sample pdf of std 11th perfect physics notes book science

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PHYSICS

Written as per the latest textbook prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune.

Printed at: India Printing Works, Mumbai

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

P.O. No. 191274TEID: 13661

Balbharati Registration No.: 2018MH0022

PERFECT

Std. XI Sci.

Written as per the new textbook

Subtopic-wise segregation for powerful concept building

Complete coverage of Textual Exercise Questions, Intext Questions, Activities and Textual

Examples (numericals illustrated in textbook)

Extensive coverage of New Type of Questions

‘Solved Examples’ offer complete numerical solution

‘Apply Your Knowledge’ section for application of concepts

‘Quick Review’ facilitates quick revision

‘Important Formulae’ at the end of every chapter compiles all formulae

‘Competitive Corner’ presents questions from prominent competitive examinations Reading Between the Lines, Enrich Your Knowledge, Gyan Guru, Connections, NCERT Corner

are designed to impart holistic education

Video links provided via QR codes for boosting conceptual retention

Salient Features

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Connections enable students to interlink concepts covered in different chapters. This is our attempt to encourage students to appreciate the subject as a whole.

Connections For points (ii) and (iv)

In chapter 3, you have studied about angularvelocity () and centripetal force.

Gyan Guru illustrates real life applications or examples related to the concept discussed. This is our attempt to link learning to the life.

GG - Gyan Guru

A slingshot is a device normally used as a toy by children. It makes use of the elastic property of rubber to hit the desired target like a fruit or a bird etc.

Enrich Your Knowledge presentsfascinating information about the concept covered.This is our attempt to create interest inthe students about the concept.

Enrich Your Knowledge

Since sizes of degrees on the Celsius scale and the Absolute scale are identical, any one of the scales can be used for temperature difference.

“Everything should be made as simple as possible, but not simpler.” - Albert Einstein. Having this vision in mind we have created “Perfect Physics: Std. XI” as per the new textbook of MaharashtraState board. It focuses on not just preparing students from examination point of view but also equipping themto understand and appreciate the beauty of Physics as a subject. Every chapter, segregated subtopic-wise, encompasses all textual content in the format of Question-Answers. Textual Exercise questions, Intext questions, ‘Can you tell’, ‘Can you recall’, ‘Try this’ and ‘Activity’ are placed aptly amongst various additional questions in accordance with the flow of subtopic. To offer studentsbetter understanding of the concept discussed in question, ‘Reading between the lines’ (not a part of the answer) has been provided as deemed necessary. Numericals along with their step-wise solutions are coveredunder heading of Solved Examples at the end of each subtopic. Quick Review has been provided to map the chapter effectively in students’ minds. Formulae covered in the chapter are compiled together as Important Formulae at the end of the chapter. Exercise and MCQ sections are added to enable students assess theirrange of preparation and knowledge of each topic. NCERT Corner and Notes are introduced to coveradditional bits of relevant information on each topic as seemed required.While ensuring complete coverage of the syllabus in an effortless and easy to grasp format, emphasis is alsogiven on active learning. To achieve this, we have infused several key features such as, Gyan Guru, Enrich YourKnowledge, Connections, Additional information and QR Codes. Also, additional sections such as Apply Your Knowledge and Competitive Corner pave the way for a robust concept building. The following screenshots will walk you through the core features of this book and elucidate how they have been carefully designed to maximize the student learning.

PREFACE

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Q.109. The Mariana trench is located in the Pacific Ocean and at one place it is nearly 11 km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench.

Apply Your Knowledge includes challenging questions. This is our attempt to take students one step further and challenge their conceptual understanding.

Apply Your Knowledge

Quick review includes tables/ flow chart to summarize the key points in chapter. This is our attempt to help students to reinforce key concepts.

Quick Review

Thermal conductivity

Steady state

is achieved due to

Temperature gradient

2. Direction of resultant vector:

= tan1 QsinP Qcos

3. Commutative law of vector addition:

P

+ Q

= Q

+ P

Important Formulae includes all of the key formulae in the chapter. This is our attempt to offer students tools of formulae handy while solving problems and last minute revision at a glance.

Important Formulae

Reading between the lines provide elaboration of concept or missing fragments of concept. This is our attempt to explain the concept without deforming the length of expected answer.

Reading between the lines

Explanation for (vi) :

Let two vectors R

and Q

are represented in magnitude by,

x y zˆ ˆ ˆR R i R j R k

and x y zˆ ˆ ˆQ Q i Q j Q k

NCERT Corner

Barrier potential under reverse bias

V0

2 1

NCERT Corner covers additional information from NCERT textbook relevant to topic. This is our attempt to bridge the gap between NCERT and State Board textbook, thereby benefitting students in their preparation of National level competitive examinations.

QR code provides access to a video in order to boost understanding of a concept or activity. This is our attempt to facilitate learning with visual aids.

[Note: Students can scan the adjacent QR code to get conceptual clarity about resolution of vectors with the aid of a linked video.]

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4. A unit vector is represented as ˆ ˆ ˆ0.8 i bj 0.4k .

Hence the value of ‘b’ must be [MHT CET 2018]

(A) 0.4 (B) 0.6 (C) 0.2 (D) 0.2

Competitive Corner presents questions from prominent competitive exams based entirely on the syllabus covered in the chapter. This is our attempt to give competitive edge to the students.

Competitive Corner

Additional Information covers interesting information covered in textbook. This is our attempt to highlight the additional information sprinkled in the chapter.

Additional Information

According to the world health organization abillion young people could be at risk of hearingloss due to unsafe listening practices. Amongteenagers and young adults aged 12-35 years(i) about 50% are exposed to unsafe levels of

Exercise includes subtopic-wise additional questions, problems and MCQs. This is our attempt to provide additional practice to students to gauge their preparation.

Exercise

2.2 Vector Analysis 1. Distinguish between scalars and vectors. Ans: Refer Q. 2 2. Define the term negative vectors. Ans: Refer Q. 5 (iii)

The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearlymissed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on: [email protected] A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants! From, Publisher

This reference book is transformative work based on textbook Physics; First edition: 2019 published by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

Disclaimer

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Chapter No. Chapter Name Page No.

1 Units and Measurements 1

2 Mathematical Methods 30

3 Motion in a Plane 63

4 Laws of Motion 101

5 Gravitation 157

6 Mechanical Properties of Solids 195

7 Thermal Properties of Matter 227

8 Sound 274

9 Optics 304

10 Electrostatics 350

11 Electric Current Through Conductors 382

12 Magnetism 414

13 Electromagnetic Waves and Communication System 432

14 Semiconductors 456

Note: 1. * mark represents Textual question.

2. # mark represents Intext question.

3. +mark represents Textual examples.

4. symbol represents textual questions that need external reference

for an answer.

CONTENTS

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Std. XI Sci.: Perfect Physics

Physics is the branch of science which deals with the study of nature and natural phenomena.

It is a quantitative science where various physical quantities are measured.

A quantity which can be measured and with the help of which, various physical happenings can be explained and expressed in the form of laws, is called a physical quantity. Examples: length, mass, time, force etc.

Q.1. What is a measurement? How is measuredquantity expressed?

Ans: i. A measurement is a comparison with

internationally accepted standard measuringunit.

ii. The measured quantity (M) is expressed interms of a number (n) followed by acorresponding unit (u) i.e., M = nu.Example:Length of a wire when expressed as 2 m, itmeans value of length is 2 in the unit of m(metre).

Q.2. State true or false. If false correct thestatement and rewrite. Different quantities are measured in different units.

Ans: True. [Note: Choice of unit depends upon its suitability for measuring the magnitude of a physical quantity under consideration. Hence, we choose different scales for same physical quantity.]

Q.3. Can you recall? (Textbook page no. 1)i. What is a unit?ii. Which units have you used in the

laboratory for measuringa. length b. massc. time d. temperature?

iii. Which system of units have you used?Ans:i. The standard measure of any quantity is called

the unit of that quantity.ii.

Physical quantity Length Mass Time Temperature

Units millimetre, centimetre, metre

gram, kilog-ram

seconds minutes

Degree celsius degree fahrenheit

iii. MKS or SI system is used mostly. At times,even CGS system is used.

Q.4. Describe briefly different types of systemsof units.

Ans: System of units are classified mainly into four types:

i. C.G.S. system:It stands for Centimetre-Gram-Second system. In this system, length, mass and time are measured in centimetre, gram and second respectively.

ii. M.K.S. system:It stands for Metre-Kilogram-Second system.In this system, length, mass and time aremeasured in metre, kilogram and secondrespectively.

Units and Measurements 1

1.1 Introduction 1.2 System of Units 1.3 Measurement of Length 1.4 Measurement of Mass 1.5 Measurement of Time

1.6 Dimensions and Dimensional Analysis 1.7 Accuracy, Precision and Uncertainty in

Measurements 1.8 Errors in Measurements 1.9 Significant Figures

Contents and Concepts

1.1 Introduction

1.2 System of Units

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iii. F.P.S. system:It stands for Foot-Pound-Second system. In thissystem, length, mass and time are measured infoot, pound and second respectively.

iv. S.I. system:It stands for System International. This systemhas replaced all other systems mentionedabove. It has been internationally accepted andis being used all over world. As the SI unitsuse decimal system, conversion within thesystem is very simple and convenient.

Q.5. What are fundamental quantities? State two examples of fundamental quantities. Write their S.I. and C.G.S. units.

Ans: Fundamental quantities: The physical quantities which do not depend on any other physical quantity for their measurements i.e., they can be directly measured are called fundamental quantities. Examples: mass, length etc.

Fundamental quantities

S.I. unit C.G.S. unit

Mass kilogram (kg) gram (g) Length metre (m) centimetre

(cm)

Q.6. What are fundamental units? State the S.I.units of seven fundamental quantities.

Ans: Fundamental units: The units used to measure fundamental quantities are called fundamental units. S.I. Units of fundamental quantities:

Fundamental quantity

SI Units Name Symbol

Length metre m Mass kilogram kg Time second s

Electric current ampere A Thermo dynamic

Temperature kelvin K

Amount of substance mole mol Luminous intensity candela cd

Q.7. State and describe the two supplementaryunits.

Ans: The two supplementary units are: i. Plane angle (d):

a. The ratio of length of arc (ds) of ancircle to the radius (r) of the circle iscalled as Plane angle (d)

i.e., d =dsr

b. Thus, d is angle subtended by the arcat the centre of the circle.

c. Unit: radian (rad)d. Denoted as c

e. Length of arc of circle = Circumference of circle = 2r.

plane angle subtended by entire circle at

its centre is = 2πr

r = 2c

ii. Solid angle (d):a. solid angle is 3-dimensional analogue of

plane angle.b. Solid angle is defined as area of a

portion of surface of a sphere to thesquare of radius of the sphere.

i.e., d = 2

dAr

c. Unit: Steradian (sr)d. Denoted as ()e. Surface area of sphere = 4r2

solid angle subtended by entire sphere at

its centre is = 2

2

4πrr

= 4 sr

Q.8. Derive the relation between radian anddegree. Also find out 1 and 1 in terms of their respective values in radian.

Ans: We know that, 2 c = 360 c = 180

1c =180 180=

π 3.1416 = 57.296

Similarly, 1 = π 3.1416=

180 180 = 1.745 102 rad

r d

dAO

r d O ds

The three systems namely CGS, MKS and FPS were used extensively till recently. In 1971, the 14th International general conference on weights and measures recommended the use of ‘International system’ of units.


