µ < 1. Forthis part, you don’t need to motivate your answer.
Part B. Exact shape of the water surface (7 points)
Curving of the water surface can be checked with high sensitiv-ity by measuring the reflection of the laser beam by the surface.We use this effect to calculate the dependence of the depth ofthe water on the horizontal position above the magnet.
i. ( 1.6 pts) Measure the dependence of the height y of thelaser spot on the screen on the caliper reading x (c.f. figure).You should cover all the usable range of caliper displacements.Fill the results into the table in the Answer Sheet.
ii. (0.7 pts) Draw the graph of the obtained dependence.
iii. ( 0.7 pts) Using the obtained graph, determine the angleα0 between the beam and the horizontal area of the watersurface.
iv. ( 1.4 pts) Please note that the slope (tan β) of the watersurface can be expressed as follows:
tan β ≈ β ≈cos2 α0
2·
y − y0 − (x − x0) tan α0
L0 + x − x0
,
where y0 is the height of the laser spot on the screen whenthe beam is reflected from the water surface at the axis of themagnet, and x0 — the respective position of the caliper.
Calculate the values of the water surface slope and fill theminto the Table on the Answer Sheet. Please note that it maybe possible to simplify your calculations if you substitute somecombination of terms in the expression from the previous taskwith a reading from the last graph.
v. ( 1.6 pts) Calculate the height of the water surface relativeto the surface far from the magnet as a function of x and fill itinto the Table on the Answer Sheet.
vi. (1 pt) Draw the graph of the latter dependence. Indicateon it the region where the beam hits the water surface directlyabove the magnet.
Part C. Magnetic permeability (2 points)
Using the results of Part B, calculate the value of µ − 1 (theso-called magnetic susceptibility), where µ is the relative mag-netic permeability of the water. Write your final formula andthe numerical result into the Answer Sheet.
— page 2 of 4 —
112
Problem E2. Nonlinear Black Box (10 points)In simple problems, electrical circuits are assumed to consistof linear elements, for which electrical characteristics are dir-ectly proportional to each other. Examples include resistance(V = RI), capacitance (Q = CV ) and inductance (V = LI),where R, C and L are constants. In this problem, however, weexamine a circuit containing nonlinear elements, enclosed intoa black box, for which the assumption of proportionality nolonger holds.The setup comprises of a multimeter (labelled “IPhO-measure”), a current source, a black box containing non-linear elements, and four test leads with stackable connectorsfor wiring. Be careful not to break the seal on the black box.
The multimeter can measure current and voltage simultan-eously. You can record with it up to 2000 data points, eachconsisting of: voltage V , current I, power P = IV , resistanceR = V/I, voltage time-derivative V , current time-derivative Iand time t. See its manual for details. If you go beyond 2000data points, the oldest data will be overwritten.
OUTIN
Multimeter
GND
A V
The constant current source supplies stable current as longas the voltage across its terminals stays between −0.6125Vand 0.6125V. When switched off, the constant current sourcebehaves as a large (essentially infinite) resistance.
Current source
+
-
I=6mA
U=-612.5mV...612.5mV
The black box contains an electric double layer capacitor(which is a slightly nonlinear high capacitance capacitor), anunknown nonlinear element and an inductor L = 10µH of negli-gible resistance, switchable as indicated on the circuit diagram.The nonlinear element can be considered as a resistance with anonlinear dependence between the voltage and the current [ Iis a continuous function of V with I(0) = 0]. Likewise, for thecapacitor, the differential capacitance C(V ) = dQ/dV is notexactly constant. We say that the voltage on the blackbox is positive when the potential on its red terminal ishigher than the potential on the black terminal. Pos-
itive voltage will be acquired when the terminals ofmatching colours on the black box and the currentsource are connected (you are allowed to use negativevoltages).
Black box
Nonlinear
device
+
-C(V)
Here it is safe to discharge the capacitor in the black box byshorting its inputs either by itself or through the IN and OUTterminals on multimeter: the internal resistance of this capa-citor is enough to keep the current from damaging anything.
You are not asked to estimate any uncertaintiesthroughout this problem.
Part A. Circuit without inductance (7 points)In this part, keep the switch on the black box closed (push “I”down), so that the inductance is shorted.
Please note that some measurements may take considerabletime, therefore it is recommended to read through all the tasksof part A to avoid unnecessary work.i. (1 pt) Confirm that the output current of the current sourceis approximately 6 mA, and determine the range within whichit varies for voltages between 0 and +480 mV. Document thecircuit diagram.ii. (1.2 pts) Show that the differential capacitance C(V ) usedin the black box is approximately 2 F by measuring its valuefor a single voltage of your choice C(V0) = C0. Document thecircuit diagram.iii. (2.2 pts) Neglecting the nonlinearity of the capacitance[C(V ) ≈ C0], determine the current–voltage characteristic ofthe nonlinear element used in the black box. Plot the I(V )curve for obtainable positive voltages on the black box ontothe answer sheet. Document the circuit diagram.iv. (2.6 pts) Using measurements taken from the whole rangeof obtainable voltages, calculate and plot the C(V ) curve forobtainable positive voltages on the black box to the answersheet. Write down the minimal and maximal values of differ-ential capacitance Cmin, Cmax. Document the circuit diagram.
Part B. Circuit with inductance (3 points)Enable the inductance by opening the switch on the black box(push “0” down). Using the same method as in pt. A-iii, meas-ure and plot the current–voltage characteristic of the nonlin-ear element. Describe any significant differences between thecurves of parts A and B and suggest a reason using qualitativearguments.
Here you need to know that the nonlinear element has actu-ally also a capacitance (ca 1 nF) which is connected in parallelto the nonlinear resistance.
— page 3 of 4 —
Problem E2. Nonlinear Black Box (10 points)In simple problems, electrical circuits are assumed to consistof linear elements, for which electrical characteristics are dir-ectly proportional to each other. Examples include resistance(V = RI), capacitance (Q = CV ) and inductance (V = LI),where R, C and L are constants. In this problem, however, weexamine a circuit containing nonlinear elements, enclosed intoa black box, for which the assumption of proportionality nolonger holds.The setup comprises of a multimeter (labelled “IPhO-measure”), a current source, a black box containing non-linear elements, and four test leads with stackable connectorsfor wiring. Be careful not to break the seal on the black box.
The multimeter can measure current and voltage simultan-eously. You can record with it up to 2000 data points, eachconsisting of: voltage V , current I, power P = IV , resistanceR = V/I, voltage time-derivative V , current time-derivative Iand time t. See its manual for details. If you go beyond 2000data points, the oldest data will be overwritten.
OUTIN
Multimeter
GND
A V
The constant current source supplies stable current as longas the voltage across its terminals stays between −0.6125Vand 0.6125V. When switched off, the constant current sourcebehaves as a large (essentially infinite) resistance.
Current source
+
-
I=6mA
U=-612.5mV...612.5mV
The black box contains an electric double layer capacitor(which is a slightly nonlinear high capacitance capacitor), anunknown nonlinear element and an inductor L = 10µH of negli-gible resistance, switchable as indicated on the circuit diagram.The nonlinear element can be considered as a resistance with anonlinear dependence between the voltage and the current [ Iis a continuous function of V with I(0) = 0]. Likewise, for thecapacitor, the differential capacitance C(V ) = dQ/dV is notexactly constant. We say that the voltage on the blackbox is positive when the potential on its red terminal ishigher than the potential on the black terminal. Pos-
itive voltage will be acquired when the terminals ofmatching colours on the black box and the currentsource are connected (you are allowed to use negativevoltages).
Black box
Nonlinear
device
+
-C(V)
Here it is safe to discharge the capacitor in the black box byshorting its inputs either by itself or through the IN and OUTterminals on multimeter: the internal resistance of this capa-citor is enough to keep the current from damaging anything.
You are not asked to estimate any uncertaintiesthroughout this problem.
Part A. Circuit without inductance (7 points)In this part, keep the switch on the black box closed (push “I”down), so that the inductance is shorted.
Please note that some measurements may take considerabletime, therefore it is recommended to read through all the tasksof part A to avoid unnecessary work.i. (1 pt) Confirm that the output current of the current sourceis approximately 6 mA, and determine the range within whichit varies for voltages between 0 and +480 mV. Document thecircuit diagram.ii. (1.2 pts) Show that the differential capacitance C(V ) usedin the black box is approximately 2 F by measuring its valuefor a single voltage of your choice C(V0) = C0. Document thecircuit diagram.iii. (2.2 pts) Neglecting the nonlinearity of the capacitance[C(V ) ≈ C0], determine the current–voltage characteristic ofthe nonlinear element used in the black box. Plot the I(V )curve for obtainable positive voltages on the black box ontothe answer sheet. Document the circuit diagram.iv. (2.6 pts) Using measurements taken from the whole rangeof obtainable voltages, calculate and plot the C(V ) curve forobtainable positive voltages on the black box to the answersheet. Write down the minimal and maximal values of differ-ential capacitance Cmin, Cmax. Document the circuit diagram.
Part B. Circuit with inductance (3 points)Enable the inductance by opening the switch on the black box(push “0” down). Using the same method as in pt. A-iii, meas-ure and plot the current–voltage characteristic of the nonlin-ear element. Describe any significant differences between thecurves of parts A and B and suggest a reason using qualitativearguments.
Here you need to know that the nonlinear element has actu-ally also a capacitance (ca 1 nF) which is connected in parallelto the nonlinear resistance.
— page 3 of 4 —
PrObLEm E2. nOnLInEar bLack bOx (10 POInTs)
OUTIN
Multimeter
GND
A V
113
Problem E2. Nonlinear Black Box (10 points)In simple problems, electrical circuits are assumed to consistof linear elements, for which electrical characteristics are dir-ectly proportional to each other. Examples include resistance(V = RI), capacitance (Q = CV ) and inductance (V = LI),where R, C and L are constants. In this problem, however, weexamine a circuit containing nonlinear elements, enclosed intoa black box, for which the assumption of proportionality nolonger holds.The setup comprises of a multimeter (labelled “IPhO-measure”), a current source, a black box containing non-linear elements, and four test leads with stackable connectorsfor wiring. Be careful not to break the seal on the black box.
The multimeter can measure current and voltage simultan-eously. You can record with it up to 2000 data points, eachconsisting of: voltage V , current I, power P = IV , resistanceR = V/I, voltage time-derivative V , current time-derivative Iand time t. See its manual for details. If you go beyond 2000data points, the oldest data will be overwritten.
OUTIN
Multimeter
GND
A V
The constant current source supplies stable current as longas the voltage across its terminals stays between −0.6125Vand 0.6125V. When switched off, the constant current sourcebehaves as a large (essentially infinite) resistance.
Current source
+
-
I=6mA
U=-612.5mV...612.5mV
The black box contains an electric double layer capacitor(which is a slightly nonlinear high capacitance capacitor), anunknown nonlinear element and an inductor L = 10µH of negli-gible resistance, switchable as indicated on the circuit diagram.The nonlinear element can be considered as a resistance with anonlinear dependence between the voltage and the current [ Iis a continuous function of V with I(0) = 0]. Likewise, for thecapacitor, the differential capacitance C(V ) = dQ/dV is notexactly constant. We say that the voltage on the blackbox is positive when the potential on its red terminal ishigher than the potential on the black terminal. Pos-
itive voltage will be acquired when the terminals ofmatching colours on the black box and the currentsource are connected (you are allowed to use negativevoltages).
Black box
Nonlinear
device
+
-C(V)
Here it is safe to discharge the capacitor in the black box byshorting its inputs either by itself or through the IN and OUTterminals on multimeter: the internal resistance of this capa-citor is enough to keep the current from damaging anything.
You are not asked to estimate any uncertaintiesthroughout this problem.
Part A. Circuit without inductance (7 points)In this part, keep the switch on the black box closed (push “I”down), so that the inductance is shorted.
Please note that some measurements may take considerabletime, therefore it is recommended to read through all the tasksof part A to avoid unnecessary work.i. (1 pt) Confirm that the output current of the current sourceis approximately 6 mA, and determine the range within whichit varies for voltages between 0 and +480 mV. Document thecircuit diagram.ii. (1.2 pts) Show that the differential capacitance C(V ) usedin the black box is approximately 2 F by measuring its valuefor a single voltage of your choice C(V0) = C0. Document thecircuit diagram.iii. (2.2 pts) Neglecting the nonlinearity of the capacitance[C(V ) ≈ C0], determine the current–voltage characteristic ofthe nonlinear element used in the black box. Plot the I(V )curve for obtainable positive voltages on the black box ontothe answer sheet. Document the circuit diagram.iv. (2.6 pts) Using measurements taken from the whole rangeof obtainable voltages, calculate and plot the C(V ) curve forobtainable positive voltages on the black box to the answersheet. Write down the minimal and maximal values of differ-ential capacitance Cmin, Cmax. Document the circuit diagram.
Part B. Circuit with inductance (3 points)Enable the inductance by opening the switch on the black box(push “0” down). Using the same method as in pt. A-iii, meas-ure and plot the current–voltage characteristic of the nonlin-ear element. Describe any significant differences between thecurves of parts A and B and suggest a reason using qualitativearguments.
Here you need to know that the nonlinear element has actu-ally also a capacitance (ca 1 nF) which is connected in parallelto the nonlinear resistance.
— page 3 of 4 —
Problem E2. Nonlinear Black Box (10 points)In simple problems, electrical circuits are assumed to consistof linear elements, for which electrical characteristics are dir-ectly proportional to each other. Examples include resistance(V = RI), capacitance (Q = CV ) and inductance (V = LI),where R, C and L are constants. In this problem, however, weexamine a circuit containing nonlinear elements, enclosed intoa black box, for which the assumption of proportionality nolonger holds.The setup comprises of a multimeter (labelled “IPhO-measure”), a current source, a black box containing non-linear elements, and four test leads with stackable connectorsfor wiring. Be careful not to break the seal on the black box.
The multimeter can measure current and voltage simultan-eously. You can record with it up to 2000 data points, eachconsisting of: voltage V , current I, power P = IV , resistanceR = V/I, voltage time-derivative V , current time-derivative Iand time t. See its manual for details. If you go beyond 2000data points, the oldest data will be overwritten.
OUTIN
Multimeter
GND
A V
The constant current source supplies stable current as longas the voltage across its terminals stays between −0.6125Vand 0.6125V. When switched off, the constant current sourcebehaves as a large (essentially infinite) resistance.
Current source
+
-
I=6mA
U=-612.5mV...612.5mV
The black box contains an electric double layer capacitor(which is a slightly nonlinear high capacitance capacitor), anunknown nonlinear element and an inductor L = 10µH of negli-gible resistance, switchable as indicated on the circuit diagram.The nonlinear element can be considered as a resistance with anonlinear dependence between the voltage and the current [ Iis a continuous function of V with I(0) = 0]. Likewise, for thecapacitor, the differential capacitance C(V ) = dQ/dV is notexactly constant. We say that the voltage on the blackbox is positive when the potential on its red terminal ishigher than the potential on the black terminal. Pos-
itive voltage will be acquired when the terminals ofmatching colours on the black box and the currentsource are connected (you are allowed to use negativevoltages).
Black box
Nonlinear
device
+
-C(V)
Here it is safe to discharge the capacitor in the black box byshorting its inputs either by itself or through the IN and OUTterminals on multimeter: the internal resistance of this capa-citor is enough to keep the current from damaging anything.
You are not asked to estimate any uncertaintiesthroughout this problem.
Part A. Circuit without inductance (7 points)In this part, keep the switch on the black box closed (push “I”down), so that the inductance is shorted.
Please note that some measurements may take considerabletime, therefore it is recommended to read through all the tasksof part A to avoid unnecessary work.i. (1 pt) Confirm that the output current of the current sourceis approximately 6 mA, and determine the range within whichit varies for voltages between 0 and +480 mV. Document thecircuit diagram.ii. (1.2 pts) Show that the differential capacitance C(V ) usedin the black box is approximately 2 F by measuring its valuefor a single voltage of your choice C(V0) = C0. Document thecircuit diagram.iii. (2.2 pts) Neglecting the nonlinearity of the capacitance[C(V ) ≈ C0], determine the current–voltage characteristic ofthe nonlinear element used in the black box. Plot the I(V )curve for obtainable positive voltages on the black box ontothe answer sheet. Document the circuit diagram.iv. (2.6 pts) Using measurements taken from the whole rangeof obtainable voltages, calculate and plot the C(V ) curve forobtainable positive voltages on the black box to the answersheet. Write down the minimal and maximal values of differ-ential capacitance Cmin, Cmax. Document the circuit diagram.
Part B. Circuit with inductance (3 points)Enable the inductance by opening the switch on the black box(push “0” down). Using the same method as in pt. A-iii, meas-ure and plot the current–voltage characteristic of the nonlin-ear element. Describe any significant differences between thecurves of parts A and B and suggest a reason using qualitativearguments.
Here you need to know that the nonlinear element has actu-ally also a capacitance (ca 1 nF) which is connected in parallelto the nonlinear resistance.
— page 3 of 4 —
Problem E2. Nonlinear Black Box (10 points)In simple problems, electrical circuits are assumed to consistof linear elements, for which electrical characteristics are dir-ectly proportional to each other. Examples include resistance(V = RI), capacitance (Q = CV ) and inductance (V = LI),where R, C and L are constants. In this problem, however, weexamine a circuit containing nonlinear elements, enclosed intoa black box, for which the assumption of proportionality nolonger holds.The setup comprises of a multimeter (labelled “IPhO-measure”), a current source, a black box containing non-linear elements, and four test leads with stackable connectorsfor wiring. Be careful not to break the seal on the black box.
The multimeter can measure current and voltage simultan-eously. You can record with it up to 2000 data points, eachconsisting of: voltage V , current I, power P = IV , resistanceR = V/I, voltage time-derivative V , current time-derivative Iand time t. See its manual for details. If you go beyond 2000data points, the oldest data will be overwritten.
OUTIN
Multimeter
GND
A V
The constant current source supplies stable current as longas the voltage across its terminals stays between −0.6125Vand 0.6125V. When switched off, the constant current sourcebehaves as a large (essentially infinite) resistance.
Current source
+
-
I=6mA
U=-612.5mV...612.5mV
The black box contains an electric double layer capacitor(which is a slightly nonlinear high capacitance capacitor), anunknown nonlinear element and an inductor L = 10µH of negli-gible resistance, switchable as indicated on the circuit diagram.The nonlinear element can be considered as a resistance with anonlinear dependence between the voltage and the current [ Iis a continuous function of V with I(0) = 0]. Likewise, for thecapacitor, the differential capacitance C(V ) = dQ/dV is notexactly constant. We say that the voltage on the blackbox is positive when the potential on its red terminal ishigher than the potential on the black terminal. Pos-
itive voltage will be acquired when the terminals ofmatching colours on the black box and the currentsource are connected (you are allowed to use negativevoltages).
Black box
Nonlinear
device
+
-C(V)
Here it is safe to discharge the capacitor in the black box byshorting its inputs either by itself or through the IN and OUTterminals on multimeter: the internal resistance of this capa-citor is enough to keep the current from damaging anything.
You are not asked to estimate any uncertaintiesthroughout this problem.
Part A. Circuit without inductance (7 points)In this part, keep the switch on the black box closed (push “I”down), so that the inductance is shorted.
Please note that some measurements may take considerabletime, therefore it is recommended to read through all the tasksof part A to avoid unnecessary work.i. (1 pt) Confirm that the output current of the current sourceis approximately 6 mA, and determine the range within whichit varies for voltages between 0 and +480 mV. Document thecircuit diagram.ii. (1.2 pts) Show that the differential capacitance C(V ) usedin the black box is approximately 2 F by measuring its valuefor a single voltage of your choice C(V0) = C0. Document thecircuit diagram.iii. (2.2 pts) Neglecting the nonlinearity of the capacitance[C(V ) ≈ C0], determine the current–voltage characteristic ofthe nonlinear element used in the black box. Plot the I(V )curve for obtainable positive voltages on the black box ontothe answer sheet. Document the circuit diagram.iv. (2.6 pts) Using measurements taken from the whole rangeof obtainable voltages, calculate and plot the C(V ) curve forobtainable positive voltages on the black box to the answersheet. Write down the minimal and maximal values of differ-ential capacitance Cmin, Cmax. Document the circuit diagram.
Part B. Circuit with inductance (3 points)Enable the inductance by opening the switch on the black box(push “0” down). Using the same method as in pt. A-iii, meas-ure and plot the current–voltage characteristic of the nonlin-ear element. Describe any significant differences between thecurves of parts A and B and suggest a reason using qualitativearguments.
Here you need to know that the nonlinear element has actu-ally also a capacitance (ca 1 nF) which is connected in parallelto the nonlinear resistance.
— page 3 of 4 —
PrObLEm E2. nOnLInEar bLack bOx (10 POInTs)
114
Problem E2. Nonlinear Black Box (10 points)In simple problems, electrical circuits are assumed to consistof linear elements, for which electrical characteristics are dir-ectly proportional to each other. Examples include resistance(V = RI), capacitance (Q = CV ) and inductance (V = LI),where R, C and L are constants. In this problem, however, weexamine a circuit containing nonlinear elements, enclosed intoa black box, for which the assumption of proportionality nolonger holds.The setup comprises of a multimeter (labelled “IPhO-measure”), a current source, a black box containing non-linear elements, and four test leads with stackable connectorsfor wiring. Be careful not to break the seal on the black box.
The multimeter can measure current and voltage simultan-eously. You can record with it up to 2000 data points, eachconsisting of: voltage V , current I, power P = IV , resistanceR = V/I, voltage time-derivative V , current time-derivative Iand time t. See its manual for details. If you go beyond 2000data points, the oldest data will be overwritten.
OUTIN
Multimeter
GND
A V
The constant current source supplies stable current as longas the voltage across its terminals stays between −0.6125Vand 0.6125V. When switched off, the constant current sourcebehaves as a large (essentially infinite) resistance.
Current source
+
-
I=6mA
U=-612.5mV...612.5mV
The black box contains an electric double layer capacitor(which is a slightly nonlinear high capacitance capacitor), anunknown nonlinear element and an inductor L = 10µH of negli-gible resistance, switchable as indicated on the circuit diagram.The nonlinear element can be considered as a resistance with anonlinear dependence between the voltage and the current [ Iis a continuous function of V with I(0) = 0]. Likewise, for thecapacitor, the differential capacitance C(V ) = dQ/dV is notexactly constant. We say that the voltage on the blackbox is positive when the potential on its red terminal ishigher than the potential on the black terminal. Pos-
itive voltage will be acquired when the terminals ofmatching colours on the black box and the currentsource are connected (you are allowed to use negativevoltages).
Black box
Nonlinear
device
+
-C(V)
Here it is safe to discharge the capacitor in the black box byshorting its inputs either by itself or through the IN and OUTterminals on multimeter: the internal resistance of this capa-citor is enough to keep the current from damaging anything.
You are not asked to estimate any uncertaintiesthroughout this problem.
Part A. Circuit without inductance (7 points)In this part, keep the switch on the black box closed (push “I”down), so that the inductance is shorted.
Please note that some measurements may take considerabletime, therefore it is recommended to read through all the tasksof part A to avoid unnecessary work.i. (1 pt) Confirm that the output current of the current sourceis approximately 6 mA, and determine the range within whichit varies for voltages between 0 and +480 mV. Document thecircuit diagram.ii. (1.2 pts) Show that the differential capacitance C(V ) usedin the black box is approximately 2 F by measuring its valuefor a single voltage of your choice C(V0) = C0. Document thecircuit diagram.iii. (2.2 pts) Neglecting the nonlinearity of the capacitance[C(V ) ≈ C0], determine the current–voltage characteristic ofthe nonlinear element used in the black box. Plot the I(V )curve for obtainable positive voltages on the black box ontothe answer sheet. Document the circuit diagram.iv. (2.6 pts) Using measurements taken from the whole rangeof obtainable voltages, calculate and plot the C(V ) curve forobtainable positive voltages on the black box to the answersheet. Write down the minimal and maximal values of differ-ential capacitance Cmin, Cmax. Document the circuit diagram.
