Secondary

PHYSICSStudent’sBookTwo

(FifthEdition)

KENYALITERATUREBUREAUP.O.Box30022-00100,NairobiWebsite:www.kenyaliteraturebureau.comE-mail:info@kenyaliteraturebureau.com

©MinistryofEducation

Allrightsreserved.Nopartofthisbookmaybereproduced,storedinaretrievalsystem,ortranscribed,inanyformorbyanymeans,electronic,mechanical,photocopying,recordingorotherwise,withoutthepriorwrittenpermissionofthepublisher.

ISBN978-9966-10-212-6

Firstpublished1988Reprinted1989(fourtimes),1990SecondEdition1994Reprinted1996,1997,1998,1999,2000,2001,2002ThirdEdition2003Reprinted2004(twice)2005(twice),2006,2007FourthEdition2007Reprinted2007(twice),2008,2009(twice)2010(twice)2011(fourtimes),2012,2013FifthEdition2013

KLB1097420m2013

PublishedandprintedbyKenyaLiteratureBureau

Contents

PrologueAcknowledgements

1.MagnetismPropertiesofmagnetismDirectionofmagneticfieldMagneticfieldpatternTheEarth’smagneticfieldExercise1.1TheDomaintheoryMagnetisationofamagneticmaterialDemagnetisationHardandSoftmagneticmaterialsApplicationofmagnetsRevisionExercise1

2.Measurement(II)Engineer’scallipersVerniercallipersExercise2.1Exercise2.2MicrometerscrewgaugeSignificantfigureStandardformPrefixesusedwithSIunitsDecimalplacesRevisionExercise2

3.TurningEffectofaForceMomentofaforceTheprincipleofmomentsMomentduetoweightofanobjectMomentsofparallelforcesEqualparallelforcesinoppositedirection

RevisionExercise3

4.EquilibriumandCentreofGravityCentreofgravityofregularshapesCentreofgravityofirregularshapesStateofequilibriumApplicationofstabilityRevisionExercise4

5.ReflectionofCurvedSurfacesTypesofcurvedsurfacesReflectionoflightbycurvedmirrorsLawofreflectionandcurvedmirrorsRaydiagramsConcavemirrorsConvexmirrorsLinearmagnificationMirrorformulaSignconventionDefectsofsphericalmirrorsRevisionExercise5

6.MagneticEffectofanElectricCurrentMagneticfieldpatternofastraightcurrentcarryingconductorUsingmagneticcompassesForceoncurrentcarryingconductorinamagneticfieldForcebetweenparallelconductorscarryingcurrentUsesofelectromagnetsRevisionExercise2

7.Hooke’sLawStretchingofmaterialsHooke’slawCompressingaspringRevisionExercise2

8.WavesCharacteristicsofwavemotionPhaseandphasedifferenceSpeedRevisionExercise8

9.SoundI

SomesourcesofsoundVoicebox(Larynx)FactorsaffectingvelocityofsoundReflectionofsoundRevisionExercise9

10.FluidFlowTypesofflowBernoulli’sprincipleRevisionExercise10

Prologue

Thisbook isprimarilymeant tocoverexhaustively theFormOnePhysicssyllabus as per the new 8-4-4 curriculum. It is by design also a versatilecompanion for those students taking related courses in technical colleges andotherinstitutions.

The book has beenmademore elaborate and in-depth theorical coverageboosted with numerous experiments to enhance a better understanding ofconceptsunderstudy.

Any student making full use of the title, and by extension the KLBSecondaryPhysicsseries,willcertainlyacquirescientificknowledgeandskillsusefulinansweringtothechallengesofdailylife.

I am grateful to the panel ofwriters and everybodywho took part in thepreparationandproductionofthisedition.

THEMANAGINGDIRECTORKenyaLiteratureBureau

Acknowledgements

The Managing Director, Kenya Literature Bureau, would like to thank thefollowingwriterswhoparticipatedintherevisionofthisbook:

OliverMinishi – KakamegaHighSchoolErastusMuni – MinistryofEducation,InspectorateHesborneOmolo – KakamegaHighSchoolGraceMwangasha – MurrayGirl’sHighSchool

Chapter1

Magnetism

Anaturallyoccurringmaterialknownas lodestone,a formof ironore,attractssome materials when they are brought near it. When a bar of lodestone issuspended freely, it settles in the north-south direction. This property oflodestonehasbeenusedtomakethepresent-daymagnets.

Magnetsattractcertainmaterialsknownasmagneticmaterials,examplesofwhich are iron, nickel, cobalt and their alloys.Magnets are made from thesemagneticmaterials.