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Chapter 1: Units and Measurements

As, 1 = 60

1 = 21.745 10

60

= 2.908 104 rad.

As, 1 = 60

1 = 42.908 10

60

4.847 106 rad.

Q.9. What are derived quantities and derivedunits? State two examples. State the corresponding S.I. and C.G.S. units of the examples.

Ans: i. Derived quantities:

Physical quantities other than fundamentalquantities which depend on one or morefundamental quantities for their measurementsare called derived quantities.

ii. Derived units:The units of derived quantities which areexpressed in terms of fundamental units fortheir measurements are called derived units.

iii. Examples and units: Derived quantity Formula S.I.

unitC.G.S.

unitVelocity Unitof displacement

Unitof timem/s cm/s

Acceleration Unitof velocityUnitof time

m/s2 cm/s2

Momentum Unit of mass Unit of velocity

kg m/s g cm/s

Q.10. Classify the following quantities intofundamental and derived quantities: Length, Velocity, Area, Electric current, Acceleration, Time, Force, Momentum, Energy, Temperature, Mass, Pressure, Magnetic induction, Density.

Ans: Fundamental Quantities: Length, Electric current, Time, Temperature, Mass. Derived Quantities: Velocity, Area, Acceleration, Force, Momentum, Energy, Pressure, Magnetic induction, Density

Q.11. Classify the following units intofundamental, supplementary and derived units: newton, metre, candela, radian, hertz, square metre, tesla, ampere, kelvin, volt, mol, coulomb, farad, steradian.

Ans:

Fundamental units

Supplementary units

Derived units

metre candela ampere kelvin mol

radian steradian

newton hertz square metre tesla volt coulomb farad

Q.12. List the conventions followed while using SIunits.

Ans: Following conventions should be followed while writing S.I. units of physical quantities:

i. Unit of every physical quantity should berepresented by its symbol.

ii. Full name of a unit always starts with smallerletter even if it is named after a person,eg.: 1 newton, 1 joule, etc. But symbol for unitnamed after a person should be in capitalletter, eg.: N after scientist Newton, J afterscientist Joule, etc.

iii. Symbols for units do not take plural form.iv. Symbols for units do not contain any full stops

at the end of recommended letter.v. The units of physical quantities in numerator

and denominator should be written as oneratio. For example the SI unit of accelerationis m/s2 or m s–2 but not m/s/s.

vi. Use of combination of units and symbols forunits is avoided when physical quantity isexpressed by combination of two. Forexample, The unit J/kg K is correct whilejoule/kg K is not correct.

vii. A prefix symbol is used before the symbol ofthe unit.a. Prefix symbol and symbol of unit

constitute a new symbol for the unitwhich can be raised to a positive ornegative power of 10.For example,1 ms = 1 millisecond = 10–3 s1 μs = 1 microsecond = 10–6 s1 ns = 1 nanosecond = 10–9 s

b. Use of double prefixes is avoided whensingle prefix is available10–6 s = 1 μs and not 1 mms10–9 s = 1 ns and not 1 mμs

viii. Space or hyphen must be introduced whileindicating multiplication of two units e.g., m/sshould be written as m s–1 or m-s–1.

[Note: Students can scan the adjacentQR code to get conceptual clarity aboutdegree and radian with the aid of alinked video.]

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+Q.13. What is the solid angle subtended by themoon at any point of the Earth, given the diameter of the moon is 3474 km and its distance from the Earth 3.84 × 108 m?

Solution: Given: Diameter (D) = 3474 km

Radius of moon (R) = 1737 km= 1.737 106 m

Distance from Earth r = 3.84 108 m To find: Solid angle (d)

Formula: d = 2

dAr

Calculation: From formula,

d = 2

2

πRr

….( cross-sectional area

of disc of moon = R2)

d =

25

28

1.737 10

3.84 10

= 6.43 10–5 sr

Ans: Solid angle subtended by moon at Earth is 6.43 10–5 sr.

[Note: Above answer is obtained substituting value of as 3.142.] Q.14. Pluto has mean diameter of 2,300 km and

very eccentric orbit (oval shaped) around the Sun, with a perihelion (nearest) distance of 4.4 109 km and an aphelion (farthest) distance of 7.3 109 km. What are the respective solid angles subtended by Pluto from Earth’s perspective? Assume that Earth’s distance from the Sun can be neglected.

Solution:

Given: Radius of Pluto, R = 2300

2 km

= 1150 km Perihelion distance rp = 4.4 109 km Aphelion distance ra = 7.3 109 km

To find: Solid angles (dp and da)

Formula: d = 2

dAr =

2

2

Rr

Calculation: From formula,

dp =

2

29

1150

4.4 10

=

2

29

3.142 1150

4.4 10

= 2.146 10–13 sr

and da =

2

29

1150

7.3 10

=

2

29

3.142 1150

7.3 10

= 7.798 10–14 sr

Ans: Solid angle at perihelion distance is 2.146 10–13 sr and at aphelion distance is 7.798 10–14 sr.

Q.15. Define a metre.Ans: The metre is the length of the path travelled by

light in vacuum during a time interval of 1/299,792,458 of a second.

Q.16. What is parallax?Ans:i. Parallax is defined as the apparent change in

position of an object due to a change inposition of an observer.

ii. Explanation: When a pencil is held in front ofour eyes and we look at it once with our left eyeclosed and then with our right eye closed,pencil appears to move against the background.This effect is called parallax effect.

Q.17. What is parallax angle?Ans:i. Angle between the two directions along which

a star or planet is viewed at the two points ofobservation is called parallax angle(parallactic angle).

ii. It is given by = bD

where, b = Separation between two points of observation,

D = Distance of source from any point of observation.

*Q.18. Star A is farther than star B. Which starwill have a large parallax angle?

Ans:

i. ‘b’ is constant for the two stars

1D

S

D D

b

1.3 Measurement of Length

Solved Examples

A

DA

DB

B

BA

b

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ii. As star A is farther i.e., DA > DB A < B.Hence, star B will have larger parallax anglethan star A.

Q.19. Explain the method to determine thedistance of a planet from the Earth.

Ans: i. Parallax method is used to determine distance

of different planets from the Earth.ii. To measure the distance ‘D’ of a far distant

planet S, select two different observatories(E1 and E2).

iii. The planet should be visible from E1 and E2observatories simultaneously i.e. at the sametime.

iv. E1 and E2 are separated by distance ‘b’ asshown in figure.

E1E2 = b

v. The angle between the two directions alongwhich the planet is viewed, can be measured.It is parallax angle, which in this case is E1SE2 =

vi. The planet is far away from the (Earth)observers, henceb D

bD 1 and ‘’ is also very small.

Hence, E1 E2 can be considered as arc of length b of circle with S as centre and D as radius.

E1S = E2S = D

=bD ….( is taken in radian)

D =bθ

Thus, the distance ‘D’ of a far away planet ‘S’ can be determined using the parallax method.

Q.20. Explain how parallax method is used tomeasure distance of a star from Earth.

Ans: i. The parallax measured from two farthest

distance points on Earth for stars will be toosmall and hence cannot be measured.

ii. Instead, parallax between two farthest points(i.e., 2 AU apart) along the orbit of Eartharound the Sun (s) is measured.

Q.21 Explain how size of a planet or star ismeasured.

Ans: i. To determine the diameter (d) of a planet or

star, two diametrically opposite points of theplanet are viewed from the sameobservatory.

ii. If d is diameter of planet or star, anglesubtended by it at any single point on theEarth is called angular diameter of planet.

iii. Let angle be angle between these twodirections.

iv. If distance between the Earth and planet or

star (D) is known, = dD

v. This relation gives, d = DThus, diameter (d) of planet or star can bedetermined.

Q.22. Name the devices used to measure verysmall distances such as atomic size.

Ans: Devices used are: Electron microscope, tunnelling electron microscope.

S

D D

E1 E2 b Earth

Orbit of Earth around Sun

B A

Star

1 AU 1 AUS

For measuring largedistances, astronomers usethe following units. The astronomical unit(AU) is the mean distancebetween the centre of theEarth and the centre of theSun. 1 astronomical unit, (AU) = 1.496 × 1011 m A light year is the distance travelled by light inone year. 1 light year = 9.46 × 1015 m A parsec (pc) is the distance from where 1 AU subtends an angle of 1 second of arc.

r =

11

c –61AU 1.496 × 10=

4.847 × 101 = 3.08 1016 m

1 parsec (pc) = 3.08 × 1016 m 3.26 light years

Sun

1 pc

1 AU

1


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Q.23. Just as large distances are measured in AU,parsec or light year, atomic or nuclear distances are measured with the help of microscopic units. Match the units given in column A with their corresponding SI unit given in column B.

Column A Column B i. Angstrom (Å) a. 1015 m ii. femtometre (fm) b. 1010 m

c. 1012 m Ans: i. – (b), ii. – (a)

+Q.24.A star is 5.5 light years away from theEarth. How much parallax in arcsec will it subtend when viewed from two opposite points along the orbit of the Earth?

Solution: Two opposite points A and B along the orbit of the Earth are 2 AU apart. The angle subtended by AB at the position of the star is

= AB

distance of the star from the Earth

= 2AU5.5 yl

= 11

15

2 1.496 105.5 9.46 10

= 5.75 10–6 rad

= 5.75 10–6 57.297 60 60 arcsec ….(converting radian into arcsecond)

= 1.186 arcsec Ans: Parallax is 1.186 arcsec

+Q.25.The moon is at a distance of 3.84 × 108 mfrom the Earth. If viewed from two diametrically opposite points on the Earth, the angle subtended at the moon is 1 54. What is the diameter of the Earth?

Solution: Given: Distance (D) = 3.84 × 108 m

Subtended angle () = 1 54 = (60 + 54) = 114 = 114 × 2.91 × 10–4 rad = 3.317 × 10–2 rad

To find: Diameter of Earth (d) Formula: d = D Calculation: From formula,

d = 3.317 10–2 3.84 108

= 1.274 107 m Ans: Diameter of Earth is 1.274 107 m.

*Q.26.When the planet Jupiter is at a distance of824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72 of arc. Calculate the diameter of the Jupiter.

Solution: Given: Angular diameter () = 35.72

= 35.72 4.847 106 rad 1.73 104 radDistance from Earth (D)= 824.7 million km= 824.7 106 km= 824.7 109 m.

To find: Diameter of Jupiter (d) Formula: d = D Calculation: From formula,

d = 1.73 104 824.7 109 = 1.428 108 m = 1.428 105 km

Ans: Diameter of Jupiter is 1.428 105 km.

Q.27. Explain the method to measure mass.Ans: Method for measurement of mass:i. Mass, until recently, was measured with a

standard mass of the international prototype ofthe kilogram (a platinum-iridium alloycylinder) kept at international Bureau ofWeights and Measures, at Serves, near Paris,France.

ii. As platinum – iridium piece was seen to pickup microparticles and found to be affected byatmosphere, its mass could no longer betreated as constant.

iii. Hence, a new definition of mass wasintroduced in terms of electric current on 20th

May 2019.

B A

Star

1 AU 1 AUS Orbit of the Earth around the Sun

1.4 Measurement of Mass

Solved Examples

To measure sizes of microscopic object using microscope, the wavelength of light to be used in microscope is so chosen that it is smaller than size of the object to be measured. Thus, visible light with wavelength range of 4000 Å to 7000 Å can measure sizes upto 4000 Å. If object size is smaller than this, smaller wavelength will be needed thereby making use of electron microscope necessary. Approximate wavelength of electrons in electron microscope is about 0.6 Å allowing measurement of atomic sizes of 1 Å.