Part B. Circuit with inductance (3 points)Enable the inductance by opening the switch on the black box(push “0” down). Using the same method as in pt. A-iii, meas-ure and plot the current–voltage characteristic of the nonlin-ear element. Describe any significant differences between thecurves of parts A and B and suggest a reason using qualitativearguments.
Here you need to know that the nonlinear element has actu-ally also a capacitance (ca 1 nF) which is connected in parallelto the nonlinear resistance.
— page 3 of 4 —
IPhO-measure: short manualIPhO-measure is a multimeter capable of measuring voltage Vand current I simultaneously. It also records their time derivat-ives V and I, their product P = V I, ratio R = V/I, and timet of the sample. Stored measurements are organized into sep-arate sets; every stored sample is numbered by the set numbers and a counter n inside the set. All saved samples are writtento an internal flash memory and can later be retrieved.
Electrical behaviourThe device behaves as an ammeter and a voltmeter connectedas follows. OUTIN
Multimeter
GND
A V
InternalRange resistance
Voltmeter 0 . . . 2 V 1 MΩVoltmeter 2 . . . 10 V 57 kΩAmmeter 0 . . . 1 A 1 Ω
Basic usage
• Push “Power” to switch the IPhO-measure on. Thedevice is not yet measuring; to start measuring, push“Start”. Alternatively, you can now start browsing yourstored data, see below.
• To browse previously saved samples (through all sets),press “Previous” or “Next”. Hold them down longerto jump directly between sets.
• While not measuring, push “Start” to start measuringa new set.
• While measuring, push “Sample” to store a datasample (with the current readings).
• While measuring, you can also browse other samples ofthe current set, using “Previous” and “Next”.
• Press “Stop” to end a set and stop measuring. Thedevice is still on, you are ready to start a new measuringsession, or browsing stored data.
• Pushing “Power” turns the device off. The device willshow the text “my mind is going . . . ”; don’t worry, all thedata measurements will be stored and you will be able tobrowse them after you switch the device on, again. Savedsamples will not be erased.
Display
A displayed sample consists of nine variables:
1. index n of the sample in the set;2. index s of the set;3. time t since starting the set;4. voltmeter output V ;5. rate of change of V (the time derivative V ); if derivative
cannot be reliably taken due to fluctuations, “+nan/s” isshown;
6. ammeter output I;7. rate of change of I (the time derivative I); if derivative
cannot be reliably taken due to fluctuations, “+nan/s” isshown;
8. product P = V I;9. ratio R = V/I.
If any of the variables is out of its allowed range, its displayshows “+inf” or “-inf”.
— page 4 of 4 —
IPhO-measure: short manualIPhO-measure is a multimeter capable of measuring voltage Vand current I simultaneously. It also records their time derivat-ives V and I, their product P = V I, ratio R = V/I, and timet of the sample. Stored measurements are organized into sep-arate sets; every stored sample is numbered by the set numbers and a counter n inside the set. All saved samples are writtento an internal flash memory and can later be retrieved.
Electrical behaviourThe device behaves as an ammeter and a voltmeter connectedas follows. OUTIN
Multimeter
GND
A V
InternalRange resistance
Voltmeter 0 . . . 2 V 1 MΩVoltmeter 2 . . . 10 V 57 kΩAmmeter 0 . . . 1 A 1 Ω
Basic usage
• Push “Power” to switch the IPhO-measure on. Thedevice is not yet measuring; to start measuring, push“Start”. Alternatively, you can now start browsing yourstored data, see below.
• To browse previously saved samples (through all sets),press “Previous” or “Next”. Hold them down longerto jump directly between sets.
• While not measuring, push “Start” to start measuringa new set.
• While measuring, push “Sample” to store a datasample (with the current readings).
• While measuring, you can also browse other samples ofthe current set, using “Previous” and “Next”.
• Press “Stop” to end a set and stop measuring. Thedevice is still on, you are ready to start a new measuringsession, or browsing stored data.
• Pushing “Power” turns the device off. The device willshow the text “my mind is going . . . ”; don’t worry, all thedata measurements will be stored and you will be able tobrowse them after you switch the device on, again. Savedsamples will not be erased.
Display
A displayed sample consists of nine variables:
1. index n of the sample in the set;2. index s of the set;3. time t since starting the set;4. voltmeter output V ;5. rate of change of V (the time derivative V ); if derivative
cannot be reliably taken due to fluctuations, “+nan/s” isshown;
6. ammeter output I;7. rate of change of I (the time derivative I); if derivative
cannot be reliably taken due to fluctuations, “+nan/s” isshown;
8. product P = V I;9. ratio R = V/I.
If any of the variables is out of its allowed range, its displayshows “+inf” or “-inf”.
— page 4 of 4 —
115
IPhO-measure: short manualIPhO-measure is a multimeter capable of measuring voltage Vand current I simultaneously. It also records their time derivat-ives V and I, their product P = V I, ratio R = V/I, and timet of the sample. Stored measurements are organized into sep-arate sets; every stored sample is numbered by the set numbers and a counter n inside the set. All saved samples are writtento an internal flash memory and can later be retrieved.
Electrical behaviourThe device behaves as an ammeter and a voltmeter connectedas follows. OUTIN
Multimeter
GND
A V
InternalRange resistance
Voltmeter 0 . . . 2 V 1 MΩVoltmeter 2 . . . 10 V 57 kΩAmmeter 0 . . . 1 A 1 Ω
Basic usage
• Push “Power” to switch the IPhO-measure on. Thedevice is not yet measuring; to start measuring, push“Start”. Alternatively, you can now start browsing yourstored data, see below.
• To browse previously saved samples (through all sets),press “Previous” or “Next”. Hold them down longerto jump directly between sets.
• While not measuring, push “Start” to start measuringa new set.
• While measuring, push “Sample” to store a datasample (with the current readings).
• While measuring, you can also browse other samples ofthe current set, using “Previous” and “Next”.
• Press “Stop” to end a set and stop measuring. Thedevice is still on, you are ready to start a new measuringsession, or browsing stored data.
• Pushing “Power” turns the device off. The device willshow the text “my mind is going . . . ”; don’t worry, all thedata measurements will be stored and you will be able tobrowse them after you switch the device on, again. Savedsamples will not be erased.
Display
A displayed sample consists of nine variables:
1. index n of the sample in the set;2. index s of the set;3. time t since starting the set;4. voltmeter output V ;5. rate of change of V (the time derivative V ); if derivative
cannot be reliably taken due to fluctuations, “+nan/s” isshown;
6. ammeter output I;7. rate of change of I (the time derivative I); if derivative
cannot be reliably taken due to fluctuations, “+nan/s” isshown;
8. product P = V I;9. ratio R = V/I.
If any of the variables is out of its allowed range, its displayshows “+inf” or “-inf”.
— page 4 of 4 —
IPhO-measure: short manualIPhO-measure is a multimeter capable of measuring voltage Vand current I simultaneously. It also records their time derivat-ives V and I, their product P = V I, ratio R = V/I, and timet of the sample. Stored measurements are organized into sep-arate sets; every stored sample is numbered by the set numbers and a counter n inside the set. All saved samples are writtento an internal flash memory and can later be retrieved.
Electrical behaviourThe device behaves as an ammeter and a voltmeter connectedas follows. OUTIN
Multimeter
GND
A V
InternalRange resistance
Voltmeter 0 . . . 2 V 1 MΩVoltmeter 2 . . . 10 V 57 kΩAmmeter 0 . . . 1 A 1 Ω
Basic usage
• Push “Power” to switch the IPhO-measure on. Thedevice is not yet measuring; to start measuring, push“Start”. Alternatively, you can now start browsing yourstored data, see below.
• To browse previously saved samples (through all sets),press “Previous” or “Next”. Hold them down longerto jump directly between sets.
• While not measuring, push “Start” to start measuringa new set.
• While measuring, push “Sample” to store a datasample (with the current readings).
• While measuring, you can also browse other samples ofthe current set, using “Previous” and “Next”.
• Press “Stop” to end a set and stop measuring. Thedevice is still on, you are ready to start a new measuringsession, or browsing stored data.
• Pushing “Power” turns the device off. The device willshow the text “my mind is going . . . ”; don’t worry, all thedata measurements will be stored and you will be able tobrowse them after you switch the device on, again. Savedsamples will not be erased.
Display
A displayed sample consists of nine variables:
1. index n of the sample in the set;2. index s of the set;3. time t since starting the set;4. voltmeter output V ;5. rate of change of V (the time derivative V ); if derivative
cannot be reliably taken due to fluctuations, “+nan/s” isshown;
6. ammeter output I;7. rate of change of I (the time derivative I); if derivative
cannot be reliably taken due to fluctuations, “+nan/s” isshown;
8. product P = V I;9. ratio R = V/I.
If any of the variables is out of its allowed range, its displayshows “+inf” or “-inf”.
— page 4 of 4 —
116
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Observing reflections from the water surface (in particular,
those of straight lines, such as the edge of a sheet of paper),
it is easy to see that the profile has one minimum and has a
relatively flat bottom, ie. the correct answer is “Option D” (full
marks are given also for Option B). This profile implies that
water is pushed away from the magnet, which means µ < 1
(recall that ferromagnets with µ > 1 are pulled).
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) The height of the spot on the screen y is tabulated
below as a function of the horizontal position x of the caliper.
Note that the values of y in millimetres can be rounded to in-
tegers (this series of measurements aimed as high as possible
precision).
x (mm) 10 15 20 25 30 32 34 36
y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5
x (mm) 38 40 42 44 46 48 50 52
y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4
x (mm) 54 56 58 60 62 64 66 68
y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6
x (mm) 70 72 74 76 78 80 85 90
y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9
ii. (0.7 pts)
On this graph, the data of to two different water levels are
depicted; blue curve corresponds to a water depth of ca 2 mm
(data given in the table above); the violet one — to 1 mm.
iii. (0.5 pts) If the water surface were flat, the dependence of x
on y would be linear, and the tangent of the angle α0 would be
given by tan α0 = ∆y
∆x, where ∆x is a horizontal displacement of
the pointer, and ∆y — the respective displacement of the spot
height. For the extreme positions of the pointer, the beam hits
the water surface so far from the magnet that there, the surface
is essentially unperturbed; connecting the respective points on
the graph, we obtain a line corresponding to a flat water sur-
face — the red line. Using these two extreme data points we
can also easily calculate the angle α0 = arctan 74.9−11.590−10
≈ 38.
iv. (1.4 pts) For faster calculations, y − y0 − (x − x0) tan α0
(appearing in the formula given) can be read from the previous
graph as the distance between red and blue line; the red line
is given by equation yr = y0 + (x − x0) tan α0. One can also
precalculate 1
2cos2 α0 ≈ 0.31. The calculations lead to the fol-
lowing table (with z = tan β · 105 ; as mentioned above, during
the competition, lesser precision with two significant numbers
is sufficient).
x (mm) 10 15 20 25 30 32 34 36
z 0 10 27 66 204 303 473 591
x (mm) 38 40 42 44 46 48 50 52
z 597 428 239 128 53 26 0 -26
x (mm) 54 56 58 60 62 64 66 68
z -72 -145 -278 -449 -606 -536 -388 -254
x (mm) 70 72 74 76 78 80 85 90
z -154 -74 -40 -20 -6 2 -2 0
— page 1 of 4 —
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Observing reflections from the water surface (in particular,
those of straight lines, such as the edge of a sheet of paper),
it is easy to see that the profile has one minimum and has a
relatively flat bottom, ie. the correct answer is “Option D” (full
marks are given also for Option B). This profile implies that
water is pushed away from the magnet, which means µ < 1
(recall that ferromagnets with µ > 1 are pulled).
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) The height of the spot on the screen y is tabulated
below as a function of the horizontal position x of the caliper.
Note that the values of y in millimetres can be rounded to in-
tegers (this series of measurements aimed as high as possible
precision).
x (mm) 10 15 20 25 30 32 34 36
y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5
x (mm) 38 40 42 44 46 48 50 52
y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4
x (mm) 54 56 58 60 62 64 66 68
y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6
x (mm) 70 72 74 76 78 80 85 90
y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9
ii. (0.7 pts)
On this graph, the data of to two different water levels are
depicted; blue curve corresponds to a water depth of ca 2 mm
(data given in the table above); the violet one — to 1 mm.
iii. (0.5 pts) If the water surface were flat, the dependence of x
on y would be linear, and the tangent of the angle α0 would be
given by tan α0 = ∆y
∆x, where ∆x is a horizontal displacement of
the pointer, and ∆y — the respective displacement of the spot
height. For the extreme positions of the pointer, the beam hits
the water surface so far from the magnet that there, the surface
is essentially unperturbed; connecting the respective points on
the graph, we obtain a line corresponding to a flat water sur-
face — the red line. Using these two extreme data points we
can also easily calculate the angle α0 = arctan 74.9−11.590−10
≈ 38.
iv. (1.4 pts) For faster calculations, y − y0 − (x − x0) tan α0
(appearing in the formula given) can be read from the previous
graph as the distance between red and blue line; the red line
is given by equation yr = y0 + (x − x0) tan α0. One can also
precalculate 1
2cos2 α0 ≈ 0.31. The calculations lead to the fol-
lowing table (with z = tan β · 105 ; as mentioned above, during
the competition, lesser precision with two significant numbers
is sufficient).
x (mm) 10 15 20 25 30 32 34 36
z 0 10 27 66 204 303 473 591
x (mm) 38 40 42 44 46 48 50 52
z 597 428 239 128 53 26 0 -26
x (mm) 54 56 58 60 62 64 66 68
z -72 -145 -278 -449 -606 -536 -388 -254
x (mm) 70 72 74 76 78 80 85 90
z -154 -74 -40 -20 -6 2 -2 0
— page 1 of 4 —
PrObLEm E1. ThE magnETIc PErmEabILITy OF waTEr (10 POInTs)
Solutions
117
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Observing reflections from the water surface (in particular,
those of straight lines, such as the edge of a sheet of paper),
it is easy to see that the profile has one minimum and has a
relatively flat bottom, ie. the correct answer is “Option D” (full
marks are given also for Option B). This profile implies that
water is pushed away from the magnet, which means µ < 1
(recall that ferromagnets with µ > 1 are pulled).
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) The height of the spot on the screen y is tabulated
below as a function of the horizontal position x of the caliper.
Note that the values of y in millimetres can be rounded to in-
tegers (this series of measurements aimed as high as possible
precision).
x (mm) 10 15 20 25 30 32 34 36
y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5
x (mm) 38 40 42 44 46 48 50 52
y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4
x (mm) 54 56 58 60 62 64 66 68
y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6
x (mm) 70 72 74 76 78 80 85 90
y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9
ii. (0.7 pts)
On this graph, the data of to two different water levels are
depicted; blue curve corresponds to a water depth of ca 2 mm
(data given in the table above); the violet one — to 1 mm.
iii. (0.5 pts) If the water surface were flat, the dependence of x
on y would be linear, and the tangent of the angle α0 would be
given by tan α0 = ∆y
∆x, where ∆x is a horizontal displacement of
the pointer, and ∆y — the respective displacement of the spot
height. For the extreme positions of the pointer, the beam hits
the water surface so far from the magnet that there, the surface
is essentially unperturbed; connecting the respective points on
the graph, we obtain a line corresponding to a flat water sur-
face — the red line. Using these two extreme data points we
can also easily calculate the angle α0 = arctan 74.9−11.590−10
≈ 38.
iv. (1.4 pts) For faster calculations, y − y0 − (x − x0) tan α0
(appearing in the formula given) can be read from the previous
graph as the distance between red and blue line; the red line
is given by equation yr = y0 + (x − x0) tan α0. One can also
precalculate 1
2cos2 α0 ≈ 0.31. The calculations lead to the fol-
lowing table (with z = tan β · 105 ; as mentioned above, during
the competition, lesser precision with two significant numbers
is sufficient).
x (mm) 10 15 20 25 30 32 34 36
z 0 10 27 66 204 303 473 591
x (mm) 38 40 42 44 46 48 50 52
z 597 428 239 128 53 26 0 -26
x (mm) 54 56 58 60 62 64 66 68
z -72 -145 -278 -449 -606 -536 -388 -254
x (mm) 70 72 74 76 78 80 85 90
z -154 -74 -40 -20 -6 2 -2 0
— page 1 of 4 —Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Observing reflections from the water surface (in particular,
those of straight lines, such as the edge of a sheet of paper),
it is easy to see that the profile has one minimum and has a
relatively flat bottom, ie. the correct answer is “Option D” (full
marks are given also for Option B). This profile implies that
water is pushed away from the magnet, which means µ < 1
(recall that ferromagnets with µ > 1 are pulled).
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) The height of the spot on the screen y is tabulated
below as a function of the horizontal position x of the caliper.
Note that the values of y in millimetres can be rounded to in-
tegers (this series of measurements aimed as high as possible
precision).
x (mm) 10 15 20 25 30 32 34 36
y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5
x (mm) 38 40 42 44 46 48 50 52
y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4
x (mm) 54 56 58 60 62 64 66 68
y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6
x (mm) 70 72 74 76 78 80 85 90
y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9
ii. (0.7 pts)
On this graph, the data of to two different water levels are
depicted; blue curve corresponds to a water depth of ca 2 mm
(data given in the table above); the violet one — to 1 mm.
iii. (0.5 pts) If the water surface were flat, the dependence of x
on y would be linear, and the tangent of the angle α0 would be
given by tan α0 = ∆y
∆x, where ∆x is a horizontal displacement of
the pointer, and ∆y — the respective displacement of the spot
height. For the extreme positions of the pointer, the beam hits
the water surface so far from the magnet that there, the surface
is essentially unperturbed; connecting the respective points on
the graph, we obtain a line corresponding to a flat water sur-
face — the red line. Using these two extreme data points we
can also easily calculate the angle α0 = arctan 74.9−11.590−10
≈ 38.
iv. (1.4 pts) For faster calculations, y − y0 − (x − x0) tan α0
(appearing in the formula given) can be read from the previous
graph as the distance between red and blue line; the red line
is given by equation yr = y0 + (x − x0) tan α0. One can also
precalculate 1
2cos2 α0 ≈ 0.31. The calculations lead to the fol-
lowing table (with z = tan β · 105 ; as mentioned above, during
the competition, lesser precision with two significant numbers
is sufficient).
x (mm) 10 15 20 25 30 32 34 36
z 0 10 27 66 204 303 473 591
x (mm) 38 40 42 44 46 48 50 52
z 597 428 239 128 53 26 0 -26
x (mm) 54 56 58 60 62 64 66 68
z -72 -145 -278 -449 -606 -536 -388 -254
x (mm) 70 72 74 76 78 80 85 90
z -154 -74 -40 -20 -6 2 -2 0
— page 1 of 4 —
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Observing reflections from the water surface (in particular,
those of straight lines, such as the edge of a sheet of paper),
it is easy to see that the profile has one minimum and has a
relatively flat bottom, ie. the correct answer is “Option D” (full
marks are given also for Option B). This profile implies that
water is pushed away from the magnet, which means µ < 1
(recall that ferromagnets with µ > 1 are pulled).
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) The height of the spot on the screen y is tabulated
below as a function of the horizontal position x of the caliper.
Note that the values of y in millimetres can be rounded to in-
tegers (this series of measurements aimed as high as possible
precision).
x (mm) 10 15 20 25 30 32 34 36
y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5
x (mm) 38 40 42 44 46 48 50 52
y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4
x (mm) 54 56 58 60 62 64 66 68
y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6
x (mm) 70 72 74 76 78 80 85 90
y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9
ii. (0.7 pts)
On this graph, the data of to two different water levels are
depicted; blue curve corresponds to a water depth of ca 2 mm
(data given in the table above); the violet one — to 1 mm.
iii. (0.5 pts) If the water surface were flat, the dependence of x
on y would be linear, and the tangent of the angle α0 would be
given by tan α0 = ∆y
∆x, where ∆x is a horizontal displacement of
the pointer, and ∆y — the respective displacement of the spot
height. For the extreme positions of the pointer, the beam hits
the water surface so far from the magnet that there, the surface
is essentially unperturbed; connecting the respective points on
the graph, we obtain a line corresponding to a flat water sur-
face — the red line. Using these two extreme data points we
can also easily calculate the angle α0 = arctan 74.9−11.590−10
≈ 38.
iv. (1.4 pts) For faster calculations, y − y0 − (x − x0) tan α0
(appearing in the formula given) can be read from the previous
graph as the distance between red and blue line; the red line
is given by equation yr = y0 + (x − x0) tan α0. One can also
precalculate 1
2cos2 α0 ≈ 0.31. The calculations lead to the fol-
lowing table (with z = tan β · 105 ; as mentioned above, during
the competition, lesser precision with two significant numbers
is sufficient).
x (mm) 10 15 20 25 30 32 34 36
z 0 10 27 66 204 303 473 591
x (mm) 38 40 42 44 46 48 50 52
z 597 428 239 128 53 26 0 -26
x (mm) 54 56 58 60 62 64 66 68
z -72 -145 -278 -449 -606 -536 -388 -254
x (mm) 70 72 74 76 78 80 85 90
z -154 -74 -40 -20 -6 2 -2 0
— page 1 of 4 —
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Observing reflections from the water surface (in particular,
those of straight lines, such as the edge of a sheet of paper),
it is easy to see that the profile has one minimum and has a
relatively flat bottom, ie. the correct answer is “Option D” (full
marks are given also for Option B). This profile implies that
water is pushed away from the magnet, which means µ < 1
(recall that ferromagnets with µ > 1 are pulled).
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) The height of the spot on the screen y is tabulated
below as a function of the horizontal position x of the caliper.
Note that the values of y in millimetres can be rounded to in-
tegers (this series of measurements aimed as high as possible
precision).
x (mm) 10 15 20 25 30 32 34 36
y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5
x (mm) 38 40 42 44 46 48 50 52
y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4
x (mm) 54 56 58 60 62 64 66 68
y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6
x (mm) 70 72 74 76 78 80 85 90
y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9
ii. (0.7 pts)
On this graph, the data of to two different water levels are
depicted; blue curve corresponds to a water depth of ca 2 mm
(data given in the table above); the violet one — to 1 mm.
iii. (0.5 pts) If the water surface were flat, the dependence of x
on y would be linear, and the tangent of the angle α0 would be
given by tan α0 = ∆y
∆x, where ∆x is a horizontal displacement of
the pointer, and ∆y — the respective displacement of the spot
height. For the extreme positions of the pointer, the beam hits
the water surface so far from the magnet that there, the surface
is essentially unperturbed; connecting the respective points on
the graph, we obtain a line corresponding to a flat water sur-
face — the red line. Using these two extreme data points we
can also easily calculate the angle α0 = arctan 74.9−11.590−10
≈ 38.
iv. (1.4 pts) For faster calculations, y − y0 − (x − x0) tan α0
(appearing in the formula given) can be read from the previous
graph as the distance between red and blue line; the red line
is given by equation yr = y0 + (x − x0) tan α0. One can also
precalculate 1
2cos2 α0 ≈ 0.31. The calculations lead to the fol-
lowing table (with z = tan β · 105 ; as mentioned above, during
the competition, lesser precision with two significant numbers
is sufficient).
x (mm) 10 15 20 25 30 32 34 36
z 0 10 27 66 204 303 473 591
x (mm) 38 40 42 44 46 48 50 52
z 597 428 239 128 53 26 0 -26
x (mm) 54 56 58 60 62 64 66 68
z -72 -145 -278 -449 -606 -536 -388 -254
x (mm) 70 72 74 76 78 80 85 90
z -154 -74 -40 -20 -6 2 -2 0
— page 1 of 4 —
PrObLEm E1. ThE magnETIc PErmEabILITy OF waTEr (10 POInTs)
v. (1.6 pts) The water height can be obtained as the integral
h =∫
tan βdx. Thus, we calculate the water height row-by-
row, by adding to the height in the previous row the product
of the horizontal displacement xi+1 −xi with the average slope1
2(tan βi+1 + tan βi).
x (mm) 10 15 20 25 30 32 34 36
−h (µm) 0 0 1 4 10 15 23 34
x (mm) 38 40 42 44 46 48 50 52
−h (µm) 46 56 63 66 68 69 69 69
x (mm) 54 56 58 60 62 64 66 68
−h (µm) 68 66 61 54 44 32 23 17
x (mm) 70 72 74 76 78 80 85 90
−h (µm) 12 10 9 8 8 8 8 8
Note that the water level height at the end of the table should
be also 0 (this corresponds also to an unperturbed region); the
non-zero result is explained by the measurement uncertainties.