PropertiesofMagnets

Magnetshaveanumberofpropertiessomeofwhichinclude:

1.MagneticPoles

Experiment1.1:TodeterminethepolesofamagnetApparatusBarmagnet,ironfilings.

Fig.1.1(a):Barmagnetinironfilings

Procedure•Dipthebarmagnetintoironfilings.Notewheretheironfilingsclingmost.•Repeattheexperimentusingahorseshoemagnet.

Fig.1.1(b):Ahorse-shoremagnetinironfilings

Observation

Theironfilingsclingmainlyaroundtheendsofthebarmagnet,seefigure1.1(a)and(b).Thisshowsthatmagneticattractionisstrongestattheendsofthebarmagnet.

Theendsofamagnetwherethepowerofattractionisgreatestarecalledthepolesofthemagnetormagneticpoles.

2.DirectionalProperty

Experiment1.2:Toinvestigatethedirectionalpropertyofamagnet

Apparatus

Barmagnet,cottonthread,andwoodenstand.

Procedure•Suspendabarmagnetwithacottonthreadfromawoodenstand,sothatthemagnet swings freely in the horizontal plane. Give it time and observe thedirectioninwhichitcomestorest.

•Fromtherestposition,turnthemagnetthroughabout90°andreleaseit.Noteagainthedirectioninwhichitcomestorest.

• Fromtherestposition,nowturn themagnet throughabout180°.Release itandnotethedirectioninwhichitcomestorest.

Observation

The suspendedmagnet alwayscomes to restwithoneendpointing roughly tothenorthdirectionandtheothertothesoutherndirectionoftheearth,seefigure1.2.

Fig.1.2:Suspendedmagnetatfinalrestingposition

Conclusion

Magnetshavetwopoles.ThepolewhichpointstowardsthenorthiscalledtheNorth-seekingor thenorth(N)pole;theotheroneis thesouth-seekingorsouth(S)pole.

Amagnetcanbe,therefore,usedasacompass.

Notethatthisexperimentcanbeusedtoidentifypolesofamagnet.

3.EffectonMagneticandNonmagneticMaterials

Experiment 1.3: To classify objects into magnetic and non-magneticmaterialsApparatusRodsofdifferentmaterials,woodenstand,cottonthread,barmagnet.

Fig.1.3:Effectofmagnetonamaterial

Procedure•Suspendarodwithathreadasshowninfigure1.3.•Bringthenorthpoleofamagnettowardsoneendoftherodandobservewhathappens.

•Repeattheexperimentusingthesouthpoleofthemagnet.•NowrepeattheexperimentusingdifferentmaterialsshowninTable1.1.Note

yourobservations.

Table1.1

Object Observations

NorthPole SouthPole

Steelneedle Attracted Attracted

Pieceofchalk

Steelpin

Aluminiumstrip

Carbonrod

Copperwire

Matchstick

Glass

Observation

Itisobservedthatsomeobjectsareattractedbyboththenorthandsouthpolesofthemagnet,whileothersarenotattracted.

Conclusion

An object which is attracted by amagnet is amagneticmaterial. An objectwhich isnot attractedbyamagnet is anonmagneticmaterial.Metals suchascobalt,ironandnickeltogetherwiththeiralloyswhicharestronglyattractedbyamagnetarecalledferromagneticmaterials.

Some examples of non-magnetic materials are copper, brass, aluminium,woodandglass.

4.AttractionandRepulsionExperiment1.4:ToestablishthelawofmagnetismApparatusTwobarmagnets,cottonthread,woodenstand,ironrod.Procedure•Suspendabarmagnetasshowninfigure1.4.

Fig.1.4:Attractionandrepulsionofmagneticpoles

• Bring the north pole of another magnet towards the north pole of thesuspendedmagnet.Observewhathappens.

•Bringthesamepoletowardsthesouthpoleofthesuspendedmagnet.•Repeattheexperimentusingthesouthpoleofthefreemagnet.•Nowrepeattheexperimentusingapieceofunmagnetisedironinsteadofthesuspendedbarmagnet.TabulateyourresultsasshowninTable1.2.

Table1.2

Northpoleofsuspendedmagnet

Southpoleofsuspendedmagnet

Northpoleofthefreemagnet

Repulsion Attraction

Southpoleofthefreemagnet

Oneendofironbar

Theotherendofironbar

Observation

Anorth pole attracts a south pole and repels a north pole,while a south polerepels a southpole.This canbe summarizedas likepolesrepel,unlikepolesattract.Thisisthebasiclawofmagnetism.

Thepolarityofamagnetcanbe testedbybringingboth itspoles, in turn,adjacent to the known poles of a suspended magnet. Repulsion only occursbetweenthelikepolesofamagnet.However,attractioncanoccurbetweentwounlike poles or between a pole and a piece of an unmagnetised magnetic

material.Thus,theonlysuretestforpolarityofamagnetisrepulsion.