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iv. Now, one kilogram mass is described in termsof amount of current which has to be passedthrough electromagnet to pull one side ofextremely sensitive balance to balance theother side which holds one standard kg mass.

v. To measure mass of small entities such asatoms and nucleus, atomic mass unit (amu) isused.

It is defined as th1

12

mass of an unexcited

atom of carbon 12 (C12). 1 amu 1.66 1027 kg.

Q. 28. What can be the reason for choosing

Carbon-12 to define atomic mass unit? Ans: i. Unlike oxygen and hydrogen, which exhibit

various isotopes in higher proportions, carbon-12 is the single most abundant (98% ofavailable carbon) isotope of carbon.

ii. It is also very stable.Hence, it makes more accurate unit of measuringmass and is used to define atomic mass unit.

Q.29. Define mean solar day. Explain the methodfor measurement of time.

Ans: i. A mean solar day is the average time interval

from one noon to the next noon.Method for measurement of time: ii. The unit of time, the second, was considered

to be 1

86400 of the mean solar day, where

a mean solar day = 24 hours= 24 60 60 = 86400 s

iii. However, this definition proved to beunsatisfactory to define the unit of timeprecisely because solar day varies graduallydue to gradual slowing down of the Earth’srotation. Hence, the definition of second wasreplaced by one based on atomic standard oftime.

iv. Atomic standard of time is now used for themeasurement of time. In atomic standard oftime, periodic vibrations of caesium atom isused.

v. One second is time required for9,192,631,770 vibrations of the radiationcorresponding to transition between twohyperfine energy states of caesium-133(Cs - 133) atom.

vi. The caesium atomic clocks are very accurate.vii. The national standard of time interval ‘second’

as well as the frequency is maintained throughfour caesium atomic clocks.

Q.30. Define dimensions and dimensional formula

of physical quantities. Give few examples of dimensional formula.

Ans: i. Dimensions:

The dimensions of a physical quantity are thepowers to which the fundamental units must beraised in order to obtain the unit of a givenphysical quantity.

ii. Dimensional formula:When any derived quantity is represented withappropriate powers of symbols of thefundamental quantities, such an expression iscalled dimensional formula.It is expressed by square bracket with nocomma in between the symbols.

iii. Examples of dimensional formula:

a. Speed =Distance

time

Dimensions of speed = LT

= [L1 M0 T–1]

[Note: As power of M is zero, it can be omitted from dimensional formula. Therefore, dimensions of speed can be written as [L1 T1]]

b. Force = Mass acceleration

= Mass 2Distance

time

Dimensions of force = 2

LM

T

= [L1 M1 T2]

c. Kinetic energy =12 mv2

= 12 m

2DistanceTime

Dimension of kinetic energy

= [M1] 2

2

LT

= [L2 M1 T2]

1.5 Measurement of Time

1.6 Dimensions and Dimensional Analysis

For measurement of small masses of atomic orsubatomic particles, mass spectrography is used.This method uses the property that mass of thecharged particle is proportional to radius of thetrajectory when particle is moving in uniformelectric and magnetic field. Large masses in the universe like planets, stars etc. can be measured by using Newton’s law of gravitation.


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d. Temperature gradient =Temperature

Distance

∴ Dimension of temperature gradient = [ ][ ]KL

= [L–1 M0 T0K1]

Some derived quantities with their SI units and dimensions:

Physical Quantities Formula S.I. units Dimensions Speed Distance

Timem/s [L1 M0 T−1]

Acceleration Change in velocityTime

m/s2 [L1 M0 T−2]

Force Mass × Acceleration N (newton) [L1 M1 T−2] Pressure Force

Areakg/m-s2 [L−1 M1 T−2]

Density MassVolume

kg/m3 [L−3 M1 T0]

Work Force × distance J (joule) [L1 M1 T−2] [L1] = [L2 M1 T−2] Energy Force × distance J (joule) [L1 M1 T−2] [L1] = [L2 M1 T−2] Power Work

TimeW (watt) [L2 M1 T−3]

Momentum Mass × Velocity kg-m/s [L1 M1 T−1] Impulse Force × Time N-s [L1 M1 T−1] Temperature (T) -- K (kelvin) [L0 M0 T0 K1] Charge Current × Time C (coulomb) [L0 M0 T1A1] Resistance Potential difference

CurrentΩ (ohm) [L2 M1 T−3A−2]

Frequency 1Time

/s [L0 M0 T–1]

Planck’s constant Energyof PhotonFrequency

J-s [L2 M1 T−1]

Electric potential V =

Wq

V (volt) [L2 M1 T−3A−1]

[Note: Students can write θ for temperature instead of K and I for current instead of A in dimensional formula.] *Q.31 . What are the dimensions of the quantity

/ gl l , l being the length and g the acceleration due to gravity?

Ans: Quantity = l × gl ….(i)

gravitational acceleration, g = velocity

time

∴ g = distancetime × time

Substituting in equation (i),

Quantity = l × 2× time

distancel

∴ Dimensional formula of quantity

= [L] × 1/ 2 2×1/ 2

1/ 2

L T

L

= [L] × [T1] = [L1T1]

[Note: When power of symbol expressing fundamental quantity appearing in dimensional formula is not given, it is taken as 1.]

In Chapter 7, you will study in detail about temperature gradient.

Connections

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Q.32. A book with many printing errors containsfour different formulae for the displacement y of a particle undergoing a certain periodic function:

i. y = a sin2 tT

ii. y = a sin v t

iii. y =aT sin

ta

iv. y = a 2 t 2 tsin cosT T2

Here, a is maximum displacement of particle, v is speed of particle, T is time period of motion. Rule out the wrong formulae on dimensional grounds.

(NCERT) Ans: The argument of trigonometrical function, i.e.,

angle is dimensionless. Now,

i. The argument, 2 tT

= TT

= 1 = [L0 M0 T0]

which is a dimensionless quantity. Hence, formula (i) is correct.

ii. The argument,[v t] = [LT–1] [T] = [L] = [L1 M0 T0]which is not a dimensionless quantity.Hence, formula (ii) is incorrect.

iii. The argument,ta

= TL

= [L–1 M0 T1]

which is not a dimensionless quantity. Hence, formula (iii) is incorrect.

iv. The argument,2 tT

= TT

= 1 = [L0 M0 T0]

which is a dimensionless quantity. Hence, formula (iv) is correct.

Q.33. State principle of homogeneity ofdimensions.

Ans: Principle of homogeneity of dimensions: The dimensions of all the terms on the two sides of a physical equation relating different physical quantities must be same.

Q.34. State the uses of dimensional analysis.Ans: Uses of dimensional analysis:i. To check the correctness of a physical

equation. (Refer Q.35. for explanation)ii. To derive the relationship between related

physical quantities. (Refer Q. 36. forexplanation)

iii. To find the conversion factor between theunits of the same physical quantity in twodifferent systems of units. (Refer Q. 37. forexplanation)

Q.35. Explain the use of dimensional analysis tocheck the correctness of a physical equation.

Ans: Correctness of a physical equation by dimensional analysis:

i. A physical equation is correct only if thedimensions of all the terms on both sides ofthat equations are the same.

ii. For example, consider the equation of motion.v = u + at .…(1)

iii. Writing the dimensional formula of everyterm, we getDimensions of L.H.S. [v] = [L1 M0 T1],Dimensions of R.H.S. = [u] + [at]= [L1 M0 T1] + [L1 M0 T2] [L1 M0 T1]= [L1M0T–1] + [L1M0T–1] [L.H.S.] = [R.H.S.]

iv. As dimensions of both side of equation issame, physical equation is dimensionallycorrect.

Q.36. Time period of a simple pendulum dependsupon the length of pendulum (l) and acceleration due to gravity (g). Using dimensional analysis, obtain an expression for time period of a simple pendulum.

Ans: Expression for time period of a simple pendulum by dimensional analysis:

i. Time period (T) of a simple pendulumdepends upon length (l) and acceleration dueto gravity (g) as follows:T l a gb

i.e., T = k l a gb .…(1) where, k = proportionality constant, which isdimensionless.

ii. The dimensions of T = [L0 M0 T1]The dimensions of l = [L1 M0 T0]The dimensions of g = [L1 M0 T2]Taking dimensions on both sides of equation(1),[L0 M0 T1] = [L1 M0 T0]a [L1 M0 T2]b

[L0 M0 T1] = [La + b M0 T2b]iii. Equating corresponding power of L, M and T

on both sides, we geta + b = 0 .…(2)and 2b = 1

b = 12

Dimensional analysis can also be used to derive the formula of a physical quantity, provided we know the factors on which the physical quantity depends.


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iv. Substituting ‘b’ in equation (2), we get

a = 12

v. Substituting values of a and b in equation (1), we have,

T = k 1 12 2g

l

T = k

1 12 2

12

= k = kg gg

l l l

vi. Experimentally, it is found that k = 2

T = 2gl

This is the required expression for time period of a simple pendulum.

Q.37. Find the conversion factor between the S.I. and the C.G.S. units of work using dimensional analysis.

Ans: Conversion factor between units of same physical quantity:

i. Let ‘n’ be the conversion factor between the units of work.

1 J = n erg ….(1) iii. Dimensions of work in S.I. system are

2 1 21 1 1L M T and in CGS system are 2 1 22 2 2L M T

iv. From (1), 1 2 1 2

1 1 1L M T = n 2 1 22 2 2L M T

n = 2 1 21 1 12 1 22 2 2

L M TL M T

= 2 1 2

1 1 1

2 2 2

L M TL M T

....(2)

v. By expressing L, M and T into its corresponding unit we have,

n = 12 2m kg second

cm g second

.…(3)

vi. Since, 1 m = 100 cm and 1 kg = 1000 g, we have,

n = 2

2100cm 1000 g (1)cm g

n = 104 103 1 = 107 Hence, the conversion factor, n = 107 Therefore, from equation (1), we have, 1 J = 107 erg. Q.38. State the limitations of dimensional

analysis. Ans: Limitations of dimensional analysis: i. The value of dimensionless constant can be

obtained with the help of experiments only. ii. Dimensional analysis cannot be used to derive

relations involving trigonometric (sin , cos , etc.), exponential (ex, 2xe , etc.), and

logarithmic functions (log x, log x3, etc) as these quantities are dimensionless.

iii. This method is not useful if constant of proportionality is not a dimensionless quantity.

iv. If the correct equation contains some more terms of the same dimension, it is not possible to know about their presence using dimensional equation.

Q.39. If two quantities have same dimensions, do

they always represent the same physical content?

Ans: When dimensions of two quantities are same, they do not always represent the same physical content.

Example: Force and momentum both have same

dimensions but they represent different physical content.

Q.40. A dimensionally correct equation need not actually be a correct equation but dimensionally incorrect equation is necessarily wrong. Justify.

Ans: i. To justify a dimensionally correct equation

need not be actually a correct equation, consider equation,

v2 = 2as Dimensions of L.H.S. = [v2] = [L2M0T2] Dimensions of R.H.S. = [as]= [L2M0T2] [L.H.S.] = [R.H.S.] This implies equation v2 = 2as is

dimensionally correct.