One can improve the result by subtracting from h a linear trend
8 µm ·x−10 mm
80 mm.
If the water level above the magnet is 1 mm, the water level
descends below its unperturbed level at the axis of the magnet
by ca 120 µm.
vi. (1 pt)
Similarly to the previous figure, blue curve corresponds to
a water depth of ca 2 mm, (data given in the table above), and
the violet one — to 1 mm.
The position of the magnet can be found by measuring the
caliper (find the positions when the laser beam hits the edges of
the magnet and determine the distance between these positions
— the result is ca 24 mm), and using the symmetry: magnet
is placed symmetrically with respect to the surface elevation
curve.
Part C. Magnetic permeability (2 points)
Water surface takes an equipotential shape; for a unit volume of
water, the potential energy associated with the magnetic inter-
action is B2
2µ0
(µ−1−1) ≈ B2 1−µ
2µ0
; the potential energy associated
with the Earth’s gravity is ρgh. At the water surface, the sum
of those two needs to be constant; for a point at unperturbed
surface, this expression equals to zero, so B2 µ−1
2µ0
+ ρgh = 0
and hence, µ − 1 = 2µ0ρgh/B2. Here, h = 120 µm stands for
the depth of the water surface at the axis of the magnet; note
that we have compensated the cumulative error as described at
the end of the previous task and obtained h as the difference
between the depth at the magnet’s axis (121 µm) and the half-
depth at the right-hand-side of the graph (1 µm). Putting in
the numbers, we obtain µ − 1 = −1.2 × 10−5 .
— page 2 of 4 —
118
v. (1.6 pts) The water height can be obtained as the integral
h =∫
tan βdx. Thus, we calculate the water height row-by-
row, by adding to the height in the previous row the product
of the horizontal displacement xi+1 −xi with the average slope1
2(tan βi+1 + tan βi).
x (mm) 10 15 20 25 30 32 34 36
−h (µm) 0 0 1 4 10 15 23 34
x (mm) 38 40 42 44 46 48 50 52
−h (µm) 46 56 63 66 68 69 69 69
x (mm) 54 56 58 60 62 64 66 68
−h (µm) 68 66 61 54 44 32 23 17
x (mm) 70 72 74 76 78 80 85 90
−h (µm) 12 10 9 8 8 8 8 8
Note that the water level height at the end of the table should
be also 0 (this corresponds also to an unperturbed region); the
non-zero result is explained by the measurement uncertainties.
One can improve the result by subtracting from h a linear trend
8 µm ·x−10 mm
80 mm.
If the water level above the magnet is 1 mm, the water level
descends below its unperturbed level at the axis of the magnet
by ca 120 µm.
vi. (1 pt)
Similarly to the previous figure, blue curve corresponds to
a water depth of ca 2 mm, (data given in the table above), and
the violet one — to 1 mm.
The position of the magnet can be found by measuring the
caliper (find the positions when the laser beam hits the edges of
the magnet and determine the distance between these positions
— the result is ca 24 mm), and using the symmetry: magnet
is placed symmetrically with respect to the surface elevation
curve.
Part C. Magnetic permeability (2 points)
Water surface takes an equipotential shape; for a unit volume of
water, the potential energy associated with the magnetic inter-
action is B2
2µ0
(µ−1−1) ≈ B2 1−µ
2µ0
; the potential energy associated
with the Earth’s gravity is ρgh. At the water surface, the sum
of those two needs to be constant; for a point at unperturbed
surface, this expression equals to zero, so B2 µ−1
2µ0
+ ρgh = 0
and hence, µ − 1 = 2µ0ρgh/B2. Here, h = 120 µm stands for
the depth of the water surface at the axis of the magnet; note
that we have compensated the cumulative error as described at
the end of the previous task and obtained h as the difference
between the depth at the magnet’s axis (121 µm) and the half-
depth at the right-hand-side of the graph (1 µm). Putting in
the numbers, we obtain µ − 1 = −1.2 × 10−5 .
— page 2 of 4 —
v. (1.6 pts) The water height can be obtained as the integral
h =∫
tan βdx. Thus, we calculate the water height row-by-
row, by adding to the height in the previous row the product
of the horizontal displacement xi+1 −xi with the average slope1
2(tan βi+1 + tan βi).
x (mm) 10 15 20 25 30 32 34 36
−h (µm) 0 0 1 4 10 15 23 34
x (mm) 38 40 42 44 46 48 50 52
−h (µm) 46 56 63 66 68 69 69 69
x (mm) 54 56 58 60 62 64 66 68
−h (µm) 68 66 61 54 44 32 23 17
x (mm) 70 72 74 76 78 80 85 90
−h (µm) 12 10 9 8 8 8 8 8
Note that the water level height at the end of the table should
be also 0 (this corresponds also to an unperturbed region); the
non-zero result is explained by the measurement uncertainties.
One can improve the result by subtracting from h a linear trend
8 µm ·x−10 mm
80 mm.
If the water level above the magnet is 1 mm, the water level
descends below its unperturbed level at the axis of the magnet
by ca 120 µm.
vi. (1 pt)
Similarly to the previous figure, blue curve corresponds to
a water depth of ca 2 mm, (data given in the table above), and
the violet one — to 1 mm.
The position of the magnet can be found by measuring the
caliper (find the positions when the laser beam hits the edges of
the magnet and determine the distance between these positions
— the result is ca 24 mm), and using the symmetry: magnet
is placed symmetrically with respect to the surface elevation
curve.
Part C. Magnetic permeability (2 points)
Water surface takes an equipotential shape; for a unit volume of
water, the potential energy associated with the magnetic inter-
action is B2
2µ0
(µ−1−1) ≈ B2 1−µ
2µ0
; the potential energy associated
with the Earth’s gravity is ρgh. At the water surface, the sum
of those two needs to be constant; for a point at unperturbed
surface, this expression equals to zero, so B2 µ−1
2µ0
+ ρgh = 0
and hence, µ − 1 = 2µ0ρgh/B2. Here, h = 120 µm stands for
the depth of the water surface at the axis of the magnet; note
that we have compensated the cumulative error as described at
the end of the previous task and obtained h as the difference
between the depth at the magnet’s axis (121 µm) and the half-
depth at the right-hand-side of the graph (1 µm). Putting in
the numbers, we obtain µ − 1 = −1.2 × 10−5 .
— page 2 of 4 —
119
v. (1.6 pts) The water height can be obtained as the integral
h =∫
tan βdx. Thus, we calculate the water height row-by-
row, by adding to the height in the previous row the product
of the horizontal displacement xi+1 −xi with the average slope1
2(tan βi+1 + tan βi).
x (mm) 10 15 20 25 30 32 34 36
−h (µm) 0 0 1 4 10 15 23 34
x (mm) 38 40 42 44 46 48 50 52
−h (µm) 46 56 63 66 68 69 69 69
x (mm) 54 56 58 60 62 64 66 68
−h (µm) 68 66 61 54 44 32 23 17
x (mm) 70 72 74 76 78 80 85 90
−h (µm) 12 10 9 8 8 8 8 8
Note that the water level height at the end of the table should
be also 0 (this corresponds also to an unperturbed region); the
non-zero result is explained by the measurement uncertainties.
One can improve the result by subtracting from h a linear trend
8 µm ·x−10 mm
80 mm.
If the water level above the magnet is 1 mm, the water level
descends below its unperturbed level at the axis of the magnet
by ca 120 µm.
vi. (1 pt)
Similarly to the previous figure, blue curve corresponds to
a water depth of ca 2 mm, (data given in the table above), and
the violet one — to 1 mm.
The position of the magnet can be found by measuring the
caliper (find the positions when the laser beam hits the edges of
the magnet and determine the distance between these positions
— the result is ca 24 mm), and using the symmetry: magnet
is placed symmetrically with respect to the surface elevation
curve.
Part C. Magnetic permeability (2 points)
Water surface takes an equipotential shape; for a unit volume of
water, the potential energy associated with the magnetic inter-
action is B2
2µ0
(µ−1−1) ≈ B2 1−µ
2µ0
; the potential energy associated
with the Earth’s gravity is ρgh. At the water surface, the sum
of those two needs to be constant; for a point at unperturbed
surface, this expression equals to zero, so B2 µ−1
2µ0
+ ρgh = 0
and hence, µ − 1 = 2µ0ρgh/B2. Here, h = 120 µm stands for
the depth of the water surface at the axis of the magnet; note
that we have compensated the cumulative error as described at
the end of the previous task and obtained h as the difference
between the depth at the magnet’s axis (121 µm) and the half-
depth at the right-hand-side of the graph (1 µm). Putting in
the numbers, we obtain µ − 1 = −1.2 × 10−5 .
— page 2 of 4 —
PrObLEm E2. nOnLInEar bLack bOx (10 POInTs)Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
It is possible to make all the measurements needed for this
problem with a single circuit as shown in the figure. While the
current source is switched on, we are charging the capacitor in
the black box, until the current I(Vmax) through the nonlinear
element equals to the output current I0 of the current source.
Vmax = 540±40mVs varies from one experimental setup to an-
other. When the current source is switched off or disconnected,
the capacitor will discharge through the nonlinear element.
Multimeter
Current source
IN OUT GND
+−Switch
O
I
Black boxSwitch
O
I
i. (1 pt) During charging of the capacitor from V = 0 to
V = Vmax we note that the output of the current source is con-
stant (I0 = 6.0 mA) close to the precision of the multimeter.
ii. (1.2 pts) Using the definition of differential capacitance,
we can calculate the current through the capacitor in the black
box from the time derivative of the voltage on the black box.
Ic =dQ
dt=
dQ
dV
dV
dt= C(V )V
There are several ways to determine the capacitance used in
the black box based on chosen voltage.
• When the voltage on the black box is close to zero, the
current through the nonlinear element is also close to
zero, because I(V = 0) = 0. After switching the current
source on, most of the input current I0 will at first go
through the capacitor.
C0 = I0/V↑(V = 0)
This can be measured more precisely after first reversing
the polarity of the current source and charging the capa-
citor backwards, because the multimeter does not display
derivatives when they change sharply (as in few moments
after switching the current source on).
Example measurements taken this way follow.
V↑(0) (mV/s) 3.51 3.32 3.55
C0 (F) 1.71 1.81 1.69
C0 = 1.74 F
• When the voltage on the black box is Vmax, the current
through the nonlinear element is I0. Switching the cur-
rent source off, we will have the capacitor discharging
with the same current.
C0 = −I0/V↓(V = Vmax)
• We can also measure the capacitance for any intermediate
voltage as in A-iv.
iii. (2.2 pts) If we neglect the nonlinearity of the capacitor,
there are (at least) two ways to obtain the current–voltage char-
acteristic of the nonlinear element in the black box.
• Applying Kirchhoff’s I law to the charging capacitor,
I(V ) = Ic − C0V↑(V ).
An I(V ) characteristic obtained by charging the capacitor
is shown on the following figure.
• Applying Kirchhoff I law to the discharging capacitor,
I(V ) = −C0V↓(V ).
0 0.1 0.2 0.3 0.4 0.50
1
2
3
4
5
6
V (V)
I (m
A)
Part APart B
iv. (2.6 pts) In order to obtain the differential capacitance,
we solve a system of linear equations by eliminating I(V ):
I0 = V↑C(V ) + I(V )
I(V ) = −V↓C(V );=⇒ C(V ) =
I0
V↑ − V↓
.
Therefore we need to take measurements during both charging
and discharging the capacitor in the black box at the same
voltages. A graph of measurement results follows.
— page 3 of 4 —
Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
It is possible to make all the measurements needed for this
problem with a single circuit as shown in the figure. While the
current source is switched on, we are charging the capacitor in
the black box, until the current I(Vmax) through the nonlinear
element equals to the output current I0 of the current source.
Vmax = 540±40mVs varies from one experimental setup to an-
other. When the current source is switched off or disconnected,
the capacitor will discharge through the nonlinear element.
Multimeter
Current source
IN OUT GND
+−Switch
O
I
Black boxSwitch
O
I
i. (1 pt) During charging of the capacitor from V = 0 to
V = Vmax we note that the output of the current source is con-
stant (I0 = 6.0 mA) close to the precision of the multimeter.
ii. (1.2 pts) Using the definition of differential capacitance,
we can calculate the current through the capacitor in the black
box from the time derivative of the voltage on the black box.
Ic =dQ
dt=
dQ
dV
dV
dt= C(V )V
There are several ways to determine the capacitance used in
the black box based on chosen voltage.
• When the voltage on the black box is close to zero, the
current through the nonlinear element is also close to
zero, because I(V = 0) = 0. After switching the current
source on, most of the input current I0 will at first go
through the capacitor.
C0 = I0/V↑(V = 0)
This can be measured more precisely after first reversing
the polarity of the current source and charging the capa-
citor backwards, because the multimeter does not display
derivatives when they change sharply (as in few moments
after switching the current source on).
Example measurements taken this way follow.
V↑(0) (mV/s) 3.51 3.32 3.55
C0 (F) 1.71 1.81 1.69
C0 = 1.74 F
• When the voltage on the black box is Vmax, the current
through the nonlinear element is I0. Switching the cur-
rent source off, we will have the capacitor discharging
with the same current.
C0 = −I0/V↓(V = Vmax)
• We can also measure the capacitance for any intermediate
voltage as in A-iv.
iii. (2.2 pts) If we neglect the nonlinearity of the capacitor,
there are (at least) two ways to obtain the current–voltage char-
acteristic of the nonlinear element in the black box.
• Applying Kirchhoff’s I law to the charging capacitor,
I(V ) = Ic − C0V↑(V ).
An I(V ) characteristic obtained by charging the capacitor
is shown on the following figure.
• Applying Kirchhoff I law to the discharging capacitor,
I(V ) = −C0V↓(V ).
0 0.1 0.2 0.3 0.4 0.50
1
2
3
4
5
6
V (V)
I (m
A)
Part APart B
iv. (2.6 pts) In order to obtain the differential capacitance,
we solve a system of linear equations by eliminating I(V ):
I0 = V↑C(V ) + I(V )
I(V ) = −V↓C(V );=⇒ C(V ) =
I0
V↑ − V↓
.
Therefore we need to take measurements during both charging
and discharging the capacitor in the black box at the same
voltages. A graph of measurement results follows.
— page 3 of 4 —
120
Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
It is possible to make all the measurements needed for this
problem with a single circuit as shown in the figure. While the
current source is switched on, we are charging the capacitor in
the black box, until the current I(Vmax) through the nonlinear
element equals to the output current I0 of the current source.
Vmax = 540±40mVs varies from one experimental setup to an-
other. When the current source is switched off or disconnected,
the capacitor will discharge through the nonlinear element.
Multimeter
Current source
IN OUT GND
+−Switch
O
I
Black boxSwitch
O
I
i. (1 pt) During charging of the capacitor from V = 0 to
V = Vmax we note that the output of the current source is con-
stant (I0 = 6.0 mA) close to the precision of the multimeter.
ii. (1.2 pts) Using the definition of differential capacitance,
we can calculate the current through the capacitor in the black
box from the time derivative of the voltage on the black box.
Ic =dQ
dt=
dQ
dV
dV
dt= C(V )V
There are several ways to determine the capacitance used in
the black box based on chosen voltage.
• When the voltage on the black box is close to zero, the
current through the nonlinear element is also close to
zero, because I(V = 0) = 0. After switching the current
source on, most of the input current I0 will at first go
through the capacitor.
C0 = I0/V↑(V = 0)
This can be measured more precisely after first reversing
the polarity of the current source and charging the capa-
citor backwards, because the multimeter does not display
derivatives when they change sharply (as in few moments
after switching the current source on).
Example measurements taken this way follow.
V↑(0) (mV/s) 3.51 3.32 3.55
C0 (F) 1.71 1.81 1.69
C0 = 1.74 F
• When the voltage on the black box is Vmax, the current
through the nonlinear element is I0. Switching the cur-
rent source off, we will have the capacitor discharging
with the same current.
C0 = −I0/V↓(V = Vmax)
• We can also measure the capacitance for any intermediate
voltage as in A-iv.
iii. (2.2 pts) If we neglect the nonlinearity of the capacitor,
there are (at least) two ways to obtain the current–voltage char-
acteristic of the nonlinear element in the black box.
• Applying Kirchhoff’s I law to the charging capacitor,
I(V ) = Ic − C0V↑(V ).
An I(V ) characteristic obtained by charging the capacitor
is shown on the following figure.
• Applying Kirchhoff I law to the discharging capacitor,
I(V ) = −C0V↓(V ).
0 0.1 0.2 0.3 0.4 0.50
1
2
3
4
5
6
V (V)
I (m
A)
Part APart B
iv. (2.6 pts) In order to obtain the differential capacitance,
we solve a system of linear equations by eliminating I(V ):
I0 = V↑C(V ) + I(V )
I(V ) = −V↓C(V );=⇒ C(V ) =
I0
V↑ − V↓
.
Therefore we need to take measurements during both charging
and discharging the capacitor in the black box at the same
voltages. A graph of measurement results follows.
— page 3 of 4 —Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
It is possible to make all the measurements needed for this
problem with a single circuit as shown in the figure. While the
current source is switched on, we are charging the capacitor in
the black box, until the current I(Vmax) through the nonlinear
element equals to the output current I0 of the current source.
Vmax = 540±40mVs varies from one experimental setup to an-
other. When the current source is switched off or disconnected,
the capacitor will discharge through the nonlinear element.
Multimeter
Current source
IN OUT GND
+−Switch
O
I
Black boxSwitch
O
I
i. (1 pt) During charging of the capacitor from V = 0 to
V = Vmax we note that the output of the current source is con-
stant (I0 = 6.0 mA) close to the precision of the multimeter.
ii. (1.2 pts) Using the definition of differential capacitance,
we can calculate the current through the capacitor in the black
box from the time derivative of the voltage on the black box.
Ic =dQ
dt=
dQ
dV
dV
dt= C(V )V
There are several ways to determine the capacitance used in
the black box based on chosen voltage.
• When the voltage on the black box is close to zero, the
current through the nonlinear element is also close to
zero, because I(V = 0) = 0. After switching the current
source on, most of the input current I0 will at first go
through the capacitor.
C0 = I0/V↑(V = 0)
This can be measured more precisely after first reversing
the polarity of the current source and charging the capa-
citor backwards, because the multimeter does not display
derivatives when they change sharply (as in few moments
after switching the current source on).
Example measurements taken this way follow.
V↑(0) (mV/s) 3.51 3.32 3.55
C0 (F) 1.71 1.81 1.69
C0 = 1.74 F
• When the voltage on the black box is Vmax, the current
through the nonlinear element is I0. Switching the cur-
rent source off, we will have the capacitor discharging
with the same current.
C0 = −I0/V↓(V = Vmax)
• We can also measure the capacitance for any intermediate
voltage as in A-iv.
iii. (2.2 pts) If we neglect the nonlinearity of the capacitor,
there are (at least) two ways to obtain the current–voltage char-
acteristic of the nonlinear element in the black box.
• Applying Kirchhoff’s I law to the charging capacitor,
I(V ) = Ic − C0V↑(V ).
An I(V ) characteristic obtained by charging the capacitor
is shown on the following figure.
• Applying Kirchhoff I law to the discharging capacitor,
I(V ) = −C0V↓(V ).
0 0.1 0.2 0.3 0.4 0.50
1
2
3
4
5
6
V (V)
I (m
A)
Part APart B
iv. (2.6 pts) In order to obtain the differential capacitance,
we solve a system of linear equations by eliminating I(V ):
I0 = V↑C(V ) + I(V )
I(V ) = −V↓C(V );=⇒ C(V ) =
I0
V↑ − V↓
.
Therefore we need to take measurements during both charging
and discharging the capacitor in the black box at the same
voltages. A graph of measurement results follows.
— page 3 of 4 —
121
Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
It is possible to make all the measurements needed for this
problem with a single circuit as shown in the figure. While the
current source is switched on, we are charging the capacitor in
the black box, until the current I(Vmax) through the nonlinear
element equals to the output current I0 of the current source.
Vmax = 540±40mVs varies from one experimental setup to an-
other. When the current source is switched off or disconnected,
the capacitor will discharge through the nonlinear element.
Multimeter
Current source
IN OUT GND
+−Switch
O
I
Black boxSwitch
O
I
i. (1 pt) During charging of the capacitor from V = 0 to
V = Vmax we note that the output of the current source is con-
stant (I0 = 6.0 mA) close to the precision of the multimeter.
ii. (1.2 pts) Using the definition of differential capacitance,
we can calculate the current through the capacitor in the black
box from the time derivative of the voltage on the black box.
Ic =dQ
dt=
dQ
dV
dV
dt= C(V )V
There are several ways to determine the capacitance used in
the black box based on chosen voltage.
• When the voltage on the black box is close to zero, the
current through the nonlinear element is also close to
zero, because I(V = 0) = 0. After switching the current
source on, most of the input current I0 will at first go
through the capacitor.
C0 = I0/V↑(V = 0)
This can be measured more precisely after first reversing
the polarity of the current source and charging the capa-
citor backwards, because the multimeter does not display
derivatives when they change sharply (as in few moments
after switching the current source on).
Example measurements taken this way follow.
V↑(0) (mV/s) 3.51 3.32 3.55
C0 (F) 1.71 1.81 1.69
C0 = 1.74 F
• When the voltage on the black box is Vmax, the current
through the nonlinear element is I0. Switching the cur-
rent source off, we will have the capacitor discharging
with the same current.
C0 = −I0/V↓(V = Vmax)
• We can also measure the capacitance for any intermediate
voltage as in A-iv.
iii. (2.2 pts) If we neglect the nonlinearity of the capacitor,
there are (at least) two ways to obtain the current–voltage char-
acteristic of the nonlinear element in the black box.
• Applying Kirchhoff’s I law to the charging capacitor,
I(V ) = Ic − C0V↑(V ).
An I(V ) characteristic obtained by charging the capacitor
is shown on the following figure.
• Applying Kirchhoff I law to the discharging capacitor,
I(V ) = −C0V↓(V ).
0 0.1 0.2 0.3 0.4 0.50
1
2
3
4
5
6
V (V)
I (m
A)
Part APart B
iv. (2.6 pts) In order to obtain the differential capacitance,
we solve a system of linear equations by eliminating I(V ):
I0 = V↑C(V ) + I(V )
I(V ) = −V↓C(V );=⇒ C(V ) =
I0
V↑ − V↓
.
Therefore we need to take measurements during both charging
and discharging the capacitor in the black box at the same
voltages. A graph of measurement results follows.
— page 3 of 4 —
122
0 0.1 0.2 0.3 0.4 0.51.7
1.75
1.8
1.85
1.9
1.95
2
2.05
2.1
V (V)
C (F
)
Part B. Circuit with inductance (3 points)
Measuring and plotting the current–voltage characteristic of
the nonlinear element in the same way as in part A-iii, we
obtain a graph that differs only in the negative differential res-
istance (I ′(V ) < 0) region, in our case 70 mV < V < 330 mV.