5.BreakingaMagnet

When a barmagnet is broken into two ormore pieces, the pieces retain theirmagnetism,eachhavinganorthpoleandasouthpoleatitsends,seefigure1.5.

Fig.1.5

6.MagneticFieldPatterns

Whenamagneticmaterial is placednearonepoleof amagnet, it is attracted.Thissuggeststhatthereisamagneticeffectinthespacearoundamagnet.Theregionorspacewherethemagneticinfluenceisfeltiscalledthemagneticfield.The field is stronger near the poles of themagnet and isweaker farther awayfromthepoles.

Experiment 1.5 To Investigate magnetic field patterns of a magnet usingIronfilingsApparatusBarmagnet,horse-shoemagnet,smoothstiffcardboard,ironfilings.

Fig.1.6:Fieldpatterns

Procedure•Placethebarmagnetonatableandputastiffcardboardoverit.•Sprinkleathinlayerofironfilingsontheboardandtapthecardboardgently.

Notetheobservations.•Repeattheexperimentwiththehorse-shoemagnetundertheboard.

Observation

Theironfilingsformpatternswhichseemtooriginatefromonepoleandendtheother,seefigure1.6(a)and(b).

Conclusion

The magnetic field around a given magnet or arrangement of magnets is adefinitepattern.

ClassActivityUsing the procedure of Experiment 1.5, determine the field patterns for thecombinationbelow:

DirectionofMagneticField

Experiment 1.6: To investigate the direction of magnetic field using amagneticneedleApparatusMagnetbar,magneticneedlefixedonacork,glasstrough.

Fig.1.7:Directionofmagneticfield

Procedure•Pourwaterintothetroughuptothree-quartersitsvolume.•Floatamagnetisedneedlewithitsnorthpoleuppermost,asshowninfigure1.7.

•Placeastrongmagnetsothatitrestsontheedgeofthetrough.•Adjustthelevelofthewatersothatthenorthpoleoftheneedleisatthesamelevelwiththeplaneofthemagnet.

•Holdtheneedlenearthenorthpoleofthemagnetandthenreleaseit.Observeitsmovement.

•Repeattheexperimentwiththesouthpoleoftheneedleuppermostandplacednearthesouthpoleofthemagnet.Noteitsmovement.

ObservationWhentheneedleisheldwithitsnorthpolenearthenorthpoleofthemagnetandreleased,theneedleisrepelledandmovestowardsthesouthpolealongacurve,asshowninfigure1.7.

Ifthesouthpoleoftheneedleisplacednearsouthpoleofthemagnetandthe needle released, the direction of motion is reversed. The direction of themotionoftheneedleisdeterminedbythepoleoftheneedlewhichisadjacenttothemagnet.

Therefore,indecidingthedirectionofthefield,oneofthesedirectionshastobechosenasastandarddirection.Byconvetion,thedirectionofmotionofthenorthpoleischosen.

Thedirectionof themagnetic fieldatapoint is thedirectionwhichafreenorthpolewouldmoveifplacedatthatpointinthefield.

Thus, amagnetic line of force can be defined as the path alongwhich anorthpolewouldmoveiffreetodoso.

Experiment1.7:ToinvestigatemagneticfieldpatternsbyuseofaplottingcompassApparatusBarmagnet,plottingcompass,whitesheetofpaper,drawingboard.

Fig.1.8:Plottingamagneticfield.

Procedure•Placethebarmagnetonthesheetofpaperfixedonadrawingboard.•Markitsoutlineandindicatethepolarityofthemagnetonthepaper.•Placeaplottingcompassnearthenorthpoleofthemagnet.•Startingfromonepoleofthemagnet,markinthepaperthenorthandsouthendsofthecompassneedlewithpencildotsatPandQasshownonfigure1.8(a).

• Move the compassuntil the southpoleof theneedle is atPmark thenewpositionofthenorthpoleatR.

•Repeattheproceduresothataseriesofdotsisobtained.Jointhesedotsusingsmoothcurves,asinfigure1.8(b).

• Repeat the aboveprocedure starting fromdifferentpointsnear themagnet.Joineachsetofdotsusingasmoothcurveasshowninthefigure.

ObservationandConclusion

Smooth parallel curves that start from one pole and end at the other pole areobtained,seefigure1.8(b).Thesecurvesrepresentthemagneticfieldlines.Magneticfieldlineshavethefollowingproperties:(i)Theyoriginatefromnorthpoleandendatthesouthpole.(ii)Theyrepeleachotherandformclosedpathsneverintersectingotherlines

offorces.(iii)Theyareclosertogetherwherethefieldisstrongest.