Explanation for point (iii): Gravitational force between two point masses is directly proportional to product of the two masses and inversely proportional to square of the distance between the two

i.e., F 1 22

m mr

F = G 1 22

m mr

The constant of proportionality 'G' is NOT dimensionless. Thus, method of dimensional analysis will not work. Explanation for point (iv):

With standard symbols, the equation s = 21at2

is dimensionally correct. However, the

complete equation is, s = ut + 21at2

Reading between the lines

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But actual equation is, v2 = u2 + 2as This confirms a dimensionally correct equation need not be actually a correct equation.

ii. To justify dimensionally incorrect equation isnecessarily wrong, consider the formula,12mv = mgh

Dimensions of L.H.S. = [mv] = [L1M1T1]Dimensions of R.H.S. = [mgh] = [L2M1T2]Since the dimensions of R.H.S. and L.H.S. arenot equal, the formula given by equation mustbe incorrect.This confirms dimensionally incorrectequation is necessarily wrong.

Q.41. State, whether all constants aredimensionless or unitless.

Ans: All constants need not be dimensionless or unitless. Planck’s constant, gravitational constant etc., possess dimensions and units. They are dimensional constants.

Q.42. If length ‘L’, force ‘F’ and time ‘T’ aretaken as fundamental quantities, what would be the dimensional equation of mass and density?

Solution: i. Force = Mass Acceleration

Mass =Force

Acceleration Dimensional equation of mass

1

1 2

FDimensions of forceDimensions of acceleration LT

= [F1L1T2]

ii. Density = Mass

Volume

Dimensional equation of density

= Dimensions of massDimensions of volume

= 1 1 2

3

F L TL

= [F1L4T2]

Ans: i. The dimensional equation of mass is [F1L1T2].

ii. The dimensional equation of density is[F1L4T2].

*Q.43. Derive the formula of kinetic energy of aparticle having mass ‘m’ and velocity ‘v’, using dimensional analysis.

Solution: Kinetic energy of a body depends upon mass (m) and velocity (v) of the body.Let K.E. mx vy

K.E. = kmxvy ….(1)where,k = dimensionless constant of proportionality.Taking dimensions on both sides of equation(1),[L2M1T2] = [L0M1T0]x [L1M0T1]y

= [L0MxT0] [LyM0Ty] = [L0 + y Mx + 0 T0y]

[L2M1T2] = [LyMxTy] .…(2) Equating dimensions of L, M, T on both sides of equation (2), x = 1 and y = 2 Substituting x, y in equation (1), we have K.E. = kmv2

+Q.44.A calorie is a unit of heat and it equals4.2 J, where 1 J = kg m2 s–2. A distant civilisation employs a system of units in which the units of mass, length and time are kg, m and s. Also J is their unit ofenergy. What will be the magnitude ofcalorie in their units?

Solution: 1 cal = 4.2 kg m2s2

S.I. system New system L1 = 1 m L2 = m M1 = 1 kg M2 = kg T1 = 1 second T2 = second

New unit of energy is J Dimensional formula of energy is [L2M1T2] According to the question,

4.2[ 21L 1

1M 21T ] = J [ 2

2L 12M 2

2T ] Hence, magnitude of calorie in the new system is given by conversion factor, ‘J’.

J = 4.22 1 2

1 1 1

2 2 2

L M TL M T

= 4.2 2 1 2

1m 1kg 1sm kg s

J = 4.2 21 2 = 2

24.2γαβ

Ans: The magnitude of a calorie in terms of the new

units J is 2

24.2γαβ

.

*Q.45.An electron with charge e enters a uniform

magnetic field B

with a velocity

v. The velocity is perpendicular to the magnetic field. The force on the charge e is given by

|F

| = B e v. Obtain the dimensions of

B .

Solution: Given: | F

| = B e v Considering only magnitude, given equation is simplified to,

Solved Examples

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12

F = B e v

B = Fev

but, F = ma = m 2

distancetime

[F] = [M1] 1

2

LT

= [L1M1T2] Electric charge, e = current time [e] = [I1T1]

Velocity v = distance

time

[v] = LT

= [L1T–1]

Now, [B] = Fev

= 1 1 2

1 1 1 1

LM TT I LT

[B] = [L0M1T2 I–1] [Note: The answer given above is calculated in accordance with textual method considering the given data.]

*Q.46.An object is falling freely under the gravitational force. Its velocity after travelling a distance h is v. If v depends upon gravitational acceleration g and distance, prove with dimensional analysis that v = k gh where k is a constant.

Solution: Given = v = k gh

Quantity Formula Dimension Velocity (v) Distance

Time LT

= [L1T1]

Height (h) Distance [L1] Gravitational acceleration (g) 2

DistanceTime

2

LT

= [L1T2]

k being constant is assumed to be

dimensionless. Dimensions of L.H.S. = [v] = [L1T1] Dimension of R.H.S. = [ g h ] = [L1T2]1/2 [L1]1/2 = [L2T2]1/2 = [L1T1] As, [L.H.S.] = [R.H.S.], v = k gh is dimensionally correct

equation.

*Q.47. v = at + 0b + v

t + cis a dimensionally valid

equation. Obtain the dimensional formula for a, b and c where v is velocity, t is time and v0 is initial velocity.

Solution: Given: v = at + 0b + v

t + c

As only dimensionally identical quantities can be added together or subtracted from each other, each term on R.H.S. has dimensions of L.H.S. i.e., dimensions of velocity.

[L.H.S.] = [v] = [L1T1] This means, [at] = [v] = [L1T1] Given, t = time has dimension [T1]

[a] =

1 1L Tt

=

1 1

1

L TT

= 1 –2L T = [L1M0T–2]

Similarly, [c] = [t] = [T1] = [L0M0T1]

1

bT

= [v] = [L1T1]

[b] = [L1T1] [T1] = [L1] = [L1M0T0] Q.48. Assume that the speed (v) of sound in air

depends upon the pressure (P) and density () of air, then use dimensional analysis to obtain an expression for the speed of sound.

Solution: It is given that speed (v) of sound in air

depends upon the pressure (P) and density () of the air.

Hence, we can write, v = k Pa b …(1) where, k is a dimensionless constant and a and

b are powers to be determined. Dimensions of v = [L1M0T1] Dimensions of P = [L1M1T2] Dimensions of = [L3M1T0] Substituting the dimensions of the quantities

on both sides of equation (1), [L1M0T1] = [L1M1T2]a [L3M1T0]b [L1M0T1] = [LaMaT2a] [L3bMbT0] [L1M0T1] = [La 3bMa + bT2a] Comparing the powers of L, M and T on both

sides, we get, 2a = 1

a = 12

Also, a + b = 0

12 + b = 0 b =

12

Substituting values of a and b in equation (1), we get

v = k P1/2 1/2

v = k P

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Q.49. Density of oil is 0.8 g cm3 in C.G.S. unit. Find its value in S.I. units.

Solution: Dimensions of density is [L3M1T0]

C.G.S unit S.I unit

Dimension = 3 1 01 1 1L M T Dimension = 3 1 0

2 2 2L M T-é ùê úë û L1 = 1 cm M1 = 1 g

L2 = 1 m = 100 cm M2 = 1 kg = 1000 g

T1 = 1 s T2 = 1 s 0.8 g cm3 = conversion factor (n) kg m3

….(1) 0.8 [L 3

1 M 1

10

1T ] = n [L 32 M 1

20

2T ]

n = 3 1 0

1 1 13 1 0

2 2 2

0.8[L M T ][L M T ]

n = 0.83 1 0

1 1 1

2 2 2

L M TL M T

= 0.8 3 1 0

1cm 1g 1s100 cm 1000g 1s

= 0.8 [103]1 [102]3 = 0.8 [103] [10]6 n = 0.8 103 Substituting the value of ‘n’ in equation (1),

we get, 0.8 g cm3 = 0.8 103 kg m3. Ans: Density of oil in S.I unit is 0.8 103 kg m3. Q.50. The value of G in C.G.S system is

6.67 108 dyne cm2 g2. Calculate its value in S.I. system.

Solution: Dimensional formula of gravitational constant = [L3M1T2]

C.G.S system S.I. system Dimension=

3 1 21 1 1[ L M T ]- -

L1 = 1 cm M1 = 1 g T1 = 1 s

Dimension =3 1 22 2 2[ L M T ]

L2 = 1 m = 100 cm M2 = 1 kg = 1000 g T2 = 1 s

6.67 108 dyne cm2 g2 = Conversion factor (n) Nm2 kg2 ....(1) 6.67 108 3 1 2

1 1 1[ L M T ] = n 3 1 22 2 2[ L M T ]

n = 6.67 108 3 1 2

1 1 1

2 2 2

L M TL M T

n = 6.67 108 3 1 2

1 cm 1 g 1 s100 cm 1000 g 1 s

n = 6.67 108 106 103 n = 6.67 1011 From equation (1), 6.67 108 dyne cm2 g2 = 6.67 1011 N–m2 kg2 Ans: Value of G in S.I. system is 6.67 1011 N–m2kg2.

Q.51. What is accuracy? Ans: Accuracy is how close a measurement is to the

actual value of that quantity. Q.52. What is precision? Ans: Precision is a measure of how consistently a

device records nearly identical values i.e., reproducible results.

Q.53. A scale in a lab measures the mass of object consistently more by 500 g than their actual mass. How would you describe the scale in terms of accuracy and precision?

Ans: The scale is precise but not accurate. Explanation: Precision measures how

consistently a device records the same answer; even though it displays the wrong value. Hence, the scale is precise.

Accuracy is how well a device measures something against its accepted value. As scale in the lab is always off by 500 g, it is not accurate.

[Note: The goal of the observer should be to get accurate as well as precise measurements.]

1.7 Accuracy, Precision and Uncertainty in measurements

[Note: Students can scan the adjacent QR code to get conceptual clarity about accuracy and precision with the aid of a linked video.]

Results of Tata MumbaiMarathon 2019 saw Kenya’s Cosmas Lagat as winner withrace time. 2:09:15. AychewBante was at second position(2:10:05) and Shumet Akalnew(2:10:14) was a close third.

GG - Gyan Guru

The result clearly marks necessity of accuratemeasurement of time. To achieve this, mosttimed races today use race bibs with a timingchip. When a runner moves past a special matat starting line, chip registers beginning timeand as runner crosses the finish line, itregisters finishing time.

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Q.54. Can you tell? (Textbook page no.8) If ten students are asked to measure the

length of a piece of cloth upto a mm, using a metre scale, do you think their answers will be identical? Give reasons.

Ans: Answers of the students are likely to be different. Length of cloth needs to be measured up to a millimetre (mm) length.

Hence, to obtain accurate and precise reading one must use measuring instrument having least count smaller than 1 mm.

But least count of metre scale is 1 mm. As a result, even smallest uncertainty in reading would vary reading significantly. Also, skill of students doing measurement may also introduce uncertainty in observation.

Hence, their answers are likely to be different. Q.55. List reasons that may introduce possible

uncertainties in an observation. Ans: Possible uncertainties in an observation may

arise due to following reasons: i. Quality of instrument used, ii. Skill of the person doing the experiment, iii. The method used for measurement, iv. External or internal factors affecting the result

of the experiment. Q.56. What is systematic error? Classify errors

into different categories. Ans: i. Systematic errors are errors that are not

determined by chance but are introduced by an inaccuracy (involving either the observation or measurement process) inherent to the system.

ii. Classification of errors: Errors are classified into following two

groups: a. Systematic errors: 1. Instrumental error (constant error), 2. Error due to imperfection in experimental

technique, 3. Personal error (human error). b. Random error (accidental error) Q.57. What is instrumental (constant) error? Ans: Instrumental error: i. It arises due to defective calibration of an

instrument.

ii. Example: If a thermometer is not graduated properly, i.e., one degree on the thermometer actually corresponds to 0.99, the temperature measured by such a thermometer will differ from its value by a constant amount.