This is the region where, when we look at small-signal oscil-
lations, the nonlinear element behaves as a negative-valued
Ohmic resistance. After enabling the inductance we have a
LC circuit whose oscillations are amplified (instead of being
dampened) by the negative differential resistance. Because the
resonant frequency ω =√
1
LCp
∼ 30 MHz (with Cp being the
capacitance of the nonlinear element) is high, we actually meas-
ure the average current through the nonlinear element, while
the real current oscillates all over the region of negative differ-
ential resistance.
— page 4 of 4 —
0 0.1 0.2 0.3 0.4 0.51.7
1.75
1.8
1.85
1.9
1.95
2
2.05
2.1
V (V)
C (F
)
Part B. Circuit with inductance (3 points)
Measuring and plotting the current–voltage characteristic of
the nonlinear element in the same way as in part A-iii, we
obtain a graph that differs only in the negative differential res-
istance (I ′(V ) < 0) region, in our case 70 mV < V < 330 mV.
This is the region where, when we look at small-signal oscil-
lations, the nonlinear element behaves as a negative-valued
Ohmic resistance. After enabling the inductance we have a
LC circuit whose oscillations are amplified (instead of being
dampened) by the negative differential resistance. Because the
resonant frequency ω =√
1
LCp
∼ 30 MHz (with Cp being the
capacitance of the nonlinear element) is high, we actually meas-
ure the average current through the nonlinear element, while
the real current oscillates all over the region of negative differ-
ential resistance.
— page 4 of 4 —
Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
It is possible to make all the measurements needed for this
problem with a single circuit as shown in the figure. While the
current source is switched on, we are charging the capacitor in
the black box, until the current I(Vmax) through the nonlinear
element equals to the output current I0 of the current source.
Vmax = 540±40mVs varies from one experimental setup to an-
other. When the current source is switched off or disconnected,
the capacitor will discharge through the nonlinear element.
Multimeter
Current source
IN OUT GND
+−Switch
O
I
Black boxSwitch
O
I
i. (1 pt) During charging of the capacitor from V = 0 to
V = Vmax we note that the output of the current source is con-
stant (I0 = 6.0 mA) close to the precision of the multimeter.
ii. (1.2 pts) Using the definition of differential capacitance,
we can calculate the current through the capacitor in the black
box from the time derivative of the voltage on the black box.
Ic =dQ
dt=
dQ
dV
dV
dt= C(V )V
There are several ways to determine the capacitance used in
the black box based on chosen voltage.
• When the voltage on the black box is close to zero, the
current through the nonlinear element is also close to
zero, because I(V = 0) = 0. After switching the current
source on, most of the input current I0 will at first go
through the capacitor.
C0 = I0/V↑(V = 0)
This can be measured more precisely after first reversing
the polarity of the current source and charging the capa-
citor backwards, because the multimeter does not display
derivatives when they change sharply (as in few moments
after switching the current source on).
Example measurements taken this way follow.
V↑(0) (mV/s) 3.51 3.32 3.55
C0 (F) 1.71 1.81 1.69
C0 = 1.74 F
• When the voltage on the black box is Vmax, the current
through the nonlinear element is I0. Switching the cur-
rent source off, we will have the capacitor discharging
with the same current.
C0 = −I0/V↓(V = Vmax)
• We can also measure the capacitance for any intermediate
voltage as in A-iv.
iii. (2.2 pts) If we neglect the nonlinearity of the capacitor,
there are (at least) two ways to obtain the current–voltage char-
acteristic of the nonlinear element in the black box.
• Applying Kirchhoff’s I law to the charging capacitor,
I(V ) = Ic − C0V↑(V ).
An I(V ) characteristic obtained by charging the capacitor
is shown on the following figure.
• Applying Kirchhoff I law to the discharging capacitor,
I(V ) = −C0V↓(V ).
0 0.1 0.2 0.3 0.4 0.50
1
2
3
4
5
6
V (V)
I (m
A)
Part APart B
iv. (2.6 pts) In order to obtain the differential capacitance,
we solve a system of linear equations by eliminating I(V ):
I0 = V↑C(V ) + I(V )
I(V ) = −V↓C(V );=⇒ C(V ) =
I0
V↑ − V↓
.
Therefore we need to take measurements during both charging
and discharging the capacitor in the black box at the same
voltages. A graph of measurement results follows.
— page 3 of 4 —
123
The 43rd International Physics Olympiad — July 2012
Grading scheme: Experiment
General rules This grading scheme describes the number of
points allotted for the design of the experiments, measurements,
plotting, and formulae used for calculations. In the case of a for-
mula, points are allotted for each term entering it. If a certain
term of a useful formula is written incorrectly, 0.1 is subtrac-
ted for a minor mistake (eg. missing non-dimensional factor);
no mark is given if the mistake is major (with non-matching
dimensionality). Points for the data measurements and calcu-
lations are not given automatically: the data which are clearly
wrong are not credited.
If the numerical data miss units, but the units can be
guessed, 25% of points will be subtracted for the corresponding
line in this grading scheme (rounded to one decimal place). The
same rule applies if there is a typo with units with a missing
or redundant prefactor (millli-, micro-, etc); no mark is given
if the mistake is dimensional (e.g Ampere instead of Volt).
No penalty is applied in these cases when a mistake is clearly
just a rewriting typo (i.e. when there is no mistake in the draft).
No penalty is applied for propagating errors unless the cal-
culations are significantly simplified (in which case mathemat-
ical calculations are credited partially, according to the degree
of simplification, with marking granularity of 0.1 pts).
— page 1 of 4 —
The 43rd International Physics Olympiad — July 2012
Grading scheme: Experiment
General rules This grading scheme describes the number of
points allotted for the design of the experiments, measurements,
plotting, and formulae used for calculations. In the case of a for-
mula, points are allotted for each term entering it. If a certain
term of a useful formula is written incorrectly, 0.1 is subtrac-
ted for a minor mistake (eg. missing non-dimensional factor);
no mark is given if the mistake is major (with non-matching
dimensionality). Points for the data measurements and calcu-
lations are not given automatically: the data which are clearly
wrong are not credited.
If the numerical data miss units, but the units can be
guessed, 25% of points will be subtracted for the corresponding
line in this grading scheme (rounded to one decimal place). The
same rule applies if there is a typo with units with a missing
or redundant prefactor (millli-, micro-, etc); no mark is given
if the mistake is dimensional (e.g Ampere instead of Volt).
No penalty is applied in these cases when a mistake is clearly
just a rewriting typo (i.e. when there is no mistake in the draft).
No penalty is applied for propagating errors unless the cal-
culations are significantly simplified (in which case mathemat-
ical calculations are credited partially, according to the degree
of simplification, with marking granularity of 0.1 pts).
— page 1 of 4 —
Grading scheme: Experiment
124
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Correct choice (B or D) — 0.5 pts;
Correct sign (µ < 1) — 0.5 pts;
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) Data
Sufficient number of reasonably accurate ±2 mm
data points: (n − 9)/10 pts but ≥ 0 and ≤ 0.9 — ≤ 0.9 pts;
(-0.1 if there is a sign error throughout the whole measure-
ments)
Sufficient range of horizontal displacements: (x − 45 mm)/50
rounded to one decimal but ≥ 0 and ≤ 0.7 — ≤ 0.7 pts.
ii. (0.7 pts) Graph
Axes supplied with units — 0.1 pts;
Data points correctly plotted — 0.4 pts;
(each clearly wrong point on graph: −0.1 pt down to total 0)
More densely spaced data points
in the regions of fast change — 0.2 pts.
iii. (0.7 pts) Angle (≈ 38 )
Idea: using the flat regions
far from the magnet — 0.5 pts
if the central part of the magnet is used: 0.1
Correct value: ±5 / ±10 / > 10 — 0.2/0.1/0 pts;
iv. (1.4 pts) Calculated table
According to the number of correctly calculated data points:
0.1n up to n = 10;
1 + 0.05(n − 10) rounded down one decimal place for n > 10
(but ≤ 1.4);
v. (1.6 pts) Height profile
Idea of calculation: integration of surface slope — 0.7 pts
(if not stated but used correctly: full marks)
∆x multiplied by mean slope at that interval — 0.3 pts
For n correctly calculated data points: — n/30 pts
(but no more than 0.6; rounded down to one decimal place).
vi. (1 pt) Graph
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape: height difference
in wings less than the 20% of maximum variation) — 0.1 pts
flat central region: the water level height variation in the
region spanning ±10 mm around the magnet’s axis
is less than 20 µm) — 0.2 pts
Magnet indicated correctly: (width 24 mm ± 2 mm) — 0.1 pts
symmetrically positioned — 0.1 pts
Part C. Magnetic permeability (2 points)
Concept of equipotentiality — 0.8 pts
Formula correctly includes magnetic energy — 0.4 pts
Formula correctly includes gravitational energy — 0.2 pts
height at the middle corrected for the integration error — 0.2 pts
Value calculated correctly from the existing data — 0.1 pts
Value: magnitude correct within 50% — 0.2 pts
(no marks if not obtained from the experimental data)
correct sign — 0.1 pts
— page 2 of 4 —
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Correct choice (B or D) — 0.5 pts;
Correct sign (µ < 1) — 0.5 pts;
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) Data
Sufficient number of reasonably accurate ±2 mm
data points: (n − 9)/10 pts but ≥ 0 and ≤ 0.9 — ≤ 0.9 pts;
(-0.1 if there is a sign error throughout the whole measure-
ments)
Sufficient range of horizontal displacements: (x − 45 mm)/50
rounded to one decimal but ≥ 0 and ≤ 0.7 — ≤ 0.7 pts.
ii. (0.7 pts) Graph
Axes supplied with units — 0.1 pts;
Data points correctly plotted — 0.4 pts;
(each clearly wrong point on graph: −0.1 pt down to total 0)
More densely spaced data points
in the regions of fast change — 0.2 pts.
iii. (0.7 pts) Angle (≈ 38 )
Idea: using the flat regions
far from the magnet — 0.5 pts
if the central part of the magnet is used: 0.1
Correct value: ±5 / ±10 / > 10 — 0.2/0.1/0 pts;
iv. (1.4 pts) Calculated table
According to the number of correctly calculated data points:
0.1n up to n = 10;
1 + 0.05(n − 10) rounded down one decimal place for n > 10
(but ≤ 1.4);
v. (1.6 pts) Height profile
Idea of calculation: integration of surface slope — 0.7 pts
(if not stated but used correctly: full marks)
∆x multiplied by mean slope at that interval — 0.3 pts
For n correctly calculated data points: — n/30 pts
(but no more than 0.6; rounded down to one decimal place).
vi. (1 pt) Graph
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape: height difference
in wings less than the 20% of maximum variation) — 0.1 pts
flat central region: the water level height variation in the
region spanning ±10 mm around the magnet’s axis
is less than 20 µm) — 0.2 pts
Magnet indicated correctly: (width 24 mm ± 2 mm) — 0.1 pts
symmetrically positioned — 0.1 pts
Part C. Magnetic permeability (2 points)
Concept of equipotentiality — 0.8 pts
Formula correctly includes magnetic energy — 0.4 pts
Formula correctly includes gravitational energy — 0.2 pts
height at the middle corrected for the integration error — 0.2 pts
Value calculated correctly from the existing data — 0.1 pts
Value: magnitude correct within 50% — 0.2 pts
(no marks if not obtained from the experimental data)
correct sign — 0.1 pts
— page 2 of 4 —
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Correct choice (B or D) — 0.5 pts;
Correct sign (µ < 1) — 0.5 pts;
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) Data
Sufficient number of reasonably accurate ±2 mm
data points: (n − 9)/10 pts but ≥ 0 and ≤ 0.9 — ≤ 0.9 pts;
(-0.1 if there is a sign error throughout the whole measure-
ments)
Sufficient range of horizontal displacements: (x − 45 mm)/50
rounded to one decimal but ≥ 0 and ≤ 0.7 — ≤ 0.7 pts.
ii. (0.7 pts) Graph
Axes supplied with units — 0.1 pts;
Data points correctly plotted — 0.4 pts;
(each clearly wrong point on graph: −0.1 pt down to total 0)
More densely spaced data points
in the regions of fast change — 0.2 pts.
iii. (0.7 pts) Angle (≈ 38 )
Idea: using the flat regions
far from the magnet — 0.5 pts
if the central part of the magnet is used: 0.1
Correct value: ±5 / ±10 / > 10 — 0.2/0.1/0 pts;
iv. (1.4 pts) Calculated table
According to the number of correctly calculated data points:
0.1n up to n = 10;
1 + 0.05(n − 10) rounded down one decimal place for n > 10
(but ≤ 1.4);
v. (1.6 pts) Height profile
Idea of calculation: integration of surface slope — 0.7 pts
(if not stated but used correctly: full marks)
∆x multiplied by mean slope at that interval — 0.3 pts
For n correctly calculated data points: — n/30 pts
(but no more than 0.6; rounded down to one decimal place).
vi. (1 pt) Graph
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape: height difference
in wings less than the 20% of maximum variation) — 0.1 pts
flat central region: the water level height variation in the
region spanning ±10 mm around the magnet’s axis
is less than 20 µm) — 0.2 pts
Magnet indicated correctly: (width 24 mm ± 2 mm) — 0.1 pts
symmetrically positioned — 0.1 pts
Part C. Magnetic permeability (2 points)
Concept of equipotentiality — 0.8 pts
Formula correctly includes magnetic energy — 0.4 pts
Formula correctly includes gravitational energy — 0.2 pts
height at the middle corrected for the integration error — 0.2 pts
Value calculated correctly from the existing data — 0.1 pts
Value: magnitude correct within 50% — 0.2 pts
(no marks if not obtained from the experimental data)
correct sign — 0.1 pts
— page 2 of 4 —
PrObLEm E1. ThE magnETIc PErmEabILITy OF waTEr (10 POInTs)
125
Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)
Correct choice (B or D) — 0.5 pts;
Correct sign (µ < 1) — 0.5 pts;
Part B. Exact shape of the water surface (7 points)
i. (1.6 pts) Data
Sufficient number of reasonably accurate ±2 mm
data points: (n − 9)/10 pts but ≥ 0 and ≤ 0.9 — ≤ 0.9 pts;
(-0.1 if there is a sign error throughout the whole measure-
ments)
Sufficient range of horizontal displacements: (x − 45 mm)/50
rounded to one decimal but ≥ 0 and ≤ 0.7 — ≤ 0.7 pts.
ii. (0.7 pts) Graph
Axes supplied with units — 0.1 pts;
Data points correctly plotted — 0.4 pts;
(each clearly wrong point on graph: −0.1 pt down to total 0)
More densely spaced data points
in the regions of fast change — 0.2 pts.
iii. (0.7 pts) Angle (≈ 38 )
Idea: using the flat regions
far from the magnet — 0.5 pts
if the central part of the magnet is used: 0.1
Correct value: ±5 / ±10 / > 10 — 0.2/0.1/0 pts;
iv. (1.4 pts) Calculated table
According to the number of correctly calculated data points:
0.1n up to n = 10;
1 + 0.05(n − 10) rounded down one decimal place for n > 10
(but ≤ 1.4);
v. (1.6 pts) Height profile
Idea of calculation: integration of surface slope — 0.7 pts
(if not stated but used correctly: full marks)
∆x multiplied by mean slope at that interval — 0.3 pts
For n correctly calculated data points: — n/30 pts
(but no more than 0.6; rounded down to one decimal place).
vi. (1 pt) Graph
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape: height difference
in wings less than the 20% of maximum variation) — 0.1 pts
flat central region: the water level height variation in the
region spanning ±10 mm around the magnet’s axis
is less than 20 µm) — 0.2 pts
Magnet indicated correctly: (width 24 mm ± 2 mm) — 0.1 pts
symmetrically positioned — 0.1 pts
Part C. Magnetic permeability (2 points)
Concept of equipotentiality — 0.8 pts
Formula correctly includes magnetic energy — 0.4 pts
Formula correctly includes gravitational energy — 0.2 pts
height at the middle corrected for the integration error — 0.2 pts
Value calculated correctly from the existing data — 0.1 pts
Value: magnitude correct within 50% — 0.2 pts
(no marks if not obtained from the experimental data)
correct sign — 0.1 pts
— page 2 of 4 —
PrObLEm E1. ThE magnETIc PErmEabILITy OF waTEr (10 POInTs)
126
PrObLEm E2. nOnLInEar bLack bOx (10 POInTs)
Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
Typical I(V ) and C(V ) curves can be found in solutions
file and sample filled-in answer sheets. Because I(V ) and C(V )
curves shape, typical capacitance, I0 and Vmax varies a bit from
one setup to another, reference curves for particular setup can
be acquired for grading when deemed necessary.
For each question, if the position of the switches on the
circuit diagram is not indicated, take -0.1 pts from the marks
for the circuit.
i. (1 pt)
Correct circuit — 0.3 pts
Measurements that cover 0 V to 480 mV — 0.3 pts
Correct value Imin (±0.4 mA) — 0.2 pts
Correct value Imax (±0.4 mA) — 0.2 pts
(unless only one data point)
In case of (single) measurement without the black box:
Circuit diagram — 0.1 pts
Measurement with the correct result — 0.1 pts
If only Imin, Imax together with a correct scheme docu-
mented, 0.7 pts in total; if only Imin, Imax, 0 pts in total; if a
scheme without the black box and a single I measurement,
0.2 pts in total.
ii. (1.2 pts) Measuring C0 at V = 0
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = I0 when V = 0 — 0.2 pts
Correct result — 0.2 pts
-0.1 for error between ±30% and ±50%
linear extrapolation to obtain V at V = 0 — 0.4 pts
Alternatively instead of the last line
Three or more measurements — 0.2 pts
Precharging the capacitor to a negative voltage — 0.2 pts
Alternate solution C0 at Vmax
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = −I0 when V = Vmax — 0.2 pts
Correct result — 0.1 pts
linear extrapolation to obtain V at V = Vmax — 0.2 pts
Three or more measurements — 0.2 pts
iii. (2.2 pts) For method:
If charging the capacitor:
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = I0 − C0V↑ — 0.2 pts
If discharging the capacitor:
Correct circuit (there are several) — 0.2 pts
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = −C0V↓— 0.2 pts
For measurements:
Total # of data correct data points
10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts
Additionaly for correct data points:
In range 0mV - 80mV at least 4/5 data pts — 0.1/0.2 pts
In range 80mV - 200mV at least 4/5 data pts — 0.1/0.2 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4/5 data pts — 0.1/0.2 pts
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape — 0.3 pts
(single maximum, single minimum with flat bottom, followed
by fast rise)
— page 3 of 4 —
127
PrObLEm E2. nOnLInEar bLack bOx (10 POInTs)
Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
Typical I(V ) and C(V ) curves can be found in solutions
file and sample filled-in answer sheets. Because I(V ) and C(V )
curves shape, typical capacitance, I0 and Vmax varies a bit from
one setup to another, reference curves for particular setup can
be acquired for grading when deemed necessary.
For each question, if the position of the switches on the
circuit diagram is not indicated, take -0.1 pts from the marks
for the circuit.
i. (1 pt)
Correct circuit — 0.3 pts
Measurements that cover 0 V to 480 mV — 0.3 pts
Correct value Imin (±0.4 mA) — 0.2 pts
Correct value Imax (±0.4 mA) — 0.2 pts
(unless only one data point)
In case of (single) measurement without the black box:
Circuit diagram — 0.1 pts
Measurement with the correct result — 0.1 pts
If only Imin, Imax together with a correct scheme docu-
mented, 0.7 pts in total; if only Imin, Imax, 0 pts in total; if a
scheme without the black box and a single I measurement,
0.2 pts in total.
ii. (1.2 pts) Measuring C0 at V = 0
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = I0 when V = 0 — 0.2 pts
Correct result — 0.2 pts
-0.1 for error between ±30% and ±50%
linear extrapolation to obtain V at V = 0 — 0.4 pts
Alternatively instead of the last line
Three or more measurements — 0.2 pts
Precharging the capacitor to a negative voltage — 0.2 pts
Alternate solution C0 at Vmax
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = −I0 when V = Vmax — 0.2 pts
Correct result — 0.1 pts
linear extrapolation to obtain V at V = Vmax — 0.2 pts
Three or more measurements — 0.2 pts
iii. (2.2 pts) For method:
If charging the capacitor:
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = I0 − C0V↑ — 0.2 pts
If discharging the capacitor:
Correct circuit (there are several) — 0.2 pts
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = −C0V↓— 0.2 pts
For measurements:
Total # of data correct data points
10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts
Additionaly for correct data points:
In range 0mV - 80mV at least 4/5 data pts — 0.1/0.2 pts
In range 80mV - 200mV at least 4/5 data pts — 0.1/0.2 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4/5 data pts — 0.1/0.2 pts
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape — 0.3 pts
(single maximum, single minimum with flat bottom, followed
by fast rise)
— page 3 of 4 —Problem E2. Nonlinear Black Box (10 points)
Part A. Circuit without inductance (7 points)
Typical I(V ) and C(V ) curves can be found in solutions
file and sample filled-in answer sheets. Because I(V ) and C(V )
curves shape, typical capacitance, I0 and Vmax varies a bit from
one setup to another, reference curves for particular setup can
be acquired for grading when deemed necessary.
For each question, if the position of the switches on the
circuit diagram is not indicated, take -0.1 pts from the marks
for the circuit.
i. (1 pt)
Correct circuit — 0.3 pts
Measurements that cover 0 V to 480 mV — 0.3 pts
Correct value Imin (±0.4 mA) — 0.2 pts
Correct value Imax (±0.4 mA) — 0.2 pts
(unless only one data point)
In case of (single) measurement without the black box:
Circuit diagram — 0.1 pts
Measurement with the correct result — 0.1 pts
If only Imin, Imax together with a correct scheme docu-
mented, 0.7 pts in total; if only Imin, Imax, 0 pts in total; if a
scheme without the black box and a single I measurement,
0.2 pts in total.
ii. (1.2 pts) Measuring C0 at V = 0
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = I0 when V = 0 — 0.2 pts
Correct result — 0.2 pts
-0.1 for error between ±30% and ±50%
linear extrapolation to obtain V at V = 0 — 0.4 pts
Alternatively instead of the last line
Three or more measurements — 0.2 pts
Precharging the capacitor to a negative voltage — 0.2 pts
Alternate solution C0 at Vmax
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = −I0 when V = Vmax — 0.2 pts
Correct result — 0.1 pts
linear extrapolation to obtain V at V = Vmax — 0.2 pts
Three or more measurements — 0.2 pts
iii. (2.2 pts) For method:
If charging the capacitor:
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = I0 − C0V↑ — 0.2 pts
If discharging the capacitor:
Correct circuit (there are several) — 0.2 pts
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = −C0V↓— 0.2 pts
For measurements:
Total # of data correct data points
10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts
Additionaly for correct data points:
In range 0mV - 80mV at least 4/5 data pts — 0.1/0.2 pts
In range 80mV - 200mV at least 4/5 data pts — 0.1/0.2 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4/5 data pts — 0.1/0.2 pts
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape — 0.3 pts
(single maximum, single minimum with flat bottom, followed
by fast rise)
— page 3 of 4 —
Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
Typical I(V ) and C(V ) curves can be found in solutions
file and sample filled-in answer sheets. Because I(V ) and C(V )
curves shape, typical capacitance, I0 and Vmax varies a bit from
one setup to another, reference curves for particular setup can
be acquired for grading when deemed necessary.