Infigure1.9(b),theleftendXhasahigherfieldstrengththanrightendY.

Fig.1.9:Magneticfieldstrength

MagneticFieldPatterns

MagneticPatternaroundaBarMagnetThefieldlinesareshowninfigure1.10.Themagneticfieldforamagnetlinesofforceoriginatefromthenorthpoleandendatthesouthpole.

ThemagneticfieldlinesaroundabarmagnetoriginatefromNorthPoleandendattheSouthPole.

Fig.1.10:Fieldpatternaroundabarmagnet

FieldPatternsbetweenUnlikePolesEachmagnethas itsownmagnetic field.The two fieldscombine toproduceasinglemagneticfield,asshowninfigure1.11.

Fig1.11.Fieldpatternsbetweenunlikepoles

For a horse-shoemagnet, the north pole to the south pole, the field linesbetweenthepolesaremoreconcentrated.

For a horse- shoe magnet, the tied lines between the poles are highlyconcentratedattheregionbetweenthepoles.

Fig.1.12:Fieldpatternofahorse-shoemagnet

FieldPatternsbetweenLikePolesWhenlikepolesareplacedadjacenttoeachothermagneticfieldfromthepolesrepeleachotherresultinginaneutralpointXasshowninfigure1.13.

Fig.1.13:Fieldpatternbetweentwolikepoles

ThereisnomagneticfieldatpointX,whichiscalledaneutralpoint.Aneutralpointcan thusbedefinedas thatpointwhere theeffectof twomagnetic fieldstotallycanceleachother.

EffectofSoftIronRodandRing

Linesofforcethemagnetgetconcentratedalongthesoftironrod,asshowninfigure 1.14.The lines emerge on the far end of the rod, thus preventing themfromreachingcertainregions.

Fig.1.14:Effectsofsoftironrod

Asoftironringconcentratesthemagneticlinesofforce,asshowninfigure1.15.This prevents lines from entering region P. Region P is thus said to bemagneticallyshieldedbytheringfrommagneticfields.

Fig.1.15:Magneticshielding

Therefore, both the iron rod and ring can be used in magnetic shielding orscreening.Someelectricalmeasuringinstrumentsandwatchesareshieldedthiswayfromstraymagneticfieldsusingalloysofhighmagneticpermeability(thepropertyofcertainmaterialstoconcentrateamagneticfield,henceleavingsomeareastotallyfreefrommagneticinfluence).

TheEarth’sMagneticField

Thefactthatasuspendedbarmagnetalignsitselfsuchthatitsnorthpolepointsroughlytothegeographicalnorthoftheearth,suggeststhattheinnercoreoftheearthbehavesasifitwereamagnet.Usingthebasiclawofmagnetism,itcanbevisualisedthat themagnetic isembeddedin theearth’smagnet isembeddedintheearth’scrustwithitssouthpolenearthegeographicalnorthanditssouthpolenearthegeographicalsouthpolesseefigure1.16.

Theoriginof theearth’smagnetism isbelieved tobe inpartassociated tothespinningliquidmetalliccoreofironandnickel.

Fig.1.16:Earth’smagneticfield.

Experiment1.8:Toplottheearth’smagneticfield

Apparatus

Drawingboard,plainwhitepaper,fourpins,plottingcompass.

Fig.1.17:Plottingtheearth’smagneticfield

Procedure•Fixaplainpaperonadrawingboardplacedonatable.•Markadotatoneendofthepaper.•Placeaplottingcompasssuchthatthesouthpoleoftheneedleisatthefirstdot.Markthepositionofthenorthpolewithaseconddot.

• Movethecompasssothat thesouthpoleoftheneedleisat theseconddot.Markthepositionofthenorthpolewithathirddot,seefigure1.17.

•Repeattheprocedureuntilaseriesofdotsisobtained.•Jointhedots.•Repeattheproceduretoobtainmorelines.•Indicatewithanarrowthedirectionthenorthpoleoftheneedlepointswhenplacedontheline.

Observation

Aseriesofparallellinesareobtained.Theselinesrepresentthedirectionoftheearth’sfield,frommagneticsouthtomagneticnorth,seefigure1.18.

Fig.1.18:Earth’smagneticfield.

Theearth’sfieldisanexampleofuniformfield.

A bar Magnet placed anywhere near the earth’s surface will have itsmagneticfieldcombiningwiththefieldoftheearth,seefigure1.19(a).

AtpointsmarkedX,themagneticfieldoftheearthandthatduetothemagnetare equal and opposite. The resultantmagnetic field is therefore zero at thesepoints,whicharecalledneutralpoints,seefigure1.19(a)and(b).