Q.58. What is error due to imperfection in experimental technique?

Ans: Error due to imperfection in experimental technique:

i. The errors which occur due to defective setting of an instrument is called error due to imperfection in experimental technique.

ii. For example the measured volume of a liquid in a graduated tube will be inaccurate if the tube is not held vertical.

Q.59. What is personal error? Ans: Personal error (Human error): i. The errors introduced due to fault of an observer

taking readings are called personal errors. ii. For example, while measuring the length of an

object with a ruler, it is necessary to look at the ruler from directly above. If the observer looks at it from an angle, the measured length will be wrong due to parallax.

Q.60. What is random error (accidental)? Ans: i. Random error (accidental): The errors which are caused due to minute

change in experimental conditions like temperature, pressure change in gas or fluctuation in voltage, while the experiment is being performed are called random errors.

ii. They can be positive or negative. iii. Random error cannot be eliminated completely

but can be minimized by taking multiple observations and calculating their mean.

Q.61. State general methods to minimise effect of systematic errors.

Ans: Methods to minimise effect of systematic errors:

i. By using correct instrument. ii. Following proper experimental procedure. iii. Removing personal error. Q.62. Define the terms: i. Arithmetic mean

*ii. Absolute error *iii. Mean absolute error *iv. Relative error *v. Percentage error Ans: i. Arithmetic mean: a. The most probable value of a large

number of readings of a quantity is called the arithmetic mean value of the quantity. This value can be considered to be true value of the quantity.

1.8 Errors in Measurements

The difference between measured value and true value of a physical quantity is called error.


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b. If a1, a2, a3,…………an are ‘n’ number of readings taken for measurement of a quantity, then their mean value is given by,

amean = n

a........aa n21

amean = 1n

n

ii 1

a

ii. Absolute error: a. For a given set of measurements of a

quantity, the magnitude of the difference between mean value (Most probable value) and each individual value is called absolute error (a) in the measurement of that quantity.

b. absolute error = |mean value measured

value| a1 = | amean a1 | Similarly, a2 = | amean a2 |, . . . . . . . . . . . . an = | amean an | iii. Mean absolute error: For a given set of measurements of a same

quantity, the arithmetic mean of all the absolute errors is called mean absolute error in the measurement of that physical quantity.

m eana = 1 2 na a ........ an

=

n

ii 1

1 an

iv. Relative error: The ratio of the mean absolute error in the measurement of a physical quantity to its arithmetic mean value is called relative error.

Relative error = mea n

m ea n

aa

v. Percentage error: The relative error represented by percentage

(i.e., multiplied by 100) is called the percentage error.

Percentage error = mean

mean

aa × 100%

[Note: Considering conceptual conventions question is modified to define percentage error and not mean percentage error.] Q.63 What does a = amean ± amean signify? Ans: a = amean ± amean signifies that the actual value

of a lies between (amean amean) and (amean + amean).

Q.64. Activity (Textbook page no.10) Perform an experiment using a Vernier

callipers of least count 0.01cm to measure the external diameter of a hollow cylinder. Take 3 readings at different positions on the cylinder and find (i) the mean diameter (ii) the absolute mean error and (iii) the percentage error in the measurement of diameter.

Ans: Given: L.C. = 0.01 cm To measure external diameter of hollow

cylinder readings are taken as follows:

Read-ing

Main Scale

Reading (M.S.R.)

cm

Vernier Scale

Reading (VSR)

(Coinciding div L.C.)

cm

Observed reading (MSR+ VSR)

cm

Mean Read-

ing (cm)

1. _______ _______ a1 amean 2. _______ _______ a2

3. _______ _______ a3 i. The mean diameter = amean = 1 2 3a + a + a

3

ii. Absolute mean error amean = 1 2 3Δa + Δa + Δa3

amean = mean 1 mean 2 mean 3a a + a a + a a3

iii. Percentage error = mean

mean

Δ a ×100a

[Note: The above table is made assuming zero error in Vernier callipers. If calliper has positive or negative zero error, the zero error correction needs to be introduced into observed reading.]

Types of zero error Corrected reading Negative: The instrument possesses –ve zero error when zero mark on Vernier scale is ahead of the main scale zero.

Original/observed reading + zero error.

Positive: The instrument possesses +ve zero error when zero mark on Vernier scale is behind the main scale zero.

Original/observed reading – zero error.

[Note: Students can scan the adjacent QR code to get conceptual clarity about how to read Vernier callipers with the aid of a linked video.]

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Q.65. What is meant by the term combination of errors?

Ans: Derived quantities may get errors due to individual errors of fundamental quantities, such type of errors are called as combined errors.

Q.66. Explain errors in sum and in difference of measured quantity.

Ans: Errors in sum and in difference: i. Suppose two physical quantities A and B have

measured values A ± ΔA and B ± ΔB, respectively, where ΔA and ΔB are their mean absolute errors.

ii. Then, the absolute error ΔZ in their sum, Z = A+B Z ± ΔZ = (A ± ΔA) + (B ± ΔB) = (A+B) ± ΔA ± ΔB ± ΔZ = ± ΔA ± ΔB, iii. For difference, i.e., if Z = A – B, Z ± ΔZ = (A ± ΔA) – (B ± ΔB) = (A – B) ± ΔAΔB ± ΔZ = ± ΔAΔB, iv. There are four possible values for ΔZ, namely

(+ΔA – ΔB), (+ΔA + ΔB), (–ΔA –ΔB), (–ΔA+ΔB). Hence, maximum value of absolute error is ΔZ = (ΔA+ΔB) in both the cases.

v. Thus, when two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities.

*67. If the measured values of two quantities are A ± A and B ± B, A and B being the mean absolute errors. What is the maximum possible error in A ± B?

Ans: Maximum possible error in (A ± B) is (A + B).

Q.68. Explain errors in product of measured quantity.

Ans: Errors in product: i. Suppose Z = AB and measured values of A

and B are (A ± ΔA) and (B ± ΔB) then, Z ± ΔZ = (A ± ΔA) (B ± ΔB) = AB ± AΔB ± BΔA ± ΔAΔB Dividing L.H.S by Z and R.H.S. by AB we get

ΔZ ΔB ΔA ΔA ΔB1 ± = 1 ± ± ±Z B A A B

Since ΔA/A and ΔB/B are very small, product is neglected. Hence, maximum relative error

in Z is ΔZ ΔA ΔB= +Z A B

ii. Thus, when two quantities are multiplied, the maximum relative error in the result is the sum of relative errors in each quantity.

Q.69. Show that if Z = AB

ΔZ ΔA ΔB= +Z A B

Ans: Errors in divisions:

i. Suppose, Z = AB and measured values of A

and B are (A ± A) and (B ± B) then,

Z ± Z = A ± ΔAB ± ΔB

Z Z1 Z

=

A 1 ± ΔA / AB 1 ± ΔB/B

1± Δ A / AA= ×B 1± Δ B / B

As, ΔBB << 1, expanding using Binomial

theorem,

Z Δ Z1 ±Z

= Z ΔA1 ±A

Δ B1B

…. A =ZB

1 ± Z

Z

= 1 ± ΔAA

B A BB A B

Ignoring term ΔAA

BB

,

Z

Z

= ± ΔAA B

B

This gives four possible values of Z

Z

as

ΔA ΔB ΔA ΔB ΔA ΔB+ , + + ,A B A B A B

and

ΔA ΔBA B

Maximum relative error of Z

Z

= ΔA ΔB+A B

ii. Thus, when two quantities are divided, the maximum relative error in the result is the sum of relative errors in each quantity.

Q.70. Explain errors due to power (index) of measured quantity.

Ans: Errors due to the power (index) of measured quantity:

i. Suppose Z = A3 = A × A × A

then, ΔZ ΔA ΔA ΔA= + +Z A A A

ii. Hence the relative error in Z =A3 is three times the relative error in A.

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iii. This means if Z = An

ΔZ ΔA= nZ A

iv. In general, if Z = p q

r

A BC

ΔZ ΔA ΔB ΔC= p + q + rZ A B C

v. This implies, the quantity in the formula which has large power is responsible for maximum error.

*Q.71. In a workshop a worker measures the length of a steel plate with a Vernier callipers having a least count 0.01 cm. Four such measurements of the length yielded the following values: 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm. Find the mean length, the mean absolute error and the percentage error in the measured value of the length.

Solution: Given: a1 = 3.11 cm, a2 = 3.13 cm, a3 = 3.14 cm, a4 = 3.14 cm Least count L.C. = 0.01 cm. To find: i. Mean length (amean) ii. Mean absolute error (amean) iii. Percentage error.

Formulae: i. amean = 1 2 3 4a a a a4

ii. an = mean na a

iii. amean = 1 2 3 4Δa + Δa + Δa + Δa4

iv. Percentage error = mean

mean

Δaa

100

Calculation: From formula (i),

amean = 3.11 3.13 3.14 3.14

4

= 3.13 cm From formula (ii), a1 = |3.13 3.11| = 0.02 cm a2 = |3.13 3.11| = 0 a3 = |3.13 3.14| = 0.01 cm a4 = |3.13 3.14| = 0.01 cm From formula (iii),

amean = 0.02 0 0.01 0.01

4

= 0.01 cm

From formula (iii),

% error =0.01 1003.13

= 0.32%

Ans: i. Mean length is 3.13 cm. ii. Mean absolute error is 0.01 cm. iii. Percentage error is 0.32 %.

[Note: As per given data of numerical, percentage error calculation upon rounding off yields percentage error as 0.32%.]

*Q.72. In Ohm’s experiments, the values of the unknown resistances were found to be 6.12 , 6.09 , 6.22 , 6.15 . Calculate the (mean) absolute error, relative error and percentage error in these measurements.

Solution: Given: a1 = 6.12 , a2 = 6.09 , a3 = 6.22 , a4 = 6.15 To find: i. Absolute error (amean) ii. Relative error iii. Percentage error

Formulae: i. amean = 1 2 3 4a a a a4

ii. an = meana a

iii. amean = 1 2 3 4Δa + Δa + Δa + Δa4

iv. Relative error = mean

mean

aa

v. Percentage error =

mean

mean

aa 100%


amean6.12 6.09 6.22 6.15

4

24.58 6.145

4 cm

From formula (ii),

a1 = |6.145 6.12| = 0.025 a2 = |6.145 6.09| = 0.055

a3 = |6.145 6.22| = 0.075

a4 = |6.145 6.15| = 0.005

From formula (iii),

amean = 0.025 0.055 0.075 0.005

4

= 0.160

4 = 0.04

From formula (iv),

Relative error = 0.046.145 = 0.0065

From formula (v), Percentage error = 0.0065 × 100 = 0.65% Ans: i. The mean absolute error is 0.04 . ii. The relative error is 0.0065 . iii. The percentage error is 0.65%. [Note: Framing of numerical is modified to reach the answer given to the numerical.]

Solved Examples

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18

+Q.73.The radius of a sphere measured repeatedly yields values 5.63 m, 5.54 m, 5.44 m, 5.40 m and 5.35 m. Determine the most probable value of radius and the mean absolute, relative and percentage errors.