For each question, if the position of the switches on the
circuit diagram is not indicated, take -0.1 pts from the marks
for the circuit.
i. (1 pt)
Correct circuit — 0.3 pts
Measurements that cover 0 V to 480 mV — 0.3 pts
Correct value Imin (±0.4 mA) — 0.2 pts
Correct value Imax (±0.4 mA) — 0.2 pts
(unless only one data point)
In case of (single) measurement without the black box:
Circuit diagram — 0.1 pts
Measurement with the correct result — 0.1 pts
If only Imin, Imax together with a correct scheme docu-
mented, 0.7 pts in total; if only Imin, Imax, 0 pts in total; if a
scheme without the black box and a single I measurement,
0.2 pts in total.
ii. (1.2 pts) Measuring C0 at V = 0
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = I0 when V = 0 — 0.2 pts
Correct result — 0.2 pts
-0.1 for error between ±30% and ±50%
linear extrapolation to obtain V at V = 0 — 0.4 pts
Alternatively instead of the last line
Three or more measurements — 0.2 pts
Precharging the capacitor to a negative voltage — 0.2 pts
Alternate solution C0 at Vmax
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = −I0 when V = Vmax — 0.2 pts
Correct result — 0.1 pts
linear extrapolation to obtain V at V = Vmax — 0.2 pts
Three or more measurements — 0.2 pts
iii. (2.2 pts) For method:
If charging the capacitor:
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = I0 − C0V↑ — 0.2 pts
If discharging the capacitor:
Correct circuit (there are several) — 0.2 pts
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = −C0V↓— 0.2 pts
For measurements:
Total # of data correct data points
10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts
Additionaly for correct data points:
In range 0mV - 80mV at least 4/5 data pts — 0.1/0.2 pts
In range 80mV - 200mV at least 4/5 data pts — 0.1/0.2 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4/5 data pts — 0.1/0.2 pts
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape — 0.3 pts
(single maximum, single minimum with flat bottom, followed
by fast rise)
— page 3 of 4 —
128
iv. (2.6 pts)
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Idea to use the reverse cylce to pt iii— 0.4 pts
Writing the lin. eq for finding C(V )— 0.4 pts
(No pts if only one Eq.)
Expressing from there C(V )— 0.1 pts
Idea to use the same voltages for both cycles— 0.4 pts
(0.3 pts if intermediate values read from graph)
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
For correct data points:
In range 0mV - 80mV at least 4 data pts — 0.1 pts
In range 80mV - 200mV at least 4 data pts — 0.1 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4 data pts — 0.1 pts
For finding Cmax and Cmin:
Finding reasonable Cmax — 0.1 pts
Finding reasonable Cmin — 0.1 pts
Part B. Circuit with inductance (3 points)
For correct data points:
In range 0mV - 80mV at least 4 data pts — 0.1 pts
In range 80mV - 200mV at least 4 data pts — 0.1 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4 data pts — 0.1 pts
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape — 0.3 pts
(two sharp falls, plateau in between)
For detecting differences:
Correct range for V :— 0.2 pts
Correct cond. for I(V ) — 0.5 pts
Noting that we have now LC-circuit— 0.2 pts
Noting that neg. resist. → instability— 0.4 pts
(or something equivalent)
Mentioning emergence of oscillations— 0.2 pts
Noting that I(V ) is the average current— 0.3 pts
— page 4 of 4 —
Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)
Typical I(V ) and C(V ) curves can be found in solutions
file and sample filled-in answer sheets. Because I(V ) and C(V )
curves shape, typical capacitance, I0 and Vmax varies a bit from
one setup to another, reference curves for particular setup can
be acquired for grading when deemed necessary.
For each question, if the position of the switches on the
circuit diagram is not indicated, take -0.1 pts from the marks
for the circuit.
i. (1 pt)
Correct circuit — 0.3 pts
Measurements that cover 0 V to 480 mV — 0.3 pts
Correct value Imin (±0.4 mA) — 0.2 pts
Correct value Imax (±0.4 mA) — 0.2 pts
(unless only one data point)
In case of (single) measurement without the black box:
Circuit diagram — 0.1 pts
Measurement with the correct result — 0.1 pts
If only Imin, Imax together with a correct scheme docu-
mented, 0.7 pts in total; if only Imin, Imax, 0 pts in total; if a
scheme without the black box and a single I measurement,
0.2 pts in total.
ii. (1.2 pts) Measuring C0 at V = 0
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = I0 when V = 0 — 0.2 pts
Correct result — 0.2 pts
-0.1 for error between ±30% and ±50%
linear extrapolation to obtain V at V = 0 — 0.4 pts
Alternatively instead of the last line
Three or more measurements — 0.2 pts
Precharging the capacitor to a negative voltage — 0.2 pts
Alternate solution C0 at Vmax
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
Realising that Ic = C0V — 0.2 pts
Using the fact that Ic = −I0 when V = Vmax — 0.2 pts
Correct result — 0.1 pts
linear extrapolation to obtain V at V = Vmax — 0.2 pts
Three or more measurements — 0.2 pts
iii. (2.2 pts) For method:
If charging the capacitor:
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = I0 − C0V↑ — 0.2 pts
If discharging the capacitor:
Correct circuit (there are several) — 0.2 pts
(-0.1 for wrong polarity of the black box)
Realising that I(V ) = −C0V↓— 0.2 pts
For measurements:
Total # of data correct data points
10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts
Additionaly for correct data points:
In range 0mV - 80mV at least 4/5 data pts — 0.1/0.2 pts
In range 80mV - 200mV at least 4/5 data pts — 0.1/0.2 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4/5 data pts — 0.1/0.2 pts
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape — 0.3 pts
(single maximum, single minimum with flat bottom, followed
by fast rise)
— page 3 of 4 —
129
iv. (2.6 pts)
Correct circuit — 0.2 pts
(-0.1 for measuring voltage on ammeter + black box)
(-0.1 for wrong polarity of the black box)
Idea to use the reverse cylce to pt iii— 0.4 pts
Writing the lin. eq for finding C(V )— 0.4 pts
(No pts if only one Eq.)
Expressing from there C(V )— 0.1 pts
Idea to use the same voltages for both cycles— 0.4 pts
(0.3 pts if intermediate values read from graph)
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
For correct data points:
In range 0mV - 80mV at least 4 data pts — 0.1 pts
In range 80mV - 200mV at least 4 data pts — 0.1 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4 data pts — 0.1 pts
For finding Cmax and Cmin:
Finding reasonable Cmax — 0.1 pts
Finding reasonable Cmin — 0.1 pts
Part B. Circuit with inductance (3 points)
For correct data points:
In range 0mV - 80mV at least 4 data pts — 0.1 pts
In range 80mV - 200mV at least 4 data pts — 0.1 pts
In range 200mV - 400mV at least 3 data pts — 0.1 pts
In range 400mV - 550mV at least 4 data pts — 0.1 pts
For plotting:
Units on axes — 0.1 pts
Data points correctly plotted — 0.4 pts
(each clearly wrong point on graph: −0.1 pt down to total 0)
Correct qualitative shape — 0.3 pts
(two sharp falls, plateau in between)
For detecting differences:
Correct range for V :— 0.2 pts
Correct cond. for I(V ) — 0.5 pts
Noting that we have now LC-circuit— 0.2 pts
Noting that neg. resist. → instability— 0.4 pts
(or something equivalent)
Mentioning emergence of oscillations— 0.2 pts
Noting that I(V ) is the average current— 0.3 pts
— page 4 of 4 —
130
131
Results
Gold MedalistsAttila Szabó, Hungary
Hengyun Zhou, People’s Republic of China
Yijun Jiang, People’s Republic of China
Chien-An Wang, Taiwan
Kai-Chi Huang, Taiwan
Paphop Sawasdee, Thailand
Eric Schneider, United States of America
Jun-Ting Hsieh, Taiwan
Siyuan Wei, People’s Republic of China
Chi Shu, People’s Republic of China
Wonseok Lee, Republic of Korea
Wenzhuo Huang, People’s Republic of China
Yu-Ting Liu, Taiwan
Wei-Jen Ko, Taiwan
Rahul Trivedi, India
Lev Ginzburg, Russia
Phot
o by
Hen
ry T
eiga
r
132
Tudor Giurgică-Tiron, Romania
Allan Sadun, United States of America
Kazumi Kasaura, Japan
Nikita Sopenko, Russia
Ding Yue, Republic of Singapore
Lam Ho Tat, Hong Kong
Sooshin Kim, Republic of Korea
Jaemo Lim, Republic of Korea
Lorenz Eberhardt, Germany
Ivan Ivashkovskiy, Russia
Michal Pacholski, Poland
Qiao Gu, Germany
Adrian Nugraha Utama, Indonesia
Albert Samoilenka, Belarus
Puthipong Worasaran, Thailand
Jaan Toots, Estonia
Nathanan Tantivasadakarn, Thailand
Phi Long Ngo, Vietnam
Yuichi Enoki, Japan
Ihar Lobach, Belarus
Veselin Karadzhov, Bulgaria
Ivan Tadeu Ferreira Antunes Filho, Brazil
Ilya Vilkoviskiy, Kazakhstan
Kevin Zhou, United States of America
Ang Yu Jian, Republic of Singapore
Fung Tsz Chai, Hong Kong
Kuan Jun Jie, Joseph, Republic of Singapore
Ngoc Hai Dinh, Vietnam
Huan Yan Qi, Republic of Singapore
gold medalists Photo by Henry Teigar
133
Silver MedalistsAlexandra Vasilyeva, Russia
Bijoy Singh Kochar, India
Jeevana Priya Inala, India
Zoltán Laczkó, Hungary
Dan-Cristian Andronic, Romania
Ramtin Yazdanian, Islamic Republic of Iran
Vladysslav Diachenko, Ukraine
Sebastian Linß, Germany
Kunal Singhal, India
Kohei Kawabata, Japan
Volodymyr Sivak, Ukraine
David Frenklakh, Russia
Vsevolod Bykov, Ukraine
Theodor Misiakiewicz, France
Woojin Kweon, Republic of Korea
Kaisarbek Omirzakhov, Kazakhstan
Huy Quang Le, Vietnam
Ondřej Bartoš, Czech Republic
Jakub Vošmera, Czech Republic
Stanislav Fořt, Czech Republic
Pongsapuk Sawaddirak, Thailand
Suyeon Choi, Republic of Korea
Yordan Yordanov, Bulgaria
Hiromasa Nakatsuka, Japan
Aliaksey Khatskevich, Belarus
Peter Kosec, Slovakia
Chan Cheuk Lun, Hong Kong
Itay Knaan Harpaz, Israel
Tasuku Omori, Japan
Vadzim Reut, Belarus
Filip Ficek, Poland
Daumantas Kavolis, Lithuania
Tanel Kiis, Estonia
Katerina Naydenova, Bulgaria
Jeffrey Yan, United States of America
Kristjan Kongas, Estonia
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Georg Krause, Germany
Yevgen Cherniavskyi, Ukraine
Jonathan Dong, France
Kaur Aare Saar, Estonia
Abdurrahman Akkas, Turkey
Jeffrey Cai, United States of America
Georgijs Trenins, Latvia
Henry Honglei Wu, Canada
Patrik Svancara, Slovakia
Oliver Edtmair, Austria
Amir Yousefi, Islamic Republic of Iran
Lubomír Grund, Czech Republic
Patrik Turzak, Slovakia
Milan Krstajić, Serbia
Atinc Cagan Sengul, Turkey
Ittai Rubinstein, Israel
Mehmet Said Onay, Turkey
Jean Douçot, France
Matias Mannerkoski, Finland
Soo Wah Ming, Wayne, Republic of Singapore
Mussa Rajamov, Kazakhstan
Lai Kwun Hang, Hong Kong
Tudor Ciobanu, Romania
Roland Papp, Hungary
Eric Wieser, United Kingdom
Nurzhas Aidynov, Kazakhstan
Adam Brown, United Kingdom
Žygimantas Stražnickas, Lithuania
Paul Kirchner, France
Atli Thor Sveinbjarnarson, Iceland
Sebastian Florin Dumitru, Romania
Péter Juhász, Hungary
Christoph Schildknecht, Switzerland
Wai Hong Lei, Macao, China
Andres Erbsen, Estonia
silver medalists Photo by Karl Veskus
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Bronze MedalistsMilan Kornjača, Serbia
Petar Tadic, Montenegro
Cristian Zanoci, Moldova
Nicholas Salmon, Australia
Oguzhan Can, Turkey
Mustafa Selman Akinci, Turkey
Volodymyr Rozsokhovatskyi, Ukraine
Richard Thorburn, United Kingdom
Andrej Vlcek, Slovakia
Xuan Hien Bui, Vietnam
Ivan Senyushkin, Kazakhstan
Chen Solomon, Israel
Martijn van Kuppeveld, Netherlands
Jonathan Lay, Australia
Mohamad Ansarifard, Islamic Republic of Iran
Supanut Thanasilp, Thailand
Jan Rydzewski, Poland
Johan Runeson, Sweden
Iiro Lehto, Finland
Mehrdad Malak Mohammadi,
Islamic Republic of Iran
Hólmfríður Hannesdóttir, Iceland
Simon Pirmet, France
Simon Blouin, Canada
Roberto Albesiano, Italy
Viet Thang Dinh, Vietnam
Pulkit Tandon, India
Sajad Khodadadian, Islamic Republic of Iran
Christoph Weis, Austria
Kacper Oreszczuk, Poland
Yigal Zegelman, Israel
Frank Bloomfield, United Kingdom
Tamara Šumarac, Serbia
Vu Phan Thanh, Germany
Federica Maria Surace, Italy
Christopher Whittle, Australia
Francisco Machado, Portugal
Konstantin Gundev, Bulgaria
Martin Stadler, Austria
Martin Raszyk, Czech Republic
Thanh Phong Lê, Switzerland
Aliaksandr Yankouski, Belarus
Sepehr Ebadi, Canada
Domen Ipavec, Slovenia
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Ruben Doornenbal, Netherlands
Yun Jia (Melody) Guan, Canada
Ion Toloaca, Moldova
Wai Pan Si, Macao, China
Eden Segal, Israel
Jose Luciano De Morais Neto, Brazil
Tobias Karg, Austria
Sergi Chalalauri, Georgia
Meylis Malikov, Turkmenistan
Virab Gevorgyan, Armenia
Tristan Downing, Canada
Luqman Fathurrohim, Indonesia
Aitor Azemar, Spain
Dinis Cheian, Moldova
Lo Hei Chun, Hong Kong
Eric Huang, Australia
Jorge Torres-Ramos, Mexico
Roberta Răileanu, Romania
Jovan Blanuša, Serbia
Michele Fava, Italy
Selver Pepic, Bosnia and Herzegovina
Vardan Avetisyan, Armenia
Maximilian Ruep, Austria
Saba Kharabadze, Georgia
Sebastian Käser, Switzerland
Giorgi Kobakhidze, Georgia
Troy Figiel, Netherlands
Áron Dániel Kovács, Hungary
Ilija Burić, Serbia
Nicoleta Colibaba, Moldova
Abdullah Alsalloum, Saudi Arabia
Haji Piriyev, Azerbaijan
Javier Mendez-Ovalle, Mexico
Ka Fai Chan, Macao, China
Guilherme Renato Martins Unzer, Brazil
Muhammad Taimoor Iftikhar, Pakistan
Emmanouil Vourliotis, Greece
Jemal Shengelia, Georgia
Lara Timbo Araujo, Brazil
Ilie Popanu, Moldova
Francesc-Xavier Gispert Sánchez, Spain
Battsooj Bayarsaikhan, Mongolia
Samuel Bosch, Croatia
Koen Dwarshuis, Netherlands
Shinjini Saha, Bangladesh
Leandro Salemi, Belgium
Chan Lon Wu, Macao, China
Adhamzhon Shukurov, Tajikistan
Peter Budden, United Kingdom
bronze medalists Photo by Henry Teigar
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Honorable MentionsJyri Maanpää, Finland
Sandro Maludze, Georgia
Siobhan Tobin, Australia
Simão João, Portugal
Tomas Čerškus, Lithuania
Osama Yaghi, Syrian Arab Republic
Koay Hui Wen, Malaysia
Kasper Tolborg, Denmark
Ramadhiansyah Ramadhiansyah, Indonesia
Ooi Chun Yeang, Malaysia
Dominic Schwarz, Switzerland
Bruno Buljan, Croatia
Bartlomiej Zawalski, Poland
Shovon Biswas, Bangladesh
Homoud Alharbi, Saudi Arabia
Bilguun Batjargal, Mongolia
Mantas Abazorius, Lithuania
Nudzeim Selimovic, Bosnia and Herzegovina
Ahmed Maksud, Bangladesh
Arttu Yli-Sorvari, Finland
Farid Mammadov, Azerbaijan
Karlo Sepetanc, Croatia
Romain Falla, Belgium
Fotios Vogias, Greece
Miha Zgubič, Slovenia
Tsogt Baigalmaa, Mongolia
Mohamed Alrazzouk, Syrian Arab Republic
Battushig Myanganbayar, Mongolia
Aram Mkrtchyan, Armenia
Munkhtsetseg Battulga, Mongolia
Michalis Halkiopoulos, Greece
Nikolaj Theodor Thams, Denmark
Federico Re, Italy
Stefan Alexis Sigurðsson, Iceland
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Martin Vlashi, Italy
Mathias Stichelbaut, Belgium
Tapio Hautamäki, Finland
Jurij Tratar, Slovenia
Wai Hei Hoi, Macao, China
Ali Alhulaymi, Saudi Arabia
Kaloyan Darmonev, Bulgaria
Pedro Paredes, Portugal
Laura Gremion, Switzerland
Yeoh Chin Vern, Malaysia
Grgur Simunic, Croatia
Matheus Marreiros, Portugal
Thijs van der Gugten, Netherlands
Salizhan Kylychbekov, Kyrgyzstan
Molte Emil Strange Andersen, Denmark
Dale Alexander Hughes, Ireland
Luka Ivanovskis, Latvia
I Made G.N. Kumara, Indonesia
Mohammad Alhejji, Saudi Arabia
David Trillo Fernández, Spain
Aleksandr Petrosyan, Armenia
Vladimir Pejovic, Montenegro
Andres Rios Tascon, Colombia
Eduardo Acosta-Reynoso, Mexico
Werdi Wedana Gunawan, Indonesia
Ghadeer Shaaban, Syrian Arab Republic
Maris Serzans, Latvia
Pétur Rafn Bryde, Iceland
Kemal Babayev, Turkmenistan
honorable mentioned students Photo by Henry Teigar
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Special PrizesIPhO 2012 absOLuTE wInnErAttila Szabó, Hungary
sPEcIaL PrIzEs FOr ThE bEsT sOLuTIOn OF ThE ExPErImEnTaLChristoph Schildknecht, Switzerland
Ivan Ivashkovskiy, Russia
Chi Shu, People’s Republic of China
sPEcIaL PrIzEs FOr ThE bEsT sOLuTIOn OF ThE ThEOrETIcaLEric Schneider, United States of America
bEsT In ExPErImEnTKai-Chi Huang, Taiwan
bEsT In ThEOryAttila Szabó, Hungary
PrIzEs OF assOcIaTIOn OF asIa PacIFIc PhysIcaL sOcIETIEsJeevana Priya Inala, India
Hengyun Zhou, People’s Republic of China
PrIzE OF EurOPEan PhysIcaL sOcIETyAttila Szabó, Hungary
sPEcIaL PrIcE FOr ThE bEsT gIrL IPhO 2012Alexandra Vassiljeva, Russia
sPEcIaL PrIcE FOr ThE bEsT EsTOnIan sTudEnTJaan Toots, Estonia
sPEcIaL PrIcE FOr ThE bEsT InnOvaTIvE sOLuTIOnLev Ginzburg, Russia
Absolute winner Attila Szabó with Jaan Kalda
Photo by Henry Teigar
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statistics of the competition resultsJaan Kalda
Academic Committee of IPhO 2012
Let us analyze the overall difficulty of the problem set using the score-rank rela-
tionship presented in Fig 1. For an ideal well-balanced problem set, the net scores
of the contestants should be equi-spaced: the point-rank graph should be a nearly
straight line connecting the winner with a maximal score and ending at the last-
place-owner with 0 points. Bearing in mind that the performance of the contest-
ants is not fully predictable, such a problem set is hardly achievable. One can also
argue that in order to determine the absolute winners and gold medalists more
reliably, it is better to have larger point differences at the small ranks.
Fig 1. Point-rank graph for the total score; blue line - before moderation;
black line - after moderation.
The graph in Fig 1 tells us that the simple questions might have been slightly too
simple: the linear trend at the middle of the graph breaks at the right-hand-side
of the graph, where the curves turn steeply down. Meanwhile, the difficult ques-
tions were difficult, indeed, and provided a good separation between the very best
contestants; this is evidenced by another steeper segment at the left-hand-side.
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It can also be of interest to compare the curves before and after the moderation.
This year, the typical distance between the two curves is less than 1 pt; without
going into details, one can say that this is quite a reasonable result. Ideally, a
smaller distance between the two curves should imply a better initial grading.
However, another factor here is the “flexibility” of the graders during the mod-
eration. For a partially solved problem, the decision of how many points should
be awarded is always slightly subjective. While the markers try to settle at the
middle of the uncertainty interval (to provide as fair a grading as possible), the
leaders tend to ask as many points as possible. (This is why the graders had been
instructed to grade “generously”: if doubting between two options, opt for more
points.) However, being too generous, i.e. giving too many marks for a partially
solved problem, will be unfair in respect to those who have solved the problem
flawlessly. Thus, regardless of how good the initial grading was, there will always
be some room for negotiation during the moderations, and a score shift is some-
times explained by the willingness of the graders to accept the arguments of the
leaders. As an example, this year there were two “compromises” due to which
the graders went through all the examination papers a second time. For all the
experimental tasks, marks were added for a correct plotting of wrong data points,
and for the Problem T3-iv and T3-v, partial credit for an incorrectly written first
law of thermodynamics was increased.
Next, let us have a look at the distribution of the theoretical and experi-
mental marks separately.
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Fig 2. Point-rank graph for the scores of the experiment and for the theory
For the theoretical examination, the linear trend extends down to the right-
most corner of the graph. Therefore, none of the questions were too simple! As for
the experiment, the graph is qualitatively very similar to the graph of the total
scores.
Let us dwell even deeper into detail and have a look at the distribution of
points for all the theoretical problems.
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Fig 3. Point-rank graphs for the scores of the Problem T1, T2 and T3.
The simple questions of problem T1 were, indeed, simple: the total cost of
these was 3x0.8=2.4 pts and ca 40% of the contestants got at least this much.
However, the supposedly medium difficulty questions, each worth 1.2 and total-
ling in 3.6 pts, were actually quite difficult: answering all the simple and medium
questions would have resulted in 6 pts, and only 6% of contestants got 6 points or
more.
The contestants with top scores are as follows (all gold medalists).
12.1: Attila Szabó (HUN);
10.9: Eric Schneider (USA);
10.2: Hengyun Zhou (CHN);
10.1: Yijun Jiang (CHN);
9.3: Ilya Vilkoviskiy (KAZ);
9.1: Paphop Sawasdee (THA), Wenzhuo Huang (CHN);
8.7: Chien-An Wang (TWN);
8.2: Wonseok Lee (KOR);
7.7: Jun-Ting Hsieh (TWN);
7.6: Ding Yue (SGP);
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7.5: Kuan Jun Jie, Joseph (SGP);
7.2: Sooshin Kim (KOR), Siyuan Wei (CHN);
7.0: Phi Long Ngo (VNM).
Next, about Problem T2. As you can see, this is a problem with a perfect balance
between simple and difficult questions: there is almost a linear line connecting
the upper left corner with the lower right corner. The contestants with top scores
are as follows (all gold medalists, unless otherwise noted).