Fig1.19:Barmagnetintheearth’smagneticfield

AtP,thefieldofthemagnetisstrongerthanthatoftheearth,whileitisweakeratQ.

Exercise1.1

1.Sketchthemagneticfieldpatternsforthearrangementsbelow:

2.ThediagrambelowshowsthemagneticfieldpatternbetweentwomagnetsPandQ:

(a)IdentifypolesAandB.(b)StatewhichofthetwomagnetsPandQisstronger.Explain.

TheDomainTheory

Magnetic materials contains small magnetic regions referred to as magneticdomains.Adomaincomprisessmallatomicmagnetscalleddipoles.

Adipolehasanorthandsouthpole.Thedipolesinaparticulardomainalignthemselvesinacommondirectiongivingthedomainanoverallnorthpoleandasouthpole.Alinecanthusbedrawn(magneticaxis)joiningthepoles,seefigure1.20.

Thetotalfieldoutsideamagneticmaterialisthesumofindividualdomainfields.

Fig.1.20:Dipolesinadomain

In the un-magnetised condition, the dipoles in neighbouring domains arrangethemselvesinclosedloops,seefigure1.19

Thedomainshowever align indifferent directiongivingno resultant fieldoutsidethematerial.

Whenamagnetisingfield isapplied to thematerial, there isadjustmentofdomainboundariescausingsomeofthedomainstoincreaseinsizeandotherstoshrink. In the process, the domain, align themselves with their magnetic axispointinginthesamedirection.Thenetexternalfieldincreasesandthematerialissaidtobemagnetised.

Fig.1.12

When all dipoles align themselves in one direction, thematerial is said to bemagnetically saturated. No further magnetisation can take place beyond thispoint

When the magnetising field is removed, domains of hard magnetic(materials which are hard tomagnetise)maintain their new orientation, whilethose of softmagneticmaterials (materialswhich are easy tomagnetise) havetheirdomainsrevertingtotheiroriginalarrangement.

MagnetisationofaMagneticMaterial

The process of making a magnet from a magnetic material is calledmagnetisation.Thecommonmethodsusedinmagnetisationare:(i)Induction.(ii)Stroking.(iii)Hammering.(iv)Electricalmethod.

InductionMethod

Experiment1.9.TomakeamagnetbyinductionmethodApparatusTwobarmagnets,officepins

Fig.1.22:Inducedmagnetism.

Procedure•Holdabarmagnetvertically.•Movethelowerpoletowardsoneendofapin.•Movethelowerendofthatpintotheendsoftheotherpins,onebyone.Notewhathappens

•Detachthetoppinfromthemagnet.Whatdoyouobserve?

Observation

Thefirstpingetsattractedtothemagnet.Theotherpinsgetattractedtothefirstpinandformachain,seefigure1.22(a).

Whenthefirstpinisdetachedfromthemagnet,thelowerpinsfalloffafterashorttime.

ExplanationTheendsofthepinsattractedtothemagnetacquireapolaritythatisoppositetothatpoleofthemagnet.Thelowerendofthepinacquiresapolaritysimilartothepoleused,seefigure1.22(c).

Thepinsbecomemagnetisedandthedipolesinthemgetalignedalongthemagneticaxisofthemagnetisingmagnet.

Thepinsintheexperimentgetmagnetisedbyinductionmethod.

StrokingMethod

Whenasteelneedleisstrokedusingoneendofastrongmagnetrepeatedly,theneedlebecomesmagnetised.Experiment1.10:Tomakeamagnetbystrokingmethod

SingleStrokeMethodApparatusAbarmagnet,cottonthread,steelbar.

Fig.1.23.Singlestrokemethod

Procedure•PlacethesteelbarXonatable.• Stroke thebarwithonepoleof themagnet fromA toB, lifting it awayasshowninfigure1.23

•Repeattheprocedureseveraltimes,keepingtheinclinationoftheroughlythesame.

•Testforthepolarityofthebarbysuspendingitfreely.Notewhathappens.

Observation

Thesuspendedbaraligns itselfwith themagnetic axisof theearth.Oneof itsendsisrepelledbytheplottingcompass.

Conclusion

Thebarismagnetisedbystrokingmethod.TheendofthebarwherethemagnetYfinishesstrokingacquiresanoppositepolaritytothatofthestrokingendofthemagnet.Sincethestrokingpoleofthemagnetinthefigureisnorth,theandBofthebarbecomesthesouthpole.

Thedisadvantageof thismethod is that itproducesmagnets inwhichonepoleisnearertheendofthemagnetisedmaterialthantheother.