Solution: Given: a1 = 5.63 m, a2 = 5.54 m, a3 = 5.44 m a4 = 5.40 m, a5 = 5.35 m, To find: i. Most probable value (Mean value) ii. Mean absolute error iii. Relative error iv. Percentage error

Formulae: i. 1 2 3 4 5mean

a a a a aa5

ii. an = mean na a iii. amean = 1 2 3 4 5a a a a a

5

iv. Relative error = mean

mean

aa

v. Percentage error = relative error 100%

Calculation: From formula (i), m eana = 5.63 5.54 5.44 5.40 5.35

5

= 27.365

= 5.472 m

From formula (ii), Absolute errors: a1 = |amean a1| = |5.472 5.63| = 0.158 a2 = |amean a2| = |5.472 5.54| = 0.068 a3 = |amean a3| = |5.472 5.44| = 0.032 a4 = |amean a4| = |5.472 5.40| = 0.072 a5 = |amean a5| = |5.472 5.35| = 0.122 From formula (ii),

amean = 0.158 0.068 0.032 0.072 0.122

5

= 0.452

5

= 0.0904 m iii. From formula (iii),

Relative error = 0.09045.472

= 1.652 10–2 (after rounding off to correct significant

digits) = 1.66 10–2 = 0.0166 Percentage error = 1.66 10–2 100 = 1.66% Ans: i. The mean value is 5.472 m. ii. The mean absolute error is 0.0904 m. iii. The relative error is 0.0166. iv. The percentage error is 1.66% [Note: Answer to relative error is rounded off using rules of significant figures and of rounding off.]

*Q.74. The masses of two bodies are measured to be 15.7 ± 0.2 kg and 27.3 ± 0.3 kg. What is the total mass of the two and the error in it?

Solution: Given: A ± A = 15.7 ± 0.2 kg and B ± B = 27.3 ± 0.3 kg. To find: Total mass (Z), and total error (Z) Formulae: i. Z = A + B ii. ± Z = ± A ± B Calculation: From formula (i), Z = 15.7 + 27.3 = 43 kg From formula (ii), ± Z = (± 0.2) + (± 0.3) = ± (0.2 + 0.3) = ± 0.5 kg Ans: Total mass is 43 kg and total error is ± 0.5 kg.

*Q.75. The distance travelled by an object in time (100 ± 1) s is (5.2 ± 0.1) m. What is the speed and it’s maximum relative error?

Solution: Given: Distance (D ± D) = (5.2 ± 0.1) m, time (t ± t) = (100 ± 1) s. To find: Speed (v), maximum relative error

Δvv

Formulae: i. v = Dt

ii. Δv ΔD Δ t+v D t


v = 5.2100 = 0.052 m/s

From formula (ii),

Δvv = ± 0.1 1

5.2 100

= ± 1 152 100

= ±19650

= ± 0.029 m/s Ans: The speed is 0.052 m/s and its maximum

relative error is ± 0.029 m/s. [Note: Framing of numerical is modified to make it specific and meaningful.]

*Q.76. If the formula for a physical quantity is

X = 4 3

1/ 3 1/ 2a b

c d and if the percentage error in

the measurements of a, b, c and d are 2%, 3%, 3% and 4% respectively. Calculate percentage error in X.

Solution:

Given: X = 4 3

1/3 1/2

a bc d

Percentage error in a, b, c, d is respectively 2%, 3%, 3% and 4%.

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Now, Percentage error in X

= Δa Δb 1 Δc 1 Δd4 + 3 + + 100%a b 3 c 2 d

= 1 14 2 3 3 3 4 100%3 2

= [8 + 9 + 1 + 2] 100% = 20% *77. Find the percentage error in kinetic energy

of a body having mass 60.0 ± 0.3 g moving with a velocity 25.0 ± 0.1 cm/s.

Solution: Given: m = 60.0 g, v = 25.0 cm/s, m = 0.3 g, v = 0.1 cm/s To find: Percentage error in E Formula: Percentage error in E

= m v2m v

100%

Calculation: From formula, Percentage error in E

= 0.3 0.1260.0 25.0

100%

= 1.3% Ans: The percentage error in energy is 1.3%.

+Q.78.In an experiment to determine the volume of an object, mass and density are recorded as m = (5 ± 0.15) kg and ρ = (5 ± 0.2) kg m–3 respectively. Calculate percentage error in the measurement of volume.

Solution: Given: M = 5 kg, M = 0.15 kg, = 5 kg/m3, = 0.2 kg/m3 To find: Percentage error in volume (V)

Formulae: i. V = M

ii. V

V

100%

Calculation: Using data,

MM

= 0.15

5 = 0.03

and

= .25

= 0.04

From formula (i),

V

V

= MM

+

….(Using result from error in division)

V

V

= 0.03 + 0.04 = 0.07

From formula (ii),

VV

100% = (0.07 100)% = 7%

Ans: The percentage error in the determination of volume is 7%.

*Q.79. If the length of a cylinder is l = (4.00 ± 0.001) cm, radius r = (0.0250 ± 0.001) cm and mass m = (6.25 ± 0.01) g. Calculate the percentage error in the determination of density.

Solution: Given: l = (4.00 ± 0.001) cm, r = (0.0250 ± 0.001) cm, m = (6.25 ± 0.01) g To find: percentage error in density Formulae:

i. Relative error in volume, ΔV 2Δr Δ= +V r

ll

….( Volume of cylinder, V = r2l)

ii. Relative error Δρ Δm ΔV= +ρ m V

….[ Density () =

mass mvolume V

]

iii. Percentage error = Relative error 100 % Calculation: From formulae (i) and (ii),

Δρρ

= Δm 2Δr Δ+ +m r

ll

= 2 0.0010.01 0.001

6.25 0.0250 4.00

= 0.08185 From formula (iii),

% error in density = Δρ ×100ρ

= 0.08185 100 = 8.185% Ans: Percentage error in density is 8.185%.

+Q.80. The acceleration due to gravity is determined by using a simple pendulum of length l = (100 ± 0.1) cm. If its time period is T = (2 ± 0.01) s, find the maximum percentage error in the measurement of g.

Solution: Given: l = 0.1 cm, l = 100 cm, T = 0.01 s,

T = 2 s To find: Percentage error

Formulae: i. T = 2g

l

ii. Percentage error = g 100g

Calculation: From formula (i), T2 = 42l/g .…(Squaring both sides)

g = 422T

l

gg =

ll

+2 TT

….(Using result from error in division)

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20

From formula (ii),

Percentage error = 2 T 100T

ll

= 0.1 2 0.01 100100 2

= (0.001 + 0.01) 100 = 1.1 % Ans: Percentage error in measurement of g is 1.1%.

Q.81.*Describe what is meant by significant figures.State the rules for determining significant figures.

Ans: Significant figures: i. Significant figures in the measured value of a

physical quantity is the sum of reliable digitsand the first uncertain digit.

ORThe number of digits in a measurement aboutwhich we are certain, plus one additional digit,the first one about which we are not certain isknown as significant figures or significant digits.

ii. Larger the number of significant figures obtainedin a measurement, greater is the accuracy of themeasurement. The reverse is also true.

iii. If one uses the instrument of smaller least count,the number of significant digits increases.Rules for determining significant figures:

i. All the non-zero digits are significant, forexample if the volume of an object is178.43 cm3, there are five significant digitswhich are 1,7,8,4 and 3.

ii. All the zeros between two nonzero digits aresignificant, eg., m = 165.02 g has 5 significantdigits.

iii. If the number is less than 1, the zero/zeroes on theright of the decimal point and to the left of thefirst nonzero digit are not significant e.g. in0.001405, the underlined zeros are not significant.Thus the above number has four significant digits.

iv. The zeros on the right hand side of the lastnonzero number are significant (but for this,the number must be written with a decimalpoint), e.g. 1.500 or 0.01500 both have 4significant figures each.On the contrary, if a measurement yieldslength L given as L = 125 m = 12500 cm =125000 mm, it has only three significant digits.

Q.82. Find the number of significant figures inthe following numbers. i. 25.42 ii. 0.004567iii. 35.320 iv. 91.000

Ans:

No. Number No. of significant figures Reason

i. 25.42 4 Rule no. i. ii. 0.004567 4 Rule no. iii. iii. 35.320 5 Rule no. iv. iv. 91.000 5 Rule no. iv.

*Q.83.Write down the number of significantfigures in the following: 0.003 m2, 0.1250 g cm–2, 6.4 106 m, 1.6 10–19 C, 9.1 10–31 kg.

Ans: Number No. of significant

figures Reason

0.003 m2 1 Rule no. iii. 0.1250 g cm2 4 Rule no. iv.

6.4 106 m 2 Rule no. i. 1.6 1019 C 2 Rule no. i. 9.1 1031 kg 2 Rule no. i.

1.9 Significant figures

NCERT Corner

Addition and Subtraction: In addition or subtraction, the final result shouldretain as many decimal places as there are in thenumber with the least decimal places.

Rules for obtaining significnt figuresupon performing arithmetic operations:

Example: the sum of the numbers536.32 g, 127.2 g and 0.301 g by mere arithmeticaddition, is 663.821 g. But the least precisemeasurement (127.2 g) is correct to only onedecimal place. The final result should, therefore,be rounded off to 663.8 g. Multiplication and Division: In multiplication or division, the final resultshould retain as many significant figures as thereare in the original number with the leastsignificant figures. Example: if the speed of light is given as3.00 108 m s1 (three significant figures) andone year (1 y = 365.25 d) has 3.1557 107 s (fivesignificant figures), the light year(3.00 108 3.1557 107) is 9.47 1015 m(three signicant figures).

Rules for rounding-off the numbers: While rounding-off numbers in measurement, following rules are applied. i. If the digit to be dropped is smaller than 5,

then the preceding digit should be leftunchanged. eg. 7.34 is rounded-off to 7.3.

ii. If the digit to be dropped is greater than 5,then the preceding digit should be raised by 1.eg. 17.26 is rounded-off to 17.3


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Q.84. *Describe what is meant by order of

magnitude. Explain with suitable examples. Ans: Order of magnitude: The magnitude of any physical quantity can be

expressed as A × 10n where ‘A’ is a number such that 0.5 ≤ A < 5 then, ‘n’ is an integer called the order of magnitude.

Examples: i. Speed of light in air = 3 × 108 m/s order of magnitude = 8 ii. Mass of an electron = 9.1 × 10 31 kg = 0.91 10–30 kg order of magnitude = –30 Q.85. Add 7.21, 12.141 and 0.0028 and express the

result to an appropriate number of significant figures.

Solution:

7 . 21

12.1410.0028

Sum 19.3538

In the given problem, minimum number of digits after decimal is 2.

Result will be rounded off upto two places of decimal.

Ans: Corrected rounded off sum is 19.35. *Q.86. The diameter of a sphere is 2.14 cm.

Calculate the volume of the sphere to the correct number of significant figures.

Ans: Volume of sphere = 34πr3

= 34 2.143.142

3 2

…. dr =2

= 43 3.142 (1.07)3

= 5.132113475 cm3

In multiplication or division, the final result should retain as many significant figures as there are in the original number with the least significant figures.