8.0 pts: Chien-An Wang (TWN), Yijun Jiang (CHN);
7.9 pts: Jun-Ting Hsieh (TWN), Tudor Giurgică-Tiron (ROU);
7.8 pts: Hengyun Zhou (CHN), Chi Shu (CHN), Rahul Trivedi (IND), David
Frenklakh (RUS, Silver), Kacper Oreszczuk (POL, Bronze),
7.7 pts: Wenzhuo Huang (CHN), Siyuan Wei (CHN), Jaemo Lim (KOR), Tanel
Kiis (EST, Silver).
Finally, Problem T3. A slight score saturation can be observed for this problem:
the curve “hits the roof” (i.e. the maximal value of 9.0 pts) at the upper left cor-
ner. There would have been probably a better balance between difficult and easy
questions if the hint about Kepler’s laws were not given in the text of the problem.
However, including the hint was the wish of the International Board, and the
problem set as a whole was difficult enough even with the hint included ...
The contestants with a full score (9.0 pts; all gold medalists): Attila Szabó
(HUN), Paphop Sawasdee (THA), Chien-An Wang (TWN), Siyuan Wei (CHN),
Yuichi Enoki (JPN), Rahul Trivedi (IND), Puthipong Worasaran (THA), Tudor
Giurgică-Tiron (ROU), Ngoc Hai Dinh (VNM), Alexandra Vasilyeva (RUS, Silver),
Volodymyr Sivak (UKR, Silver), Bijoy Singh Kochar (IND, Silver), Nurzhas Aidynov
(KAZ, Silver), Cristian Zanoci (MDA, Bronze).
Next, let us have a look at the experimental problems.
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Fig 4. Point-rank graphs for the scores of the Problems E1 and E2.
Problem E1 has a nice distribution at the upper left corner, but too steep a
fall-off at the right edge – the simplest tasks of this problem were perhaps too
simple. The contestants with top scores are as follows (all gold medalists, unless
otherwise noted).
10: Jaan Toots (EST);
9.9: Kai-Chi Huang (TWN), Ivan Ivashkovskiy (RUS);
9.8: Wei-Jen Ko (TWN);
9.7: Attila Szabó (HUN), Siyuan Wei (CHN);
9.6: Allan Sadun (USA);
9.5: Hengyun Zhou (CHN), Chien-An Wang (TWN);
9.4: Wonseok Lee (KOR);
9.3: Jun-Ting Hsieh (TWN);
9.2: Eric Schneider (USA), Wenzhuo Huang (CHN);
9.1: Ngoc Hai Dinh (VNM), Alexandra Vasilyeva (RUS, Silver), Chi Shu (CHN).
Meanwhile, Problem E2 was intended to be a difficult problem, aimed at
finding the winner of the best experimentalist’s prize. And difficult it was: it
had actually no easy tasks, as evidenced by a concave shape of the curve. The con-
testants with top scores are as follows (all gold medalists, unless otherwise noted).
8.8: Chi Shu (CHN);
8.5: Kai-Chi Huang (TWN);
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8.3: Christoph Schildknecht (CHE, Silver);
8.1: Ivan Ivashkovskiy (RUS);
7.7: Attila Szabó (HUN);
7.5: Hengyun Zhou (CHN), Huan Yan Qi (SGP);
7.4: Lev Ginzburg (RUS);
7.2: Abdurrahman Akkas (TUR, Silver);
7: Kristjan Kongas (EST, Silver);
6.9: Yu-Ting Liu (TWN), Adam Brown (GBR, Silver);
6.7: Kevin Zhou (USA), Frank Bloomfield (GBR, Bronze).
And now, it is time to have a look at the most difficult questions (tasks). Let us
start with the three parts of Problem 1.
Fig 5. Point-rank graphs for the scores of Tasks T1A , T1B, and T1C.
I was quite sure that question iii of Part A would be very difficult for the con-
testants, and question iii of Part C would be extremely difficult, and I was not
mistaken. However, I did believe that question iii of Part B was not that difficult
(just difficult, not “very” or “extremely”), and my colleagues from the Academic
Committee agreed. However, here we were mistaken: Part B turned out to be the
most difficult part!
Part 1A: only ca 20% of students were able to figure out the correct shape
of the trajectory. Meanwhile, there was also a considerable number of those
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who got everything correctly done, including q. iii! This is an interesting case
because in order to be able to solve this problem only a moderate physical educa-
tion is needed. This is evidenced by the fact that among the best solvers of Part
1A, there are several students whose overall results were not so good. One can
only hypothesize that had they passed a full course of physics covering all the
Syllabus of the IPhO, they would have been able to get gold medals. The contest-
ants with top scores are as follows (all gold medalists, unless otherwise noted).
4.5 pts: Attila Szabó (HUN), Hengyun Zhou (CHN), Eric Schneider (USA), Wenzhuo
Huang (CHN), Yijun Jiang (CHN), Rahul Trivedi (IND), Ilya Vilkoviskiy (KAZ),
Kuan Jun Jie (SGP), Joseph Ramadhiansyah Ramadhiansyah (IDN, Honourable
Mention);
4.4 pts: Paphop Sawasdee (THA), Jeffrey Cai (USA, Silver), Puthipong Worasaran
(THA), Nathanan Tantivasadakarn (THA);
4.2 pts: Michele Fava (ITA, Bronze);
4.1 pts: Ding Yue (SGP);
4 pts: Hakon Tásken (NOR, Participation Certificate).
Part 1B: the list of top-solvers is shorter than before because all the others just
did not get enough marks for q. iii to be qualified as someone who really solved
this problem. As usual, everyone below got a gold medal, unless otherwise noted.
3.9 pts: Jun-Ting Hsieh (TWN);
3.8 pts: Attila Szabó (HUN);
3.6 pts: Sooshin Kim (KOR);
3.5 pts: Yijun Jiang (CHN);
3.4 pts: Siyuan Wei (CHN), Kai-Chi Huang (TWN), Wonseok Lee (KOR);
3.3 pts: Ihar Lobach (BLR);
3.1 pts: Wenzhuo Huang (CHN);
3 pts: Georgijs Trenins (LVA, Silver), Karlo Sepetanc (HRV, Honourable Mention).
Finally, Part 1C: note that about half of those who performed very well here
(listed below) lost 0.2 in q. i for drawing too curved field lines. There were only
four contestants who got the idea of magnetic charges and realized it flawlessly
(these are the first three in the list below, and Kunal Singhal). However, owing
to the fact that there is also another way of calculating the force (via integrating
over dipoles), several contestants got a correct estimate of the force, and, thereby,
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collected enough marks to be listed below
4.5 pts: Ilya Vilkoviskiy (KAZ);
4.3 pts: Chien-An Wang (TWN);
4.2 pts: Paphop Sawasdee (THA);
4 pts: Wonseok Lee (KOR), Wei-Jen Ko (TWN);
3.9 pts: Eric Schneider (USA);
3.8 pts: Attila Szabó (HUN), Yuichi Enoki (JPN), Hengyun Zhou (CHN);
3.7 pts: Phi Long Ngo (VNM);
3.6 pts: Kazumi Kasaura (JPN);
3.5 pts: Kunal Singhal (IND, Silver);
3.4 pts: Yu-Ting Liu (TWN);
3 pts: Jun-Ting Hsieh (TWN), Siyuan Wei (CHN), Ivan Tadeu Ferreira Antunes
Filho (BRA).
The rest of the theoretical test was not that difficult (q iii of Problem T3 would
have been quite difficult, but with the hint inserted by the International Board it
no longer was). So we won’t dwell more on the theoretical results, and we’ll switch
to the really tricky experimental tasks: A-iv and B of Problem E2.
Fig 6. Point-rank graphs for the scores of the Tasks E2B and E2A-iv.
In the case of Task A-iv, the number of those who really got the correct idea
how to measure C(V) was really small - essentially only those who are listed below.
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2.6 pts: Kuan Jun Jie, Joseph (SGP), Kai-Chi Huang (TWN), Lev Ginzburg (RUS),
Ivan Ivashkovskiy (RUS);
2.5 pts: Chi Shu (CHN), Qiao Gu (DEU), Kevin Zhou (USA), Allan Sadun (USA)
2.4 pts: Yu-Ting Liu (TWN), Kristjan Kongas (EST, Silver), Adrian Nugraha
Utama (IDN), Tudor Giurgică-Tiron (ROU), Sebastian Linß (DEU, Silver).
Finally, Task B. There was a surprisingly small number of those contestants
who noticed that the difference in the graphs of Part A and Part B is localized to the
region of negative differential resistance. As for the explanation of the phenome-
non (which consists of three key elements), none of the contestants managed to
list all the key elements flawlessly, and the only one to get it almost done (with
some omissions in the average current part) was Christoph Schildknecht; the
next two in the list below mentioned one key element. And so, the best results
for Task 2B:
2.9 pts: Christoph Schildknecht (CHE, Silver);
2.3 pts: Attila Szabó (HUN);
2.1 pts: Chi Shu (CHN);
2 pts: Kai-Chi Huang (TWN), Luka Ivanovskis (LVA, Honourable Mention).
For those who want to go beyond this statistical analysis, there is also an Excel
file http://www.ioc.ee/~kalda/ipho/Results_for_web.xlsx (the names of those who got less
than 12.4 pts have been stripped).
The problems of the 43rd IPhO have been thought to be difficult, and it has
even been stated that the problem set was the most difficult one during the last
20 years. In order to make a comparative study about how difficult the problems
actually were several types of data are needed, which are not freely available for all
the Olympiads. Still, I managed to get more or less what is needed (overall number
of participants, number of medals, medal boundaries in points, the scores of the
absolute winners) for the period covering 1994–2012. The graph is shown below.
(The last two digits of the year are shown alongside the curve – except for some
curves in the central densely populated region; note that the curves which are
based only on the number of medals and on the medal boundaries are interpolated
and smooth.)
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Fig 7. Point-rank graphs for the overall scores of the last nineteen IPhO-s
(for those years for which the complete data were unavailable, the curves are
interpolated between the datapoints corresponding to the medal boundaries).
Even with these data, the IPhOs apart as much as 19 years are not fully compa-
rable. I have a feeling that, in average, the preparation level of the leading group of
contestants has risen significantly. So, the graph here does not allow comparison
of the absolute difficulties of the problems, but only relative ones – relative to the
preparation level of the students. One should also bear in mind that the scores of
absolute winners have a high intrinsic variability (there is essentially no statistical
averaging); c.f. this year: the first and second places were separated by a huge
margin of 3 pts.
Therefore, the conclusion is that the claim, about this being the most difficult
problem set in the last 20 years, was slightly exaggerated. The problems in Beijing
in 1994 were even more difficult, at least in relative terms, and at least when
leaving out the contestants with ranks from 2 to 6.
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International boardMinutesmInuTEs OF ThE mEETIngs OF ThE InTErnaTIOnaL bOard durIng ThE 43rd InTErnaTIOnaL PhysIcs OLymPIad In TaLLInn, EsTOnIa, JuLy 15Th – 24Th, 2012
1. A total number of 378 contestants from the following 80 countries were
present at the 43rd International Physics Olympiad:
Albania, Armenia, Australia, Austria, Azerbaijan, Bangladesh, Belarus,
Belgium, Bolivia, Bosnia & Herzegovina, Brazil, Bulgaria, Canada, China,
Colombia, Croatia, Cyprus, Czech Republic, Denmark, El Salvador, Estonia,
Finland, France, Georgia, Germany, Greece, Hong Kong, Hungary, Iceland, India,
Indonesia, Iran, Ireland, Israel, Italy, Japan, Kazakhstan, Kuwait, Kyrgyzstan,
Latvia, Liechtenstein, Lithuania, Macau, Macedonia, Malaysia, Mexico,
Moldova, Mongolia, Montenegro, Netherlands, Nigeria, Norway, Pakistan,
Poland, Portugal, Puerto Rico, Republic of Korea, Romania, Russia, Saudi Arabia,
Serbia, Singapore, Slovakia, Slovenia, South Africa*, Spain, Sri Lanka, Suriname,
Sweden, Switzerland, Syria, Taiwan, Tajikistan, Thailand, Turkey, Turkmenistan,
Ukraine, United Kingdom, USA, Vietnam.
* : new country invited by the Organizing Committee to the IPhO this year.
2. Results of marking the papers by the organizers were presented.
The best score (45.8 points) was achieved by Mr. Attila Szabó from Hungary (the
overall winner of the 43rd IPhO). The following limits for awarding the medals and
the honorable mention were established according to the Statutes:
Gold Medal 31.0 points
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Silver Medal 23.9 points
Bronze Medal 17.2 points
Honorable Mention 12.4 points
According to the above limits, 45 Gold Medals, 71 Silver Medals, 92 Bronze
Medals, and 63 Honorable Mentions were awarded. The grade lists of the awardees
were distributed to all the delegation leaders in print.
3. In addition to the regular prizes, the following special prizes were awarded:
The Overall Winner
Mr. Attila Szabó, Hungary
Best in Theory
Mr. Attila Szabó, Hungary
Best in Experiment
Mr. Kai-Chi Huang, Taiwan
Special Prize for the Best Solution of the Theoretical
Mr. Eric Schneider, United States of America
Special Prizes for the Best Solution of the Experimental
Mr. Christoph Schildknecht, Switzerland
Mr. Ivan Ivashkovskiy, Russia
Mr. Chi Shu, People`s Republic of China
Special Prize for the best innovative solution
Mr. Lev Ginzburg, Russia
Special Prize for the best Estonian student
Mr. Jaan Toots, Estonia
Special Prize for the Best Girl IPhO 2012
Ms. Alexandra Vassiljeva, Russia
Prize of European Physical Society
Mr. Attila Szabó, Hungary
Prizes of Association of Asia Pacific Physical Societies
Ms. Jeevana Priya Inala, India
Mr. Hengyun Zhou, People`s Republic of China
4. The following three leaders were designated by the International Board
to serve as consultants to the local Academic Committee for grading the
examination papers :
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Prof. Niels Hartling (Denmark), Prof. Suwan Kusamran (Thailand), and Prof.
Helmuth Mayr (Austria)
5. The International Board discussed a proposal, presented by the President of
the IPhO, Dr. Hans Jordens, of establishing a legal body of a fund-raising
bank account, named “the Foundation of International Physics Olympiad”,
which would be affiliated with the IPhO. The objective of the non-profit
Foundation is to raise funds for helping economy-weak countries to send
students to participate in the IPhO and for providing financial support to
future hosting countries in case of need. The Statutes of the Foundation
which were examined and revised by the Advisory Committee had been
disseminated to all leaders three months prior to the IB meeting. The
proposal was accepted by the International Board with an overwhelming
majority in the affirmative. The Board of the Foundation comprises at least
3 members. In addition to the IPhO President and Secretary, a Treasurer
will be nominated and elected in the next IPhO.
6. The term of the current President of the IPhO will expire in 2013. Since
no specific regulation on the election of presidency is stated in the IPhO
Statutes, the International Board unanimously agreed to adopt the same
procedures as for the electing of the IPhO Secretary, as specified in the
Regulations to the Article #8, for the election of a new president for the
term 2013–2018.
7. The International Board discussed a proposal submitted by Icelandic
leader, Dr. Martin Swift, which asked to modify the regulations on using
“non-programmable” pocket calculators as stated in the Article #5 of the
IPhO Statutes. The International Board approved to form a specific commit-
tee to deliberate on revising the concerned regulations. The IPhO Secretariat
has invited the following three leaders to form the calculator-committee:
Dr. Martin Swift (Iceland), Dr. Eli Raz (Israel), and Dr. Matthew Verdon
(Australia).
8. The IPhO Secretariat received an official letter from the Korean Physical
Society which stated that the Republic of Korea would prefer to host the
154
58th IPhO in 2027 instead of the 63th IPhO in 2032 as previously proposed.
The request was accepted by the International Board.
9. The President of the IPhOs, Prof. Hans Jordens, acting on behalf of all the
participants, expressed his gratitude to the Minister of Education and
Research, Prof. Dr. Jaak Aaviksoo, and deep thanks to Mrs Ene Koitla (Head
of the Organizing Committee), Prof. Jaak Kikas (Head of the Academic
Committee), Dr. Jaan Kalda (Co-head of the Academic Committee), Dr.
Viire Sepp (Academic Secretary of the Steering Committee) and all other
members involved in organizing the IPhO2012 for excellent preparation
and conduction of the 43rd International Physics Olympiad. Deep thanks
were also conveyed to Tallinn University of Technology, the University of
Tartu, the Estonian Physical Society, and the Ministry of Education and
Research of Estonia, all the sponsors, graders, guides and other people who
contributed to the success of the Olympiad.
10. The Danish delegates disseminated printed materials about the 44th IPhO
in 2013 to all the delegations and described the present state of the prepara-
tory works to ensure smooth organization of the next Olympiad.
11. At the Closing Ceremony of the Olympiad, on behalf of the organizers of
the next International Physics Olympiad, Prof. Niels Hartling announced
that the 44th International Physics Olympiad would be organized in
Copenhagen, Denmark from July 7th to 15th, 2013 and cordially invited
all the participating countries to attend the competition.
(signature)
......................
Prof. Hans Jordens
President of the IPhO
(signature)
......................
Prof. Ming-Juey Lin
Secretary of the IPhO
(signature)
......................
Prof. Jaak Kikas
Head of the Academic
Committee of IPhO2012
Tallinn, Estonia
July 22nd, 2012
155Photos by Merily Salura, Henry Teigar and Karl Veskus
156
157
StatutessTaTuTEs OF ThE InTErnaTIOnaL PhysIcs OLymPIads
Version accepted in 1999 in Padova (Italy)
Revised: 2000 - Leicester (Great Britain)
2001 - Antalya (Turkey)
2002 - Bali (Indonesia)
2004 - Pohang (Korea)
2006 - Singapore
2008 - Hanoi (Vietnam)
§1
In recognition of the growing significance of physics in all fields of science and
technology, and in the general education of young people, and with the aim of
enhancing the development of international contacts in the field of school educa-
tion in physics, an annual physics competition has been organised for secondary
school students. The competition is called the International Physics Olympiad and
is a competition between individuals.
§2
The competition is organised by the Ministry of Education, the Physical Society
or another appropriate institution of one of the participating countries on whose
territory the competition is to be conducted. The organising country is obliged
to ensure equal participation of all the delegations, and to invite teams from all
those countries that participated during the last three years. Additionally, it has
the right to invite other countries. The list of such new countries must be pre-
sented to Secretariat of the IPhOs (# 8) at least six months prior to the competition.
Within two months the Secretariat has the right to remove, after consultations
with the Advisory Committee (# 8), from the suggested list the teams that in
opinion of Secretariat or Advisory Committee do not meet the criteria of partici-
pation in the IPhOs. The new countries not accepted by the Secretariat or Advisory
Committee may, however, participate as “guest teams” but such participation
does not create any commitments with respect to inviting these countries to the
158
next competition(s).
No country may have its team excluded from participation on any political
reasons resulting from political tensions, lack of diplomatic relations, lack of rec-
ognition of some country by the government of the organising country, imposed
embargoes and similar reasons. When difficulties preclude formal invitation of
the team representing a country students from such a country should be invited
to participate as individuals.
The competition is conducted in the friendly atmosphere designed to promote
future collaborations and to encourage the formation of friendship in the scien-
tific community. Therefore all possible political tensions between the participants
should not be reflected in any activity during the competition. Any political activ-
ity directed against any individuals or countries is strictly prohibited.
§3
Each participating country shall send a delegation, normally consisting of five stu-
dents (contestants) and two accompanying persons (delegation leaders) at most.
The contestants shall be students of general or technical secondary schools
i.e. schools which cannot be considered technical colleges. Students who have
finished their school examinations in the year of the competition can be members
of the team as long as they have not commenced their university studies. The age
of the contestants should not exceed twenty years on June 30th of the year of the
competition.
The delegation leaders must be specialists in physics or physics teachers, capa-
ble of solving the problems of the competition competently. Each of them should
be able to speak English.
§4
The Organisers of the Olympiad determine in accordance to the programme the
day of arrival and the day of departure as well as the place in their country from
which the delegations are supposed to arrive and depart. The costs for each dele-
gation as a result of activities connected to the Olympiad from the day of arrival
till the day of departure are covered by the Organising Committee.
§5
The competition shall be conducted over two days, one for the theoretical exam-
ination and one for the experimental examination. There will be at least one full
159
day of rest between the examinations.
The theoretical examination shall consist of three theoretical problems and
shall be of five hours total duration.
The experimental examination shall consist of one or two problems and shall
be of five hours total duration.
Contestants may bring into the examination drawing instruments and non-pro-
grammable pocket calculators. No other aids may be brought into the examination.
The theoretical problems should involve at least four areas of physics taught at
secondary school level, (see Syllabus). Secondary school students should be able
to solve the competition problems with standard high school mathematics and
without extensive numerical calculation.
The competition tasks are chosen and prepared by the host country and have
to be accepted by the International Board (§ 7).
The host country has to prepare at least one spare problem, which will be pre-
sented to the International Board if one of the first three theoretical problems is
rejected by two thirds of members of the International Board. The rejected problem
cannot be considered again.
§6
The total number of marks awarded for the theoretical examination shall be 30 and
for the experimental examination 20. The competition organiser shall determine
how the marks are allocated within the examinations.
After preliminary grading (prior to discussion of the grading with the dele-
gation leaders) the organizers establish minima (expressed in points) for Gold
Medals, Silver Medals, Bronze Medals and Honourable Mentions according to the
following rules:
Gold Medals should be awarded to 8% of the contestants (rounded up the
nearest integer).
Gold or Silver Medals should be awarded to 25% of the contestants (rounded
up the nearest integer).
Gold, Silver or Bronze Medals should be awarded to 50% of the contestants
(rounded up the nearest integer).
An Olympic Medal or Honourable Mention should be awarded to 67% of the
contestants (rounded up the nearest integer).
The minima corresponding to the above percentages should be expressed
160
without rounding. The suggested minima shall be considered carried if one
half or more of the number of the Members of the International Board cast
their vote in the affirmative.
Results of those candidates who only receive a certificate of participation
should strictly remain to the knowledge of the Members of the International
Board and persons allowed to attend its meetings.
§7
The governing body of the IPhO is the International Board, which consists of the
delegation leaders from each country attending the IPhO.
The chairman of the International Board shall be a representative of the organ-
ising country when tasks, solutions and evaluation guidelines are discussed and
the President of the IPhO in all other topics.
A proposal placed to the International Board, except Statutes, Regulations and
Syllabus (see§ 10), shall be considered carried if more than 50% of all delegation
leaders present at the meeting vote in the affirmative. Each delegation leader
is entitled to one vote. In the case of equal number of votes for and against, the
chairman has the casting vote. The quorum for a meeting of the International
Board shall be one half of those eligible to vote.
The International Board has the following responsibilities:
to direct the competition and supervise that it is conducted according to
the regulations;
to ascertain, after the arrival of the competing teams, that all their mem-
bers meet the requirements of the competition in all aspects. The Board will
disqualify those contestants who do not meet the stipulated conditions;
to discuss the Organisers’ choice of tasks, their solutions and the suggested
evaluation guidelines before each part of the competition. The Board is author-
ised to change or reject suggested tasks but not to propose new ones. Changes
may not affect experimental equipment. There will be a final decision on the
formulation of tasks and on the evaluation guidelines. The participants in the
meeting of the International Board are bound to preserve secrecy concerning
the tasks and to be of no assistance to any of the participants;
to ensure correct and just classification of the students. All grading has to
be accepted by the International Board;
to establish the winners of the competition and make a decision concerning
161
presentation of the medals and honourable mentions. The decision of the
International Board is final;
to review the results of the competition;
to select the countries which will be assigned the organisation of future
competitions;
to elect the members of the Secretariat of the IPhO.
§8
The long-term work involved in organising the Olympiads is co-ordinated by a
Secretariat for the International Physics Olympiads. This Secretariat consists of the
President and Secretary. They are elected by the International Board for a period of
five years when the chairs become vacant.