Thiscanbeavoidedbyuseofthedoublestrokemethod.

DoubleStrokeMethodApparatusSteelbar,twobarmagnets,cottonthread,plottingcompass.Procedure•StrokethesteelbarZfromthecentrerepeatedlyinoppositedirections,using

oppositepolaritiesoftwobarmagnetsPandQ,seefigure1.24.

Fig.1.24Magnetisationbydoublestrokemethod

•Testforthepolarityofthesteelbar.ObservationTheendAofthebaracquiresanorthpolewhileendBacquiresasouthpole.ConclusionThesteelbarhasbeenmagnetisedbydoublestroking.Ifasteelbarismagnetisedbythedoublestrokemethodusingnorthpolesoftwomagnets,thebaracquiresasouthpoleateachendandadoublenorthpoleatthecentre,seefigure1.25

Fig.1.25:Makingamagnetwithconsequentpolesatthecentre

HammeringMethod

Thismethodmakesuseoftheinfluenceofearth’smagneticfield.Asteelbartobemagnetised isplaced inclined to thehorizontal inanorth–southdirectionatanangleθandtheupperendhammeredseveraltimes.

Fig.1.26Hammeringintheearth’sfield

Note:θisreferredtoastheangleofdip.Itvariesaccordingtopositionontheearth.

Theprocessresultsintotheformationofaweakmagnet.

ElectricalMethod

Thisisthebestandquickestmethodofmakingamagnet.Itutilisesthemagneticeffectofanelectriccurrent.Themethodisusedwidelyinindustrialproductionofmagnets.

Acoilwithmanyturnsofinsulatedcopperwire,calledasolenoid,isused.Adirectcurrent(d.c.)ispassedthroughthesolenoid.

Experiment1.11:Tomagnetiseasteelbarbyelectricalmethod

Apparatus

Steelbar,asolenoid,battery,ammeter,switch,connectingwires,cotton threadandplottingcompass.

Fig.1.27:Magnetisingbyelectricalmethod

Procedure•PlaceasteelbarABinsidethesolenoidconnectedinserieswiththebattery.

•Switchonthecurrentforsometime,thenswitchoff.•Testforthepolarityofthesteelbar.

Observation

Thebar ismagnetised.Thepolarityof themagnetdependson thedirectionoftheelectriccurrent.

ThepolesAandBcanbeidentifiedusingtheclockrule,whichstatesthatif,onviewingoneendofthebar,thecurrentflowsinclockwisedirection,then that end is a southpole. If anticlockwise, then it is a northpole, seefigure1.28.

Fig.1.28:Theclockrule

The polesA andB can also be identified using the right-hand grip rule for acurrent-carryingcoil,seefigure1.29.

Fig.1.29:Right-handgriprule

The right-handgrip rule states that ifacoilcarryingacurrent isgrasped intherighthandsuchthatthefingerspointinthedirectionofthecurrentinthecoil,thenthethumbpointsinthedirectionofnorthpole.

Allowingthecurrenttoflowforalongtimedoesnotincreasetheextendofmagneticsaturation.Itonlycausesoverheatingofthesolenoid,whichadverselyaffectsmagnetism.

Demagnetisation

Themagneticpropertiesofamagnetcanbedestroyedbyhammeringintheeast-westdirection,heatingorusingelectricity.

HammeringandHeating

Amagnetisdemagnetisedbyheatingorhammeringitwhenplacedineast-westdirection.The magnetism is lost because hammering or heating disorients magneticdipoles.

ElectricalMethod

ThisisthemosteffectivemethodofdemagnetisationAna.cvoltageisconnectedinserieswithasolenoidwhichisplacedwithitsaxispointingeast-west,asshowninfigure1.30.Thebarmagnettobedemagnetisedisplacedinsidethesolenoidandthealternatingcurrentswitchedon.Afterafewminutes,itiswithdrawnslowlyfromthesolenoid.

Fig.1.30:Demagnetisingusingelectriccurrent

Whensuspendedfreely,itdoesnotsettleinanyparticulardirection.

Thisshowsthatithaslostmagnetism.

Themagnetlosesitspowerbecausealternatingcurrentreversesmanytimespersecond,disorientingthemagneticdipoles.Note:Intheabovemethods, themagnetsareplacedintheeast-westdirectionsothattheydon’tretainsomemagnetismduetotheearth’smagneticfield.

HardandSoftMagneticMaterials

Whenabarofsoftironisplacedinsideasolenoidandacurrentswitchedon,itbecomesastrongmagnet.Whenthecurrentisswitchedoff,itlosesmagnetism.

Magnetic materials such as soft iron which are magnetised easily but do notretaintheirmagnetismaresaidtobesoftmagneticmaterials.