Volume in correct significant figures = 5.13 cm3

*Q.87.The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Solution: Given: l = 4.234 m, b = 1.005 m, t = 2.01 cm = 2.01 102 m = 0.0201 m To find: i. Area of sheet to correct

significant figures (A) ii. Volume of sheet to correct

significant figures (V) Formulae: i. A = 2(lb + bt + tl) ii. V = l b t Calculation: From formula (i), A = 2(4.234 1.005 + 1.005

0.0201 + 0.0201 4.234) A = 8.721 m2 In correct significant figure, A = 8.72 m2

From formula (ii), V = 4.234 1.005 0.0201 = 0.085528917 m3 In correct significant figure (rounding

off), V = 0.086 m3 Ans: i. Area of sheet to correct significant

figures is 8.72 m2. ii. Volume of sheet to correct significant

figures is 0.086 m3. [Note: The given solution is arrived to by considering a rectangular sheet.] Q.88. The mass of a box measured by a grocer’s

balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (i) the total mass of the box? (ii) the difference in the masses of the pieces to correct significant figures? (NCERT)

Solution: i. Total mass of the box = (2.3 + 0.02017 + 0.02015) kg = 2.34032 kg

Solved Examples

iii. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit should be raised by 1.

eg. 7.351, on being rounded-off to first decimal becomes 7.4.

iv. If the digit to be dropped is 5 or 5 followed by zero, then the preceding digit is not changed if it is even.

eg. 3.45, on being rounded-off becomes 3.4. v. If the digit to be dropped is 5 or 5 followed

by zeros, then the preceding digit is raised by 1 if it is odd. eg. 3.35, on being rounded-off becomes 3.4.

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22

Since, the last number of significant figure is 2, therefore, the total mass of the box = 2.3 kg

ii. Difference of mass = (20.17 20.15) = 0.02 g Since, there are two significant figures so the

difference in masses to the correct significant figures is 0.02 g.

Ans: i. The total mass of the box to correct significant figures is 2.3 kg.

ii. The difference in the masses to correct significant figures is 0.02 g.

*Q.89. Nuclear radius R has a dependence on the mass number (A) as R =1.3 × 10–16A1/3 m. For a nucleus of mass number A = 125, obtain the order of magnitude of R expressed in metre.

Ans: R = 1.3 1016 A1/3 m For A = 125 R = 1.3 1016 (125)1/3 = 1.3 1016 5 = 6.5 1016 = 0.65 1015 m Order of magnitude = 15 *Q.90.A large ball 2 m in radius is made up of a

rope of square cross section with edge length 4 mm. Neglecting the air gaps in the ball, what is the total length of the rope to the nearest order of magnitude?

Ans: Volume of ball = Volume enclosed by rope.

43 (radius)3 = Area of cross-section of rope

length of rope.

length of rope l = 34 πr

3A

Given: r = 2 m and Area = A = 4 4 = 16 mm2 = 16 106 m2

l = 3

66

4 3.142 2 3.142 2 103 16 10 3

m

2 106 m. Total length of rope to the nearest order of

magnitude = 106 m = 103 km Q.91. Write the dimensions of a and b in the relation

2b xE=a

Where E is energy, x is distance and t is time.

Ans: The given relation is 2b xE

a

As x is subtracted from b, dimensions of b are x2 ;

i.e., b = [L2]

We can write equation as 2LE

a

Or 2La

E =

2

2 2

L[L MT ] = [L0M1T2]

Q.92. What is the difference between 6.0 and

6.00? which is more accurate? Ans: 6.0 indicates the measurement is correct up to

first decimal place, whereas 6.00 indicates that the measurement is correct up to second decimal place. Thus, 6.00 is a more accurate value than 6.0.

Q.93. A child walking on a footpath notices that the width of the footpath is uneven. He reported this to his school principal and the complaint was forwarded to the municipal officer.

i. What is the possible error encountered? ii. What is the relative error in width of

footpath if width of footpath in 10 m length are noted as 5 m, 5.5 m, 5 m, 6 m and 4.5 m?

Ans: i. The error encountered is personal error. ii. Mean value of widths

wmean = 1 2 3 4 5w w w w w5

= 5 5.5 5 6 4.5

5

= 265 = 5.2 m

Mean absolute error of widths

wmean = 1 2 3 4 5w w w w w5

= 0.2 0.3 0.2 0.8 0.7

5

= 0.44 m

Relative error = mean

mean

ww =

0.445.2 = 0.084

The relative error in width of footpath is 0.084. Q.94. A factory owner kept five identical spheres

between two wooden blocks on a ruler as shown in figure. He called all his workers and told them to take reading, to check their efficiency and knowledge.

Apply Your Knowledge

cm

Spheres Wooden block

SAMPLE C

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23


i. What is the area of central sphere? ii. What is the absolute error in reading of

diameter of second sphere? Ans: i. From above diagram radius of central sphere is

r = 1 cm Area = r2 = 3.14 (1)2 = 3.14 cm2 The area of central sphere is 3.14 cm2. ii. Mean value of all reading of diameters

dmean = 1 2 3 4 5d d d d d5

=

2 2 2 2 25

= 105 = 2 cm

Absolute error in reading of second sphere. d2 = |dmean d2| = 2 2 = 0 The absolute error in reading of diameter of

second sphere is zero. Q.95. A potential difference of V = 100 ± 2 volt,

when applied across a resistance R gives a current of 10 ± 0.5 ampere. Calculate percentage error in R given by R = V/I.

Ans: Here, V = 100 ± 2 volt and I = 10 ± 0.5 ampere Expressing limits of error as percentage error, We have

V = 100 volt ± 2

100 100% = 10 volt ± 2%

and I = 10 ampere ± 0.510 100%

= 10 ampere ± 5%

R = VI

% error in R = % error in V + % error in I = 2% + 5% = 7% Q.96. Internet my friend (Textbook page no.12) i. videolectures.net/mit801f99_lewin_lec01/ ii. hyperphysics.phy-astr.gsu.edu/hbase/

hframe.html [Students can use links given above as reference and collect information about units and measurements.]

Quick Review

possess

lead to

Should be done considering

require

ideally should be devoid of

estimated as

measuredusing

give rise toUnits

need

Instrumental

Derived

Dimensions

Dimensional analysis

Significant figures & order of magnitude

Accuracy, Precision, Knowledge of uncertainty in measurement

Errors

Combination of errors

Random

Systematic Personal error

Errors due to imperfectionin experimental

Derived Fundamental

System ofunits

FPS

CGS

MKS

SI

Relative error mean

mean

aa

Percentage error mean

mean

aa × 100%

Absolute error an = mean na a

Mean absolute error

m eana = 1 2 na a ........ an

= n

ii 1

1 an

Physical Quantities

Measurement of quantities

metre (Length) kilogram (Mass) second (Time) ampere (Current) kelvin (Temperature) mole (Amount of substance) candela (Luminous intensity)

Fundamental quantities

Length Mass Time Current Temperature Amount of substance Luminous intensity

types

SAMPLE C

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24

1. Measure of physical quantity: M = nu where, n = numerical value, u = unit 2. Relation between numerical value and size

of unit: n1u1 = n2u2 3. Conversion factor of a unit in two system of

units:

n = a

1

2

MM

b

1

2

LL

c

1

2

TT

4. Plane angle: d = dsr

5. Solid angle: d = 2

dAr

6. Parallax angle: = bD

7. Diameter of planet/star: d = D.

8. Average value or mean value:

amean = 1 2 3 na a a .. an

=

1n i 1

n

ia

9. Absolute error: = Average value Measured value | an | = | amean an | 10. Mean absolute error:

amean = 1 2 na a ... an

=

1n

n

i 1ia

11. Relative (fractional) error: = mean

mean

aa

12. Percentage error: = mean

mean

a 100a

% 13. If Z = A B, then maximum error: Z = (A + B)

14. If Z = AB or Z = AB then,

ΔZZ = ± ΔA ΔB+

A B

15. If Z = Am Bn, then error in measurement:

Z

Z

= m A

A

+ n BB

Various prefixes to express a physical

quantity: Prefix Symbol Power of

10 Prefix Symbol Power of 10

Tera T 1012 micro 106 Giga G 109 nano n 109 Mega M 106 angstrom Å 1010 Kilo k 103 pico p 1012 milli m 103 femto f 1015 1.1 Introduction 1. Define unit of physical quantity. Ans: Refer Q.3. (i). 1.2 System of Units 2. Define F.P.S system. Ans: Refer Q.4. 3. Define C.G.S system. Ans: Refer Q.4. 4. Compare plane angle and solid angle. Ans: Refer Q.7. 1.3 Measurement of Length 5. Explain the method to determine the distance of

a planet from the Earth. Ans: Refer Q.19. 1.4 Measurement of Mass 6. What is atomic mass unit (amu)? Why is

element carbon used to define it? Ans: Refer Q.27 (v) and Q.28. 1.5 Measurement of time 7. Why was solar day rejected as a unit of time? Ans: Refer Q.29. 1.6 Dimensions and Dimensional Analysis 8. What are the dimensions of power? Ans: [L2M1T3] 9. What are the dimensions of frequency? Ans: [L0M0T1] 10. Find the conversion factor between S.I. and

CGS units of force using dimensional analysis. Ans: Conversion factor = 105 i.e., 1 N = 105 dyne. 11. Find the dimensional correctness of

kinematical equations: v = u + at Ans: Refer Q.35.

Important Formulae

Exercise

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25


12. Check the correctness of T = 2gl

Ans: Refer Q.36. 13. Find the conversion factor between the S.I. and

the C.G.S. units of work using dimensional analysis.

Ans: Refer Q.37. 14. The value of G in C.G.S system is

6.67 108 dyne cm2 g2. Calculate its value in S.I. system.

Ans: 6.67 × 1011 Nm2/kg2 1.7 Accuracy, Precision and Uncertainty in

Measurements 15. What are the reasons that may introduce

possible uncertainties in an observation? Ans: Refer Q.55. 1.8 Errors in Measurements 16. Explain: i. Arithmetic mean ii. absolute error iii. relative error iv. percentage error v. mean absolute error Ans: Refer Q.62. 17. Error in the measurement of radius of a sphere

is 1%. Then error in the measurement of volume will be?

Ans: 3% 18. The length of a rod as measured in an

experiment was found to be 2.48 m, 2.46 m, 2.49 m, 2.50 m and 2.48 m. Find the mean absolute error, relative error and percentage error.

Ans: (i) 0.01m (ii) 0.004 m (iii) 0.4% 19. The length of a metal plate was measured

using a Vernier callipers of least count 0.01 cm. The measurement made were 4.11 cm,4.13 cm, 4.21 cm and 4.09 cm. Find the mean length, the mean absolute error, relative error and the percentage error in the measurement of length.

Ans: (i) 4.135 cm, (ii) 0.0375 cm, (iii) 9.068×103, (iv) 0.906 %

1.9 Significant figures 20. Define significant figures. Ans: Refer Q.81. (only definition) 21. What are the rules for determining significant

figures? Ans: Refer Q.81. (only rules)

22. What is order of magnitude? Explain with two examples.

Ans: Refer Q.84. 23. Add 3.8 106 and 4.2 105 with due

regards to significant figures. Ans: 4.6 × 105 24. Two different masses are determined as

(23.7 0.5) g and (17.6 0.3) g. What is the sum of their masses?

Ans: 41.3 ± 0.8 g 25. If three measurements made are 18.425 cm,

7.21 cm and 5.0 cm. Write its addition up to proper significant figures.

Ans: 31cm 26. The length, breadth and thickness of a

rectangular sheet of metal are 4.658 m, 1.356 m and 2.04 cm respectively. Give the area and volume of the sheet to correct significant figures.

Ans: (i) 6.316 m2, (ii) 0.31 m3 27. The acceleration due to gravity at a place is

9.8 ms2. Find its value in km h2 and its order of magnitude for that value.