The President and the Secretary of the IPhO should be invited to the Olympiads
as the members and heads of the International Board, their relevant expenses
should be paid by the organizers of the competition. The President and the
Secretary should not be leaders of any national team.
There shall be an Advisory Committee convented at the President of the IPhOs.
The Advisory Committee consists of:
1. The President,
2. The Secretary,
3. The host of the past Olympiad,
4. The hosts of the next two Olympiads,
5. Such other persons appointed by the President.
§9
The working language of the IPhO is English.
The competition problems should be presented to the International Board in
English, Russian, German, French and Spanish.
The solutions to the problems should be presented in English.
It is the responsibility of the delegation leaders to translate the problems into
languages required by their students.
These statutes and other IPhO-documents shall be written in English.
Meetings of the International Board shall be held in English.
§10
These statutes are supplemented by Regulations concerning the details of the
162
organisation the Syllabus mentioned in § 5.
Proposals for amendment to these Statutes and the supplementing documents
may be submitted to the president or his nominee no later than December 15th
prior to consideration.
The President shall circulate, no later than March 15th, all such proposals
together with the recommendation of the President’s Advisory Committee, to
the last recorded address of each delegation leader who attended at the last IPhO.
Such proposals shall be considered by a meeting of the International Board at the
next IPhO and shall be considered carried if
in case of Statutes and Syllabus two thirds or more and
in case of Regulations more than one half
of the number of the members of the International Board present at the
meeting cast their vote in the affirmative. Such changes shall take effect
from the end of the current IPhO and cannot affect the operation of the
competition in progress. The vote can only take place if at least 2/3 of the
all leaders are present at the meeting.
§11
Participation in an International Physics Olympiad signifies acceptance of the
present Statutes by the Ministry of Education or other institution responsible for
sending the delegation.
163
rEguLaTIOns assOcIaTEd wITh ThE sTaTuTEs OF ThE InTErnaTIOnaL PhysIcs OLymPIads
Regulations to §2
The Ministry of Education, or the institution organising the competition, allots
the task of preparation and execution of the Competition to an appropriate body.
Official invitations to the participating countries should be sent at least six
months before the Olympiad. They normally are sent to the national institution
that sent the delegation to the previous Olympiad. Copies of the invitation are also
sent to the previous years’ delegation leaders. The invitation should specify the
place and time of the Competition plus the address of the organising secretariat.
Countries wishing to attend the current IPhO must reply to the invitation before
March 15, nominating a contact person. Each participating country must in addition
supply the host country with the contestants’ personal data (surname, given name,
sex, address, date of birth and address of school) by May 15 or as soon as possible.
The host country is only obliged to invite delegations from countries that partici-
pated in one of the last three competitions. It may refuse
applications for participation from any other country
applications from participating countries not belonging to the delegation
as defined in §3 (observers, guests).
Each country should, within five years of entry, declare its intention to host for
a future Olympiad, suggesting possible years. A country that is unable to organise
the competition may be prevented from participating in IPhOs by decision of the
International Board.
Regulations to §3
The accompanying persons are considered by the organisers of the next Olympiad
and by the Secretariat of the IPhO (§ 8) as contact persons until the next Olympiad
(unless new accompanying persons or other contact persons are nominated by the
participating country).
Each participating country must ensure that the contestants are all secondary
school pupils when they announce the names of the members of their delegations.
In addition to the delegations, teams may be accompanied by observers and guests.
Observers may attend all Olympiad meetings, including the meetings of the
International Board. However they may not vote or take part in the discussions.
Guests do not attend the meetings of the International Board.
164
If possible, the host country should accept as observers any of the following
persons:
the organiser(s), or nominee(s), from the host country in the subsequent
three years
a representative of any country expressing an intention to participate in
the following IPhO.
Regulations to §4
The host country must pay for organisation of IPhO, food, lodging, transport and
excursions of the delegations plus prizes.
However it is not responsible for medical costs and sundry expenses of the
participants. Observers and guests may be asked to pay the full cost of their stay
plus an attendance fee.
The host country may ask the delegations for a voluntary contribution to the
obligatory costs.
Regulations to §5
It is recommended that the Competition should last 10 days (including arrival and
departure days).
The host country is obliged to ensure that the Competition is conducted accord-
ing to the Statutes. It should provide full information for participating countries,
prior to their arrival, concerning venue, dates, accommodation, transport from
airports, ports and railway stations. The addresses, telephone, fax, e-mail of all
IPhO officers should be provided, together with information concerning relevant
laws and customs of the host country.
A program of events during the IPhO should be prepared for the leaders and
contestants. It should be sent to the participating countries, prior to the Olympiad.
The organisers of the IPhO are responsible for devising all the problems. They
must be presented in English and the other official languages of the Olympiad as
indicated in § 9. The examination topics should require creative thinking and knowl-
edge contained within the Syllabus. Factual knowledge from outside the Syllabus
may be introduced provided it is explained using concepts within the Syllabus.
Everyone participating in the preparation of the competition problems must
not divulge their content.
The standard of problems should attempt to ensure that approximately half
165
the students obtain over half marks.
The International Board shall be given time to consider the examination papers.
It may change, or reject, problems. If a problem is rejected, the alternative prob-
lem must be accepted. The host country will be responsible for grading the exam-
ination papers. The delegation leaders shall have an opportunity to discuss with
the examiners the grading of their students’ papers. If an agreement, between
graders and leaders, to the final marks cannot be reached, the International Board
has to decide.
The organisers shall provide the delegation leaders with copies of their stu-
dents’ scripts and allow at least 12 hours for them to mark the scripts.
The host country shall provide medals and certificates in accordance with the
Statutes. They must also produce a list of all contestants receiving awards with their
marks and associated award. The awards are presented at the Closing Ceremony.
The host country is obliged to publish the Proceedings of the Competition, in
English, in the subsequent twelve months. A free copy of the Proceedings should
be sent to all delegation leaders and competitors.
Regulations to §6
Special prices may be awarded. The participant who obtains the highest score
should receive a special prize.
Regulations to §7
During the meeting of the graders where the final and most detailed version of the
grading scheme is set, 3 members of the International Board will be present. They
have the right to give advice to the group of graders in order to keep the grading
scheme within the tradition of the IPhOs.
If it is found that leaders, observers or students from a country have been in col-
lusion to cheat in one of the International Olympiad examinations, the students
concerned should be disqualified from that Olympiad. In addition, the leaders,
observers and students involved should not be allowed to return to any future
Olympiad. Appropriate decisions are taken by the International Board.
166
Regulations to §8
Election of secretary
The secretary has to have been for the five years prior to the nomination:
a member of the International Board for at least three of these years,
or an observer or member of the International Board, who has attended all
these five IPhOs
The secretary will hold office for a period of five years commencing at the con-
clusion of the final meeting of the International Board at which the secretary has
been chosen.
The secretary and the president must not be appointed at the same IPhO. If this
is the case, however, the period of the secretary will have to be shortened in such
a way that the two elections can be held at different IPhOs.
The secretary and the president must not come from the same delegation.
If the term of the secretary comes to an end, the International Board has to
be informed one year in advance that there will be the ballot of a new secretary
during the following IPhO. In addition to that, the secretariat is responsible to
send a letter to all leaders of the last three IPhOs with this information and with
the question if any leader will be ready to act as secretary for the coming period
by 31st January. This is normally done by e-mail.
If someone is willing to be a candidate for the secretary-ballot, he or she will
have to tell this to the current secretary by 31st March, normally by e-mail.
A nominee has to send his/her curriculum vitae up to 31st March. A nomination
may not be made by a person from the same country as the current president.
The secretariat is responsible to collect all these answers and has to make a list
with all the names.
If the current secretary is willing to continue his/her activity as secretary, he
or she has to enter his/her name in this list and has to follow the same rules as all
the other candidates.
The list with the candidates for the new secretary has to be published on the IPhO-
home-page and the home page of the IPhO during which the ballot will be held.
If there is just one candidate, the secretary has to inform the president about
that. In that case this candidate is accepted as secretary.
The secretariat and the organisers of the IPhO during which the election will
be held are responsible for a democratic, secret ballot of the secretary during the
last meeting of the International Board.
167
If the current secretary resigns or becomes incapable of continuing his/her
work as a secretary, the president shall appoint a replacement to act as provisional
secretary up to the next IPhO. The ballot of the new secretary has to be made as
soon as possible.
168
SyllabusAppendix to the Statutes of the International Physics Olympiads
General
a. The extensive use of the calculus (differentiation and integration) and the
use of complex numbers or solving differential equations should not be
required to solve the theoretical and practical problems.
b. Questions may contain concepts and phenomena not contained in the Syllabus
but sufficient information must be given in the questions so that candidates
without previous knowledge of these topics would not be at a disadvantage.
c. Sophisticated practical equipment likely to be unfamiliar to the candidates
should not dominate a problem. If such devices are used then careful
instructions must be given to the candidates.
d. The original texts of the problems have to be set in the SI units.
A. Theoretical Part
The first column contains the main entries while the second column contains
comments and remarks if necessary.
1. Mechanics
a) Foundation of kinematics of a
point mass
Vector description of the position of
the point mass, velocity and acceler-
ation as vectors
b) Newton’s laws, inertial systems Problems may be set on changing mass
c) Closed and open systems, momen-
tum and energy, work, power
d) Conservation of energy, conserva-
tion of linear momentum, impulse
e) Elastic forces, frictional forces, the
law of gravitation, potential energy
and work in a gravitational field
Hooke’s law, coefficient of friction
(F/R = const), frictional forces, static
and kinetic, choice of zero of poten-
tial energy
f) Centripetal acceleration, Kepler’s laws
169
2. Mechanics of Rigid Bodies
a) Statics, center of mass, torque Couples, conditions of equilibrium
of bodies
b) Motion of rigid bodies, transla-
tion, rotation, angular velocity,
angular acceleration, conservation
of angular momentum
Conservation of angular momen-
tum about fixed axis only
c) External and internal forces, equa-
tion of motion of a rigid body around
the fixed axis, moment of inertia,
kinetic energy of a rotating body
Parallel axes theorem (Steiner’s
theorem), additivity of the moment
of inertia
d) Accelerated reference systems,
inertial forces
Knowledge of the Coriolis force
formula is not required
3. Hydromechanics
No specific questions will be set on this but students would be expected to know
the elementary concepts of pressure, buoyancy and the continuity law.
4. Thermodynamics and Molecular Physics
a) Internal energy, work and
heat, first and second laws of
thermodynamics
Thermal equilibrium, quantities
depending on state and quantities
depending on process
b) Model of a perfect gas, pressure
and molecular kinetic energy,
Avogadro’s number, equation of
state of a perfect gas, absolute
temperature
Also molecular approach to such
simple phenomena in liquids and
solids as boiling, melting etc.
c) Work done by an expanding gas
limited to isothermal and adiabatic
processes
Proof of the equation of the adiaba-
tic process is not required
d) The Carnot cycle, thermodynamic
efficiency, reversible and irrevers-
ible processes, entropy (statistical
approach), Boltzmann factor
Entropy as a path independent
function, entropy changes and
reversibility, quasistatic processes
170
5. Oscillations and waves
a) Harmonic oscillations, equation
of harmonic oscillation
Solution of the equation for har-
monic motion, attenuation and
resonance -qualitatively
b) Harmonic waves, propagation of
waves, transverse and longitudinal
waves, linear polarization, the clas-
sical Doppler effect, sound waves
Displacement in a progressive wave
and understanding of graphical
representation of the wave, measure-
ments of velocity of sound and light,
Doppler effect in one dimension only,
propagation of waves in homogene-
ous and isotropic media, reflection
and refraction, Fermat’s principle
c) Superposition of harmonic
waves, coherent waves, interfer-
ence, beats, standing waves
Realization that intensity of wave
is proportional to the square of its
amplitude. Fourier analysis is not
required but candidates should have
some understanding that complex
waves can be made from addition of
simple sinusoidal waves of different
frequencies. Interference due to
thin films and other simple systems
(final formulae are not required),
superposition of waves from second-
ary sources (diffraction)
6. Electric Charge and Electric Field
a) Conservation of charge,
Coulomb’s law
b) Electric field, potential, Gauss’
law
Gauss’ law confined to simple sym-
metric systems like sphere, cylinder,
plate etc., electric dipole moment
c) Capacitors, capacitance, dielec-
tric constant, energy density of
electric field
7. Current and Magnetic Field
171
a) Current, resistance, internal
resistance of source, Ohm’s law,
Kirchhoff’s laws, work and power
of direct and alternating currents,
Joule’s law
Simple cases of circuits containing
non-ohmic devices with known V-I
characteristics
b) Magnetic field (B) of a current,
current in a magnetic field, Lorentz
force
Particles in a magnetic field, simple
applications like cyclotron, mag-
netic dipole moment
c) Ampere’s law Magnetic field of simple symmetric
systems like straight wire, circular
loop and long solenoid
d) Law of electromagnetic induc-
tion, magnetic flux, Lenz’s law,
self-induction, inductance, perme-
ability, energy density of magnetic
field
e) Alternating current, resistors,
inductors and capacitors in
AC-circuits, voltage and current
(parallel and series) resonances
Simple AC-circuits, time constants,
final formulae for parameters of
concrete resonance circuits are not
required
8. Electromagnetic waves
a) Oscillatory circuit, frequency of
oscillations, generation by feedback
and resonance
b) Wave optics, diffraction from one
and two slits, diffraction grating,re-
solving power of a grating, Bragg
reflection,
c) Dispersion and diffraction spec-
tra, line spectra of gases
d) Electromagnetic waves as trans-
verse waves, polarization by reflec-
tion, polarizers
Superposition of polarized waves
172
e) Resolving power of imaging
systems
f) Black body, Stefan-Boltzmanns law Planck’s formula is not required
9. Quantum Physics
a) Photoelectric effect, energy and
impulse of the photon
Einstein’s formula is required
b) De Broglie wavelength,
Heisenberg’s uncertainty principle
10. Relativity
a) Principle of relativity, addition of
velocities, relativistic Doppler effect
b) Relativistic equation of motion,
momentum, energy, relation
between energy and mass, conser-
vation of energy and momentum
11. Matter
a) Simple applications of the Bragg
equation
b) Energy levels of atoms and
molecules (qualitatively), emission,
absorption, spectrum of hydrogen
like atoms
c) Energy levels of nuclei (quali-
tatively), alpha-, beta- and gam-
ma-decays, absorption of radiation,
halflife and exponential decay,
components of nuclei, mass defect,
nuclear reactions
173
B. Practical Part
The Theoretical Part of the Syllabus provides the basis for all the experimental
problems. The experimental problems given in the experimental contest should
contain measurements.
Additional requirements:
Candidates must be aware that instruments affect measurements.
Knowledge of the most common experimental techniques for measuring
physical quantities mentioned in Part A.
Knowledge of commonly used simple laboratory instruments and devices
such as calipers, thermometers, simple volt-, ohm- and ammeters, poten-
tiometers, diodes, transistors, simple optical devices and so on.
Ability to use, with the help of proper instruction, some sophisticated
instruments and devices such as double-beam oscilloscope, counter, ratem-
eter, signal and function generators, analog-to-digital converter connected
to a computer, amplifier, integrator, differentiator, power supply, univer-
sal (analog and digital) volt-, ohm- and ammeters.
Proper identification of error sources and estimation of their influence on
the final result(s).
Absolute and relative errors, accuracy of measuring instruments, error of a
single measurement, error of a series of measurements, error of a quantity
given as a function of measured quantities.
Transformation of a dependence to the linear form by appropriate choice
of variables and fitting a straight line to experimental points.
Proper use of the graph paper with different scales (for example polar and
logarithmic papers).
Correct rounding off and expressing the final result(s) and error(s) with
correct number of significant digits.
Standard knowledge of safety in laboratory work. (Nevertheless, if the
experimental set-up contains any safety hazards the appropriate warnings
should be included into the text of the problem.)
174
175
AppendicesTartu – The World Capital of PhysicsPrOgram
20th of July10:00 – 16:30 Career day of the universities
12:00 – 12:30 Declaration of Tartu as the World Capital of Physics
at the Town Hall Square
12:30 – 16:30 Workshops
DETAILED TIMESCHEDULE
10:00 – 16:30 Career day presentations by universities in the conference hall
of the University of Tartu history museum Address: Lossi 25
10:15 - 11:00 University of Tartu
11:10 - 11:40 University of Oxford
13:00 - 14:00 National University of Singapore
14:10 - 15:10 Massachusetts Institute of Technology
15:20 - 16:20 Tallinn University of Technology
12:00 – 12:30 Declaration of Tartu as the World Capital of Physics
at the Town Hall Square
WORKSHOPS
12:30 – 16:30 Workshops at the Cathedral ruins and the history museum
of the University of Tartu Address: Lossi 25
CONTINUOUS WORKSHOPS
200-year-old physics office
Physics equipment belonging to the physics office of the University of Tartu from
the 19th century will be introduced.
176
Making of magic disks
Make a popular optical toy from the 19th century.
Colours to the T-shirt
Design a unique T-shirt for yourself by using a chromatographic method.
Micro and macro worlds in nature
Study various animals and stones with a microscope.
Finding the thief
A fun game introducing genetics.
How special are you?
A game designed for Tallinn TV Tower based on the database of the Estonian
Genome Centre where people can test how similar or different they are compared
to other Estonians.
Circles of the woods
One tree can tell endless stories if you get to know it up close.
Making of tar
How to make tar from a block of wood?
Timber preservation workshop
People have used timber for a very long time and old buildings have preserved well.
But nowadays logs or floors tend to break up in a few years. We will find out why.
Various ways of inoculating and grafting fruit cultures
Organic farming: growth, cultures, breeds, foods
Measuring a person
Find out your height, weight, and body mass index.
Let’s go inside a head.
We will go inside a head and study how the brain works: eyesight, reading, deci-
sion-making. Will be possible to make a back-up brain and solve a brain puzzle.
Feel the dimensions
Ethnographic experiments: units of length, weight and volume in modern times.
Inventions from the University of Tartu
Inventions by university scientists that have been or will be implemented in
everyday life: glass that changes transparency, a new photosensitive material,
instrument for diagnosing glaucoma, ME-3 bacterium, myometer, robots.
Interesting facts from the botanic kingdom
Finding close or distant relatives in the botanic kingdom: which trunks and cones
match.
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Workshops with definite start and end times
12:30 - 13:15 Glance at the past of the University of Tartu
Guided tour in the history museum13:30 - 14:15
14:30 - 15:15
15:30 - 16:15
Workshops in the Tartu observatory and in its proximity Address: Lossi 40
CONTINUOUS WORKSHOPS
Science bus
Various experiments and workshops related to physics, chemistry and biology.
Workshops with definite start and end times
12:30 - 13:00 Planetarium show
During the planetarium show “Night sky”, the planets,
their movement during the year will be shown together
with more interesting constellations from both the
northern and southern hemisphere.
13:15 - 13:45
14:00 - 14:30
14:45 - 15:15 Planetarium StarLab
15:30 - 16:00 StarLab planetarium with an inflatable tent as a dome
will be used for the planetarium show.
12:30-13:15 Guided tours in the planetarium
Observation of the sun and a workshop for making a
sun clock
Workshop for making a quadrant
Weather measurement
13:15-14:00
14:00-14:45
14:45-15:30
15:30-16:15
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yEar OF scIEncE In EsTOnIa and TarTu — ThE wOrLd caPITaL OF PhysIcsViire Sepp
IPhO 2012 Steering Committee Secretary
Estonia hosting the IPhO 2012 became the incentive for proclaiming the period
between September 2011 and September 2012 a Year of Science in our country.
The objective was to raise youth interest in exact, natural and technical sciences
and towards career choices related to these subject areas. Under the auspices of
the Year of Science, hundreds of events have taken place in schools, towns and
counties organized by the partners of the Year of Science – schools, universities,
associations, societies, etc.
One of the main events took place on the 20th of July when Tartu was proclaimed
the World Capital of Physics within the IPhO 2012. The objectives of proclaiming
the host city as the World Capital of Physics were to highlight the importance of
natural and exact sciences and valuation of education, creativity and purposeful
self-development. Numerous public events took place during that day in Tartu
– the first “World Capital of Physics”. At the same time, the annual Hanseatic
days also commenced in Tartu. At 12 o’clock, the ceremony declaring Tartu the
World Capital of Physics took place in the Town Hall Square. The IPhO delega-
tion handed over the Declaration announcing Tartu the Capital of Physics to the
Mayor of Tartu. Subsequently, everyone was welcomed to the historic Toome Hill
where the Science Town was set up during the Hanseatic days. The Science Town
included tens of hands-on workshops, shows by the science bus of the Estonian
Physical Society, and the 2007 winner of the EU prize for Science Communication.
In addition, there was a possibility to visit the 200-year-old observatory which was
affirmed in the world science history by its director F.G.F. Struve who was the first
to measure the distance between a star and Earth in 1835. Tartu Observatory has
belonged to the UNESCO World Heritage since 2005. In the history museum of the
University of Tartu it was possible to get acquainted with the university’s history
and contemporary situation. Also, the University Fair (so called “a career day”)
featured presentations on enrolment and study opportunities by Massachusetts
Institute of Technology, University of Oxford, National University of Singapore,
Tallinn University of Technology and University of Tartu. The honorary guest of
the IPhO, 1996 Nobel Prize winner Sir Harold Kroto, gave an academic lecture at
179
the Vanemuise Concert Hall in the evening and the day was concluded with a
reception by the Mayor of Tartu.
Since the IPhO is a relatively closed event, it is essential to acknowledge the
bridge between the IPhO and the host city, make the IPhO more visible in the
urban space, strengthen international reputation, and advance potential inter-
national relations of the hosting city. The idea that the host city could bear that
title was beforehand introduced to the IPhO president Dr Hans Jordens who found
the initiative worthwhile. The Statute and the Declaration of the World Capital
of Physics were prepared by organizers of IPhO2012 and were presented to team
leaders of participating countries in order to make this event traditional in future
IPhOs. These documents are printed herewith.
sTaTuTE OF ThE IPhO wOrLd caPITaL OF PhysIcs In order to highlight cities that host the International Physics Olympiad (IPhO)
and to intensify the impact of the IPhO in extending international contacts of
these cities, IPhO2012 Estonia launches the tradition of nominating the hosting
city of the IPhO, for one day of the Olympiad, as the World Capital of Physics.
The first IPhO World Capital of Physics is Tartu (Estonia), which was the host-
ing city of the 43rd International Physics Olympiad in 2012.
The statute establishes the aims of nomination and the principles of co-op-
erative activities of the IPhO secretariat (President), Organizing Committee and
hosting city and regalia of the World Capital of Physics.
The aims of nomination are to:
Attract public and primarily youth attention to the significance of the
natural and exact sciences, promote education, creativity and purposeful
self-actualization
Acknowledge the venue of the IPhO
Extend the international reputation of the hosting city whilst promoting
its potential international relations
The nomination of the IPhO hosting city as the World Capital of Physics takes
place in a public urban space.