Analloyofnickelandironisanotherexampleofasoftmagneticmaterial.Itcanbemagnetisedbyveryweakfields,suchastheearth’smagneticfield.Thesemagneticmaterialsareusedinmakingelectromagnets,transformercoresandformagneticshielding.

On the other hand, steel is difficult tomagnetise, but oncemagnetised, itretainsthemagnetismforalongtime.Magneticmaterialssuchassteelaresaidto behardmagneticmaterials. Hardmagneticmaterials are used inmakingpermanentmagnets.

StoringMagnets

Abarmagnet tends to becomeweakerwith time due to self-demagnetisation.Thisiscausedbythepolesattheendwhichtendtoupsetthealignmentofthedomains inside it. To prevent this,magnets are stored in pairswith small softironbars,calledkeepers,placedacrosstheirends.Unlikepolesofthemagnetsareplacedadjacenttoonanotherasinfigure1.31.

Fig.1.31:Magnetickeepers

Thekeepersacquirepolaritiesasshownin thefigureso that thedipoles in themagnetandthekeepersformcompleteloops.Thedipolesthusretaintheirorientationandmagnetismismaintained.

ApplicationofMagnets

Magnetshaveawiderangeofusessomeofwhichareelaboratedbelow:(i)Loudspeaker/Microphones

Loudspeakers convert electrical signals into sound.Magnet in the speakerprovidesamagneticfieldwhichinteractswiththefieldofvaryingelectricalcurrent fed into the speaker and results in a vibration of the coil of thespeaker.

Fig1.32:Speakers

In microphones sound waves are converted into electrical signals with amagnetplayingasignificantrole.

Fig.1.33:Microphone

(ii)ComputerHardDisk/RecordingandreadingHead/VideoTapes.

The head of hard disk of a computer records data on magnetic tapes orstripesinthedisk.Thisisdonethroughmagnetisationofmagneticmaterialsin thosestripescorrespondingto thedata input.Thereadingheadtogetherwithencodingsystemin thecomputeroncommandby theuser reproducethisinformationformanipulation.

Fig.1.34:Computerharddisk

(iii)InMedicine

Hospitals are employing theuseofmagnets from the simple extractionofmetallic objects from patients, body (e.g., eye) to Magnetic ResonanceImaging (MRI).Figure1.35showsanexampleofM.R. Imachinewhichuses powerful magnet whose fields enable doctors to take pictures ofinternalpartsofhumanbody.ThisproducesphotographswithmoredetailsthanwhenX-raysareused.

Fig1.35:AnMRImachine

Magnets are also used in what is referred to as magnetic therapy wherespecialmagnetsareusedtoreliefpainandrelaxthebody.

(iv)Sorting/Separating

Magnetsareusedtosortorseparatemagneticmaterialsasinquarrieswheremetal is separated from ore. Very strong magnets are used to recovermetallicobjectsfromoceanbed.

(v)MagneticScrewDrivers

In lifting small screws or putting them in place during repair work,techniciansmakeuseofmagneticscrewdrivers.Seefigure1.36below.

Fig.1.36:Magneticscrewdriver

RevisionExercise1

1.Describetwomethodsofmagnetisingasteelrod.2.Comparemagneticpropertiesofsteelandiron.3.Explainwhyamagneticmaterialisattractedbyamagnet.4.Explainwhyironfilingsarenotsuitableforplottinglinesofforceofaweak

magneticfield.5.Explainwhysoftironcannotbeusedtomakepermanentmagnets.6.Describehowyouwouldverifythebasiclawofmagnetism.7.Explainthemeaningofthefollowing:

(a)Magneticfield.(b)Magneticlinesofforce.

8.Describehowyouwouldshieldamagneticmaterialfromamagneticfield.Stateoneapplicationofmagneticshielding.

9.Describethreemethodsofdemagnetisingapermanentmagnet.10. Use domain theory to explain the difference betweenmagnetic and non-

magneticmaterials.11.AcoilofinsulatedwireiswoundaroundaU-shapedsoftironcoreXYand

connectedtoabatteryasshowninthefigurebelow.

StatethepolaritiesofendsXandY.12.Explain,withthehelpofadiagram,whataneutralpointofamagneticfield

is.13.(a)Withtheaidofdiagrams,explainhowyouwouldmagnetiseasteelbar

soastoobtainasouthpoleatamarkedendofthebarby:(i)Usingapermanentmagnet.

(ii)Usinganelectriccurrent.(b)Statewhichoftheabovemethodsproducesthestrongermagnet.