Ans: 127008 km/h2, 5 28. Find the order of magnitude of following data. i. Height of a tower 4325 m ii. Weight of a car 789 kg iii. Charge on electron 1.6 1019 C Ans: (i) 3, (ii) 3, (iii) 19 29. Round off the following numbers as indicated i. 15.654 upto 3 digits ii. 1426 upto 5 digits iii. 5.996 105 upto 3 digits iv. 0.06218 upto 4 digits v. 2.663 upto 3 digits vi. 6.221 upto 2 digits Ans: (i) 15.6, (ii) 1426.0, (iii) 6.00 × 105, (iv) 0.062, (v) 2.66, (vi) 6.2 30. Find the number of significant figures in the

following numbers. i. 25.42 ii. 0.004567 iii. 35.320 iv. 4.56 × 108

v. 1.609 × 1019 vi. 91.000 Ans: (i) 4, (ii) 4, (iii) 5, (iv) 3, (v) 4, (vi) 5 31. What will be the kinetic energy of body if its

mass is 2 kg and moving with a velocity of 2 m/s? Write its order of magnitude and significant figures.

Ans: (i) 4 J, (ii) 0, (iii) 1

SAMPLE C

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26


26

1. A physical quantity may be defined as (A) the one having dimension. (B) that which is immeasurable. (C) that which has weight. (D) that which has mass. 2. Which of the following is the fundamental unit? (A) Length, force, time (B) Length, mass, time (C) Mass, volume, height (D) Mass, velocity, pressure 3. Which of the following is NOT a fundamental

quantity? (A) Temperature (B) Electric charge (C) Mass (D) Electric current *4. Which of the following is not a fundamental

unit? (A) cm (B) kg (C) centigrade (D) volt 5. The distance of the planet from the earth is

measured by _______. (A) direct method (B) directly by metre scale (C) spherometer method (D) parallax method

6. The two stars S1 and S2 are located at distances d1 and d2 respectively. Also if d1 d2 then following statement is true.

(A) The parallax of S1 and S2 are same. (B) The parallax of S1 is twice as that of S2 (C) The parallax of S1 is greater than

parallax of S2 (D) The parallax of S2 is greater than

parallax of S1 7. Which of the following is NOT a unit of time? (A) Hour (B) Nano second (C) Microsecond (D) parsec *8. Light year is a unit of (A) time (B) mass (C) distance (D) luminosity 9. An atomic clock makes use of _______.

(A) cesium-133 atom (B) cesium-132 atom (C) cesium-123 atom (D) cesium-131 atom

10. S.I. unit of energy is joule and it is equivalent to (A) 106 erg (B) 107 erg (C) 107 erg (D) 105 erg

11. [L1M1T1] is an expression for _______. (A) force (B) energy (C) pressure (D) momentum 12. Dimensions of sin is (A) [L2] (B) [M] (C) [ML] (D) [M0L0T0] *13. [L1M1T–2] is the dimensional formula for (A) velocity (B) acceleration (C) force (D) work *14 Dimensions of kinetic energy are the same as

that of (A) force (B) acceleration (C) work (D) pressure 15. Accuracy of measurement is determined by (A) absolute error (B) percentage error (C) human error (D) personal error 16. Zero error of an instrument introduces _______. (A) systematic error (B) random error (C) personal error (D) decimal error 17. The diameter of the paper pin is measured

accurately by using _______. (A) Vernier callipers (B) micrometer screw gauge (C) metre scale (D) a measuring tape

*18. The error in the measurement of the sides of a rectangle is 1%. The error in the measurement of its area is

(A) 1% (B) 1/2% (C) 2% (D) None of the above. 19. The number of significant figures in

11.118 106 is (A) 3 (B) 4 (C) 5 (D) 6 20. 0.00849 contains ______ significant figures. (A) 6 (B) 5 (C) 3 (D) 2 21. 3.310 102 has ______ significant figures. (A) 6 (B) 4 (C) 2 (D) 1 22. The Earth’s radius is 6371 km. The order of

magnitude of the Earth’s radius is (A) 103 m (B) 109 m (C) 107 m (D) 102 m 23. ________ is the smallest measurement that

can be made using the given instrument (A) Significant number (B) Least count (C) Order of magnitude (D) Relative error

Multiple Choice Questions

SAMPLE C

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27


1. (A) 2. (B) 3. (B) 4. (D) 5. (D) 6. (D) 7. (D) 8. (C) 9. (A) 10. (C) 11. (D) 12. (D) 13. (C) 14. (C) 15. (B) 16. (A) 17. (B) 18. (C) 19. (C) 20. (C) 21. (B) 22. (C) 23. (B) 18. A = l b

ΔA Δ Δb= +A b

ll

= 1% + 1% = 2%

1. In an experiment, the percentage of error

occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X,

where X = 2

2

133

1AB

C D, will be:[NEET (UG) 2019]

(A) –10 % (B) 10 %

(C) 313

% (D) 16 %

Hint: Given: X = 2

2

133

1AB

C D

Error contributed by A = 2 A 100A

= 2 1% = 2%

Error contributed by B = 12 B 100

B

= 12 2% = 1%

Error contributed by C = 13 C 100

C

= 13 3% = 1%

Error contributed by D = 3 D 100D

= 3 4 = 12% Percentage error in x is given as,

xx

100 = (error contributed by A) +

(error contributed by B) + (error contributed by C) + (error contributed by D)

= 2% + 1% + 1% + 12% = 16%

2. The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n – 1) divisions of main scale. The least count of the vernier callipers is,

[NEET (Odisha) 2019]

(A) 1n(n 1)

cm (B) 1(n 1)(n 1)

cm

(C) 1n cm (D) 2

1n cm

Hint: 1 V.S.D. = (n 1)

n

M.S.D.

L.C. = 1 M.S.D. 1 V.S.D.

= 1 M.S.D. (n 1)

n

M.S.D.

= 1nM.S.D.

= 1n

1ncm

L.C. = 2

1n cm

3. A student measures time for 20 oscillations of

a simple pendulum as 30 s, 32 s, 35 s and 31 s. If the minimum division in the measuring clock is 1 s, then correct mean time in second is [MHT CET 2019]

(A) 32 + 3 (B) 32 + 1 (C) 32 + 2 (D) 32 + 5

Hint: Mean (t) = 30 32 35 31 32

4

Mean error (t) = 1 2 3 4t t t t4

= 2 0 3 1

4

= 64 = 1.5

Hence rounding off, t = 2 s t t = 32 2 s 4. A student measured the diameter of a small

steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of – 0.004 cm, the correct diameter of the ball is

[NEET (UG) 2018] (A) 0.521 cm (B) 0.525 cm (C) 0.053 cm (D) 0.529 cm Hint: Least count of screw gauge = 0.001 cm = 0.01 mm Main scale reading = 5 mm, Zero error = – 0.004 cm = – 0.04 mm Zero correction = +0.04 mm

Answers to Multiple Choice Questions

Competitive Corner

Hints to Multiple Choice Questions

SAMPLE C

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28

Observed Mainscale

(Division least count)reading reading

Observed reading = 5 + (25 0.01) = 5.25 mm

Corrected Observed zeroreading reading correction

Corrected reading = 5.25 + 0.04 = 5.29 mm = 0.529 cm 6. The density of the material in the shape of a

cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:

[JEE (Main) 2018] (A) 4.5% (B) 6% (C) 2.5% (D) 3.5%

Hint: Density () = 3mass m

Volume l

…(for cube V = l3) Percentage relative error in density will be,

100

=

m 100 3 100m

ll

= 1.5 + (3 1) = 1.5 + 3 = 4.5% 5. Let x =

2 2a bc

be the physical quantity. If the

percentage error in the measurement of physical quantities a, b and c is 2, 3 and 4 percent respectively then percentage error in the measurement of x is [MHT CET 2018]

(A) 7% (B) 14% (C) 21% (D) 28%

Hint: Given: x = 2 2a bc

Percentage error is given by,

x 2 a 2 b c

x a b c

= (2 2) + (2 3) + 4 = 4 + 6 + 4 = 14

x% 14%

x

7. A physical quantity of the dimensions of

length that can be formed out of c, G and 2

0

e4

is [c is velocity of light, G is universal constant of gravitation and e is charge]:

[NEET (UG) 2017]

(A)

1/22

20

1 eGc 4πε (B)

1/222

0

ec G4

(C) 1/22

20

1 ec G4

(D) 2

0

1 eGc 4

Hint: Let the physical quantity formed of the dimensions of length be given as,

[L] = [c]x [G]y z2

0

e4

….(i)

Now, Dimensions of velocity of light [c]x = [LT1]x Dimensions of universal gravitational constant

[G]y = [L3T2M1]y

Dimensions of z2

0

e4

= [ML3T2]z

Substituting these in equation (i) [L] = [LT1]x [M1L3T2]y [ML3T2]z = Lx + 3y + 3z My + z Tx 2y 2z Solving for x, y, z x + 3y + 3z = 1 y + z = 0 x + 2y + 2z = 0 Solving the above equation,

x = 2, y =12, z =

12

L = 2

1c

122

0

eG

8. The following observations were taken for

determining surface tension T of water by capillary method:

diameter of capillary, D = 1.25 10–2 m rise of water, h = 1.45 10–2 m. Using g = 9.80 m/s2 and the simplified relation

T = 3rhg 102 N/m, the possible error in surface

tension is closest to: [JEE (Main) 2017] (A) 0.15 % (B) 1.5 % (C) 2.4 % (D) 10 % Hint: D = 1.25 10–2 m; h = 1.45 10–2 m The maximum permissible error in D = D = 0.01 10–2 m The maximum permissible error in h = h = 0.01 10–2 m g is given as a constant and is errorless.

T = 3rhg 102 N/m = 3dhg 10

4 N/m

% error T d h

T d h

T d h100 100 100

T d h

= 2 2

2 2

0.01 10 0.01 10 1001.25 10 1.45 10

= 100 100125 145

T

T

= 0.8 % + 0.7 % = 1.5 %

SAMPLE C

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29


For example, The newly defined unit kilogram uses Planck constant, the metre and the second, while the kilogram itself is used in defining the kelvin and the candela. This definition relates the kilogram to the equivalent mass of the energy of a photon given its frequency, via the Planck constant. Given below, refers to the corresponding definitions before 20 May 2019 to offer quick comparision to know which definitions are modified and how.

0 A

m

c s

K

TTPW

cd

mol m(C)

Kcd

kg

MIPK

Cs

Old SI

Additional Information

Till May 20, 2019 the kilogram did not have a definition; it was mass of the prototype cylinder kept under controlled conditions of temperature and pressure at the SI museum at Paris. A rigorous and meticulous experimentation has shown that the mass of the standard prototype for the kilogram has changed in the course of time. This shows the acute necessity for standardisation of units. The new definitions aim to improve the SI without changing the size of any units, thus ensuring continuity with existing measurements. As per new SI units, each of the seven fundamental units (metre, kilogram, etc.) uses one of the following 7 constants which are proposed to be having exact values as given below: The Planck constant (h), The elementary charge (e), The Boltzmann constant (kB), The Avogadro constant (number) (NA), The speed of light in vacuum (c), The ground state hyperfine structure transition frequency of Caesium-133 atom (Cs). The luminous efficacy of monochromatic radiation of frequency 540 × 1012 Hz (KCd) The figures show the dependency of various units upon their respective constants and other units (wherever used). The arrows arriving at that unit refer to the constant and the fundamental unit (or units, wherever used) for defining that unit. The arrows going away from a unit indicate other units which use this unit for their definition.

e A

m

c s

K

kB

cd

mol NA

Kcd

kg

h

Cs

New SI

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