Regalia of the World Capital of Physics:
Parchment (Declaration of nomination)
Appendix (good wishes from the participants of the IPhO to the hosting city)
180
Symbol of the World Capital of Physics (the Chronicle Book) launched by
Tartu (Estonia, IPhO2012)
Local Organizing Committee of the Olympiad:
Arranges and conducts the World Capital of Physics nomination ceremony
in co-operation with the hosting city and IPhO secretariat (President)
Organizes Parchment/Declaration compilation and its delivery to the
Mayor of the hosting city
Arranges perpetuation of the hosting city in the Chronicle Book of the
World Capital of Physics
Hosting city:
Guarantees the presence of the Mayor or his/her representative in the nom-
ination ceremony of the World Capital of Physics
Accepts the Parchment (Declaration)
Delivers the Chronicle Book of the World Capital of Physics to the IPhO
President (or his/her representative)
IPhO secretariat/President:
Follows the continuity of the tradition of the World Capital of Physics
Delivers the Parchment to the hosting city
Accepts the Symbol (the Chronicle Book) and delivers it to the organiz-
ers of the next IPhO (at the IB meeting or at the closing ceremony of the
Olympiad)
dEcLaraTIOn Date and place
We, the participants of the …. (number) International Physics Olympiad:
Appreciating the beauty and power of science, that can take us far beyond the
realm of everyday life - from microworlds to the distant cosmos;
Understanding the importance of physics as one of the cornerstones of modern
technological society that has given birth to numerous inventions, which are of
benefit to mankind;
Acknowledging science as an essential collective effort, uniting peoples and
181Photos by Henry Teigar, Karl Veskus and Siim Pille
Tartu town hall square
geniuses in red
President of IPhO hans Jordens declares Tartu as the world capital of Physics
Estonian student kaur aare saar reads the declaration of Tartu as the world capital of Physics choir named E sTuudio presenting Estonian folk music
182
countries in solving the crucial problems that society faces, in the name of peace
and common progress;
Recognizing the everlasting curiosity of people of all ages, never stopping
asking questions and searching for answers, as the main driving force of science;
Committing to the hard work one has to do in mastering science in order to
exploit its potentialities to the fullest extent;
Being indebted to our teachers for the knowledge and excitement they provided
us with, together with the responsibility to carry this on to the next generation;
We declare the City of …….(name of city ) on the occasion of …..(number)
International Olympiad of Physics to be the World Capital of Physics for
…….(date).
We challenge everyone to discover the power and beauty of physics in order to
contribute to the progress and welfare of mankind and to share our love of science!
Signature Signature
IPhO President IPhO Secretary
183
CircularsFIrsT cIrcuLar
Dear Colleagues,The Republic of Estonia has the honour of being the host of the 43rd International
Physics Olympiad (IPhO 2012), which will be held from the 15th to the 24th of July 2012.
The Olympiad will be held in two locations, in the capital Tallinn (team lead-
ers), and in the oldest university town in Estonia, Tartu (students).
The IPhO2012 is organized under the auspices of the Estonian Ministry of
Education and Research by the Estonian Information Foundation (main organizing
institution) jointly with the University of Tartu, Tallinn University of Technology,
the National Institute of Chemical Physics and Biophysics, the Estonian Academy
of Sciences, Archimedes Foundation, and the Estonian Society of Physics. The
Steering Committee, comprizing the representatives of these institutions, governs
the activities of the Academic and Organizing Committees of the IPhO 2012.
InvitationOn behalf of the organizers of the 43rd International Physics Olympiad, it is our
pleasure and honor to invite and welcome your country to send a delegation to take
part in the IPhO 2012 in Estonia. The International Physics Olympiad is the most
prestigious international physics competition for individual secondary school stu-
dents aimed to stimulate young people’s interest in physics, to propagate natural
and exact sciences amongst school students, and to promote science education
throughout the world by means of international contacts.
Requirements for the participating teamsAccording to the IPhO Statutes, each team will comprise up to five students and two
leaders. The contestants shall be students of general or technical secondary schools,
i.e. schools which cannot be considered technical colleges. Students who have fin-
ished their school graduating examinations in the year of the competition can be
members of the team as long as they have not commenced their university studies.
The age of the students should not exceed twenty years, as of the 30th of June 2012.
The team leaders should be physics teachers or specialists who can speak English.
184
Fees and ExpensesThe IPhO2012 Organizing Committee will provide meals, accommodation and
transportation free of charge to all the participants during the duration of the
Olympiad period. However, travel expenses of each participating country will be
the liability of each participating country. Arrangements for extra accommodation
before and/or after the official period will be provided, though the related expenses
are to be covered by the participating country.
As an established custom in the International Physics Olympiad, each par-
ticipating country is asked to contribute a voluntary fee of € 3,500.Observers and
visitors are welcome, but they are expected to cover their own expenses. Observers
may attend all Olympiad meetings, including the meetings of the International
Board. However, they may not vote or take part in the discussions. Visitors do not
attend the meetings of the International Board. The fee of each observer is € 1,200
and for visitor € 1,100.
Transportation Transportation from Tallinn International Airport and Port of Tallinn to hotels
and vice versa will be arranged by the local Organizing Committee at no addi-
tional cost.
Information and contactsFor the detailed information of IPhO 2012 and the most recent update, please visit
our official homepage www.ipho2012.ee
You may contact us via e-mail: [email protected]
Phone number: +372 628 5800
Address: 43rd IPhO Organizing Committee
Estonian Information Technology Foundation
Raja 4C, 12616 Tallinn
Estonia
185
Preliminary confirmation of participationAs the first step, please complete the attached form: PRELIMINARY CONFIRMATION
FORM OF IPhO 2012 PARTICIPATION.
This information is needed for sending official letter of invitation to participate
in IPhO2012 to your Minister of Education or the equivalent official in your country.
In addition, this information is necessary for communication with the contact
person in your country.
The form should be filled in and sent as an e-mail attachment to the 43rd IPhO
Organizing Committee ([email protected]) no later than: 20th of December 2011.
Estonian people and the organizers of the Olympiad are looking forward to
meeting young physicists and their supervisors from all over the world, and to
introducing them to the innovative country, which values education, has a rich
cultural heritage and beautiful land.
Welcome to IPhO2012 in Estonia!
Sincerely,
Prof Jaak Kikas Dr Viire Sepp
Head of the Academic Committee Academic secretary of the
Steering Committee
186
Second CircularDear Colleagues,It is our pleasure to inform you about the progress regarding the 43rd International
Physics Olympiad in Tallinn/Tartu, Estonia during 15th – 24th of July 2012.
Confirmation of participation to this dateUntil the present date, the following 86 teams have confirmed their participation
in the 43rd IPhO:
Albania
Armenia
Australia
Austria
Azerbaijan
Bangladesh
Belarus
Belgium
Bolivia
Bosnia & Herzegovina
Brazil
Bulgaria
Cambodia
Canada
People`s Republic of China
Colombia
Croatia
Cuba
Cyprus
Czech Republic
Denmark
Ecuador
El Salvador
Estonia
Finland
France
Georgia
Germany
Ghana
Greece
Hong-Kong
Hungary
Iceland
India
Indonesia
Islamic Republic of Iran
Ireland
Israel
Italy
Japan
Jordan
Kazakhstan
Republic of Korea
Kuwait
Kyrgyzstan
Lao PDR
Latvia
Liechtenstein
Lithuania
Macao, China
187
Macedonia
Malaysia
Mexico
Moldova
Montenegro
Netherlands
Nigeria
Norway
Pakistan
Philippines
Poland
Portugal
Puerto Rico
Qatar
Romania
Russia
Saudi Arabia
Serbia
Republic of Singapore
Slovakia
Slovenia
South Africa
Spain
Sri Lanka
Suriname
Sweden
Switzerland
Syria
Taiwan
Tajikistan
Thailand
Ukraine
United Kingdom
United States of America
Uzbekistan
Vietnam
Official invitationDuring on beginning of February official invitation letters were sent to the
Ministers of Education or equivalent officials of the teams that had already returned
the Confirmation Form to the IPhO2012 Organizing Committee. Pdf copies of the
original invitation letters were also sent to the ministries and equivalent officials.
If you have not received these letters, please contact us ([email protected]).
Online registrationThe online registration will be available at the IPhO 2012 website: http://www.
ipho2012.ee from February 20, 2012. The contact person of each country will receive an
auto-generated e-mail message with USERNAME, PASSWORD and instructions.
The contact person is requested to register all participants in his/her respective
team (students, leaders, visitors and observers).
188
VisaComplete information on visa application and issuances is available at the website
of Estonian Ministry of Foreign Affairs: http://www.vm.ee/?q=en/node/53
Who does not need a visa to visit Estonia?
Nationals of the member states of the European Union (EU) and the European
Economic Area (EEA) and any third-country national who is a holder of a residence
permit of a Schengen state do not need a visa to enter Estonia.
EU and EEA member states are Austria, Belgium, Bulgaria, Cyprus, Czech
Republic, Denmark, Estonia, Finland, France, Germany, Greece, Hungary,
Iceland, Ireland, Italy, Latvia, Liechtenstein, Lithuania, Luxembourg, Malta,
Norway, Poland, Portugal, Romania, Slovakia, Slovenia, Spain, Sweden, The
Netherlands, United Kingdom and Switzerland.
Schengen States are Austria, Belgium, Czech Republic, Denmark, Estonia,
Finland, France, Germany, Greece, Hungary, Iceland, Italy, Latvia, Liechtenstein,
Lithuania, Luxembourg, Malta, The Netherlands, Norway, Poland, Portugal,
Slovakia, Slovenia, Spain, Sweden, Switzerland.
! Although documents are not checked when one crosses an internal Schengen
border, it is still necessary to carry a passport or ID card. Authorities (police,
immigration officials) in Schengen states do have the right to check identifying
documents, if necessary.
The holders of passports of the following countries do not need a visa to enter a
Schengen region (incl Estonia) for stays of no more than 90 days in a 6 month period:
Albania*, Andorra, Antigua and Barbuda, Argentina, Australia, Bahamas,
Barbados, Bosnia and Herzegovina*, Brazil, Brunei, Canada, Chile, Costa
Rica, Croatia, El Salvador, Guatemala, Holy See, Honduras, Hong Kong Special
Administrative Region, Israel, Japan, Macao Special Administrative Region,
Macedonia*, Malaysia, Mauritius, Mexico, Monaco, Montenegro*, New Zealand,
Nicaragua, Panama, Paraguay, San Marino, Serbia*, Seychelles, Singapore, South
Korea, St Kitts-Nevis, Taiwan**, United States of America, Uruguay, Vatican City,
Venezuela.
* Only for the biometrical passports holders
** Passports issued by Taiwan which include an identity card number
Countries requiring visa for entry to Estonia, please start the application pro-
cess early. It will take approximately 2–3 weeks for the applications to be processed.
189
Please take into consideration that it is impossible to buy the Estonian visa
at the order.
If you have questions concerning visas or need additional documents, please
contact the Organizing Committee ([email protected]) .
Fees and paymentAll participating teams (5 students + 2 leaders) are encouraged to contribute a
voluntary fee of EUR 3,500. The fee will cover: meals, accommodation and official
social events during the official period of the Olympiad. Observers and visitors are
welcome, and they are expected to cover their expenses, visitors EUR 1,100 and
observers EUR 1,200. Observers may attend all Olympiad meetings, including the
meetings of the International Board. However, they may not vote or take part in
discussions. Visitors do not attend the meetings of the International Board.
If the participating team from your country is smaller than 5 people, then the
participation fee is 500 euros per person and the total amount of the invoice will
depend on the number of people.
We kindly ask you to pay the invoice before arrival to Estonia, within 10 days
upon receiving it by e-mail. In case you wish to pay the fee on site, please inform
the Organizing Committee well in advance ([email protected])
TransportationIPhO 2012 will start and end in Tallinn. Please make all your travel arrangements
with arrival to and departure from Tallinn Airport (http://www.tallinn-airport.ee/eng)
For teams travelling via Helsinki (Finland), sea transportation from
Helsinki to Tallinn could be the alternative (http://www.portofhelsinki.fi/passengers/
departure_times_and:terminals).
Transportation from the airport and the seaport to the hotels and vice versa
will be provided by the Organizing Committee at no additional cost.
AccommodationIn Tallinn students will be staying at the Sokos Hotel Viru. During the student’s
programme in Tartu from the 16th to the 21st of July contestants will be staying at
the following hotels: Hotel Dorpat, Tartu Hotel, Hotel Pallas.Leaders, visitors
and observers will be accommodated in Radisson Blu Hotel Olümpia in Tallinn.
190
All hotels are situated in the very heart of Tallinn and Tartu. Two persons will
share a room.
If leaders, visitors and observers would like to stay in a single room, please let
us know when you register for 43rd IPhO; there is an additional charge of EUR 400
per room for 9 nights during the official period of the Olympiad.
If you plan to arrive in Estonia earlier than July 15th 2012 or leave Tallinn later
than July 24th 2012, please let us know well in advance (by noting it in the online
registration and by sending us an e-mail, indicating the number of persons and
type of the room) so that we can arrange your accommodation well in advance,
because July is the tourism high season in Estonia and the hotels need to be booked
as early as possible. The extra charge per night for Tallinn hotel double rooms are
EUR 70 for the students’ hotel and EUR 75 for the leaders’ hotel.
43rd IPhO Website and ContactsThe circulars and all information regarding the 43rd IPhO can be found at our web-
site at http://www.ipho2012.ee which will be updated regularly.
Organizers of the 43rd IPhO are looking forward to meeting you in Estonia!
Prof Jaak KIkas Dr Viire Sepp
Head of the Academic Committee Academic Secretary of the
Steering Committee
191
Third CircularDear Colleagues,We are only 20 days before the beginning of the 43rd International Physics Olympiad
in Estonia. It is our pleasure to inform you about the progress regarding the prepa-
ration and necessary information for your teams.
RegistrationThe online registration is about to complete, the following 86 teams have con-
firmed their participation in the 43rd IPhO:
1. Albania
2. Armenia
3. Azerbaijan
4. Australia
5. Austria
6. Bangladesh
7. Belarus
8. Belgium
9. Bolivia
10. Bosnia & Herzegovina
11. Brazil
12. Bulgaria
13. Canada
14. Colombia
15. Croatia
16. Czech Republic
17. Cuba
18. Cyprus
19. Denmark
20. Ecuador
21. El Salvador
22. Estonia
23. Finland
24. France
25. Georgia
26. Ghana
27. Germany
28. Greece
29. Hong Kong
30. Hungary
31. Iceland
32. India
33. Indonesia
34. Ireland
35. Islamic Republic of Iran
36. Israel
37. Italy
38. Japan
39. Kazakhstan
40. Kuwait
41. Kyrgyzstan
42. Latvia
43. Liechtenstein
44. Lithuania
45. Macao, China
46. Macedonia
47. Malaysia
48. Mexico
192
49. Moldova
50. Mongolia
51. Montenegro
52. Nepal
53. Netherlands
54. Nigeria
55. Norway
56. Pakistan
57. People`s Republic of China
58. Poland
59. Portugal
60. Puerto Rico
61. Qatar
62. Republic of Korea
63. Republic of Singapore
64. Romania
65. Russia
66. Saudi Arabia
67. Serbia
68. Slovakia
69. Slovenia
70. South Africa
71. Spain
72. Sri Lanka
73. Suriname
74. Sweden
75. Switzerland
76. Syria Arab Republic
77. Taiwan
78. Tajikistan
79. Thailand
80. Turkey
81. Turkmenistan
82. Ukraine
83. United Kingdom
84. United States of America
85. Uzbekistan
86. Vietnam
Fees and paymentAll participating teams (5 students + 2 leaders) are encouraged to contribute with a
voluntary fee of EUR 3,500. The fee will cover: meals, accommodation and official
social events during the official period of the Olympiad. Observers and visitors are
welcome, and they are expected to cover their expenses, visitors EUR 1,100 and
observers EUR 1,200.
Observers may attend all Olympiad meetings, including the meetings of
the International Board. However they may not vote or take part in discussions.
Visitors do not attend the meetings of the International Board.
If the participating team from your country is smaller than 5 people, then the
participation fee is 500 euros per person and the total amount of the invoice will
depend on the number of people.
We kindly ask you to pay the invoice before arrival to Estonia, within 10 days
upon receiving it by e-mail. In case you wish to pay the fee on site, please inform
the Organizing Committee well in advance ([email protected])
193
TransportationIPhO 2012 will start and end in Tallinn. Please make all your travel arrangements
with arrival and departure at Tallinn Airport (http://www.tallinn-airport.ee/eng)
For teams travelling via Helsinki (Finland), sea transportation from
Helsinki to Tallinn could be the alternative (http://www.portofhelsinki.fi/passengers/
departure_times_and:terminals).
Transportation from the airport and the seaport to the hotels and vice versa
will be provided by the Organizing Committee at no additional cost.
Please insert your delegation travel plans into IPhO 2012 registration environ-
ment by June 29th.
AccommodationIn Tallinn student will stay at Sokos Hotel Viru. In Tartu student will stay at the
following hotels: Hotel Dorpat, Tartu Hotel, Hotel Pallas and Hotel London.
Leaders, visitors and observers will be accommodated in Radisson Blu Hotel
Olümpia. All hotels are situated in the very heart of Tallinn and Tartu. Two persons
will share a room.
If leaders, visitors and observers would like to stay in single rooms, please let
us know when you register for the 43rd IPhO; there is an additional charge of EUR
400 for 9 nights during the official period of the Olympiad.
If you plan to arrive in Estonia earlier than July 15th, 2012 or leave Tallinn later
than July 24th, 2012, please let us know well in advance (by noting it in the online
registration and by sending us an e-mail, indicating the number of persons and
type of the room) so that we can arrange your accommodation well in advance,
because July is the tourism high season in Estonia and the hotels need to be booked
as early as possible. The extra charge per night and per person in twin rooms
in Tallinn is EUR 35 for the students’ hotel and EUR 37.50 for the leaders’ hotel.
Breakfast is included.
Between the 16th and the 19th of July, there will be no phones in the student’s
hotel rooms. Because of that it is recommended that the students take along their
own alarm clocks.
Computing facilities for team leadersComputing facilities, including laptops and printers, will be available in the
International Board Meeting Room.
194
You can bring your own laptop/notebook. Standard power supply in Estonia
is 220V/50 Hz.
CalculatorsNotice regarding the calculators which can be brought into the examination room.
According to the Statutes of the IPhO, the only tools which can be brought into
the examination room by competitors are drawing instruments and a non-pro-
grammable pocket calculator. [§5] In order to give a clear guideline about which
calculators are allowed, the Academic Committee of the IPhO 2012 provides the
following definition.
A non-programmable calculator is not allowed to have:
more than 9 memory slots for recording intermediate data
the possibility of entering user-defined functions for repetitive calculations
pre-stored physical constants
the possibility of plotting graphs
The calculator needs to be a commercial product, complete with an English
manual (printed or downloadable); it is allowed to have standard mathemati-
cal functions (trigonometric, hyperbolic etc.) and standard statistical functions
(mean, standard deviation, etc.).
All the calculators of the contestants will be checked prior to the competition
and the allowed ones will be labeled. A limited amount of replacement calculators
will be available from the organizers.
Clothing and weather During the opening and closing ceremonies and the farewell party it is required
that the boys wear long trousers (please do not wear shorts unless these are part
of the national costume).
Please take comfortable clothing and shoes for the excursions (there will be
some walking and climbing).
There will be a football tournament for the students on 22nd of July. Participating
students need to have sports clothing with them.
Temperatures in the summer vary from +15 to +25 Celsius. The weather is
changeable; shiny and rainy days are intermittent so it is advisable to prepare for
both. Please bring your umbrella or raincoat. You can find more information about
Estonian weather here: http://www.weather.ee/ ; http://www.emhi.ee/index.php?nlan=eng.
195
The Chemistry Olympiad in the United States of AmericaSome IPhO students are taking part in the Chemistry Olympiad in the USA.
Countries that need separate transportation for their student(s) to the airport,
please let us know by writing to [email protected]
Information leak during the competitionThe organizers of IPhO 2012 have taken some measures to minimize the possibility
that information about the problems will leak. We hope for your understanding
and count on your co-operation to execute the following measures:
The students are not allowed to possess electronic equipment that has the
possibility to communicate with the outer world until after the last test
(July 19th). The equipment has to be delivered to the organization of IPhO
2012 in Tallinn and will be stored in special boxes.
From July 16th at 08:00 AM till July 19th at 08:00 PM the students are not
allowed to use internet facilities or phones at the hotel, nor are they allowed
to use such facilities outside the hotel.
All communication between the students and their leaders, parents etc.
during the above indicated period has to go via the guides.
Leaders and observers are not allowed to communicate with the outer world
from the start of the International Board meetings where the discussions of
the problems take place up to the beginning of the respective examination
(in the case of the experimental round, up to the beginning of the second
shift of the experimental competition). In case of emergency the commu-
nication needs to go via the IPhO 2012 office.
All violations against what is stated above will be regarded as cheating as mentioned
in the IPhO Regulations §7 and appropriate measures will be taken accordingly.
Here we quote from the Regulations Associated with the Statutes of the
International Physics Olympiads:
If it is found that leaders, observers or students from a country have been in col-
lusion to cheat in one of the International Olympiad examinations, the students
concerned should be disqualified from that Olympiad. In addition, the leaders,
observers and students involved should not be allowed to return to any future
Olympiad. Appropriate decisions are taken by the International Board.
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43rd IPhO Website and ContactsThe circulars and all information regarding the 43rd IPhO can be found on our
website http://www.ipho2012.ee, which will be updated regularly.
The Academic and Organizing Committee of the 43rd International Physics
Olympiad looks forward to welcoming all delegates in Estonia.
Yours sincerely,
Prof Jaak Kikas Dr Viire Sepp
Head of the Academic Committee Academic Secretary of the
Steering Committee
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NewsletterIPhO 2012 newsletter was called “Physics is in The Air” and aimed to give a daily
update about the Olympiad. The newsletter consisted of articles, interviews, photo
galleries and so on. There were 8 issues on paper handed to all participants each
morning and 10 online numbers published on website paralelly. Online numbers
always concisted of more interviews and details whereas the paper version had
its limits on characters.
Get all the articles and pictures on the newsletter website: www.ipho2012.ee/newsletter
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The process of organising the IPhO 2012 started 3 years ago. The task was highly
prioritized by the Estonian Ministry of Education and Research. Eight institutions
formed the steering committee for organisational aspects – the Estonian Ministry
of Education and Research, Archimedes Foundation, the Estonian Academy of
Sciences, the Estonian Physical Society, the National Institute of Chemical and
Biophysics, the Tallinn University of Technology, the University of Tartu and the
Estonian Information Technology Foundation. The Estonian Information Technology
Foundation was responsible for the organisational side of the IPhO 2012. Eight
people comprised the Organizing Committee: Ene Koitla - head of the Organizing
Committee, Marily Hendrikson - IPhO 2012 project manager, Annika Vihul - head of
accounting, transportation and accommodation, Malle Tragon - head of events and
catering, Kerli Kusnets - head of media, Eneli Sutt - head of information technology,
Julia Šmakova and Anna Gureeva - heads of group leaders.
623 people from 80 countries participated in the IPhO 2012, 378 of these were
contestants. 177 volunteers contributed to the success of the event.
The biggest challenge facing the IPhO 2012 Organising Committee was holding
the event in two cities – Tallinn and Tartu. Issues such as accommodation, trans-
portation and having examinations posed the biggest problems. Estonia is a small
country where such a number of participants can easily create logistical liabilities.
There was also plenty of planning concerning the events and activities during free
time. The IPhO 2012 featured several innovative IT solutions assisting in organizing
the event. All teams registered by using the special registration system. Due to the
distance between the two cities, Tallinn and Tartu, there were limitations to do
some operations in a usual way. To get the translated problem sheets to the students’
desk, the leaders had to upload ready-made files to the special IT system in Tallinn
and the sheets were printed out in Tartu. The same system was used to transfer the
solutions back to the leaders and markers. Each leader and marker could decide if
he/she wanted to print them out or download the solutions. It was made possible for
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the leaders and markers to use the computer whenever
they wanted. All marks were inserted in the web-based
evaluation system. The students’ meetings with leaders
were held using the Skype call.
We are very honoured that we had the chance to host
the 2012 International Physics Olympiad contestants,
leaders, observers and visitors from all over the world.
We hope that the participants have enjoyed the physics
contest and we managed to demonstrate Estonian cul-
ture and variety of nature.
Organizing Committee
IPhO 2012
Photo by Merily Salura
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