Giveareason.14.Youaregiventhreesteelbars.Oneismagnetisedwithoppositepolesatits

ends. Another is magnetised with consequent poles. The third is notmagnetised.Describe an experimentwhichyouwouldperform to identifyeach.

15.Thegraphsbelowarefortwomagneticmaterials:

State:(a)whichmaterialiseasiertomagnetise.(b)whichmaterialformsastrongermagnet.(c)oneapplicationofeach.

16.Brieflydescribehowyoucanconstructasimpleplottingcompass.17.Indiagrams(a)and(b)below,twopinsareattachedtomagnetsasshown:

Explainthebehaviourofthepinsineachcase.18.Describefourpropertiesofmagnets.19.Statefiveusesofmagnets.

Chapter2

MeasurementII

InMeasurementI,thebasicinstrumentformeasuringlengthwasthemeterrule.Inthischapter,measurementsoflengthssuchasdiameterofawireandthatofatest- tube,which cannot beobtaineddirectlyusing anordinary ruler, aredoneusingcalipersandmicrometerscrewgauge.Thetwotypesofcallipersusedareengineer’scalipersandverniercalipers.

Engineer’sCallipers

Figure2.1(a)showsapairofengineer’scallipersusedformeasuringlengthsonsolidobjectswhereanordinarymetrerulecannotbeuseddirectly.Itconsistsofpair of hinged steel jaws which are closed until they touch the object at thedesiredposition.

Diameters of round objects can be measured using outside and insidecalipersshowninfigure2.1(b)and(c)respectively.

Onekindischangedtotheotherbyturningthejawsround.

Fig.2.1:Engineer’scallipers

Whenusingthecallipers,thejawsareopenedjusttoslippastthecylinderorthewidestpartofthesphere.Thedistancebetweenthejawsisthentransferredandreadonanordinarymillimetrescale,asshowninfigure2.2.

Fig.2.2:Measuringlengthbetweenthejawsofcallipers

VernierCallipers

Averniercallipersconsistsofasteelframewithafixedjawandaslidingjaw,asshowninfigure2.3.Thesteelframecarriesthemainscalewhichisgraduatedincentimetresbutalsohasmilimetredivisions.

Fig.2.3:Verniercalipers

Theslidingjawcarriesthevernierscale,whichhastenequaldivisions.Thelengthofthevernierscaleis0.9cm.,eachdivisionofthevernierscale

is 0.09 cm. The difference in length between themain scale division and thevernierscaledivisionisknownastheleastcount.

Leastcount = (0.1–0.09)cm

Leastcount = (0.1–0.09)cm = 0.01cm

Mostverniercallipershaveboth insideandoutside jaws.Theoutside jawsareusedtomeasureexternaldiameterswhiletheinsidejawsareformeasuringtheinternaldiameters.

Fig.2.4(a):

Vernier callipers have small steel metal protruding at the back for measuringdepths(depthprobe)seefigure2.4(a)above.Digitalverniercallipersareavailableinthemarket,seefigure2.2(b)below.

Fig.2.4(b):Digitalverniercallipers

UsingVernierCallipers•Placetheobjectwhosediameteristobemeasuredbetweentheoutsidejaws.Closethejawstilltheyjustgriptheobject,seefigure2.5.

Fig.2.5Usingverniercallipers

•Recordthereadingonthemainscale,oppositeandtotheleftofthezeromarkofthevernierscale.Fromfigure2.5,itis2.3cm.

• Read thevernierscalemark thatcoincidesexactlywithamainscalemark.Thefourthmarkofthevernierscalecoincidesexactlywithamainscalemark.Thisgivesareadingoftheleastcount

=4×0.01=0.04cmThesumofthevernierscalereadingandthemainscalereadinggivesthediameteroftheballbearing.Therefore,thediameterofballbearing=(2.3+0.04)cm=2.34cm

When the readingsare taken incentimetres, the reading from thevernier scalegivestheseconddecimalplace.Example1

Whatisthediameterofthecirculardiscinthefigurebelow?

Fig.2.6

Solution

Mainscalereading=2.1cm

Thevernierscalemarkthatcoincideswithmainscalemarkisthefourthone.

So,

vernierscalereading=0.02cm

Diameterofthedisc=(2.1+0.02)cm

=2.12cm

Exercise2.1

1.Describehowyouwouldmeasuretheinternaldiameterof100cm3beakerusingverniercallipers.

2.Writedowntheverniercallipersreadingsindiagram(a),(b)and(c)below(diagramsnotdrawntoscale)

3. List down some objects both small and big, whose diameters cannot bemeasuredusingverniercallipers.

ClassActivityModelofaVernierCallipers•Usingmanilapaper,makeyourownverniercallipers.Themainscaleshouldrunfrom0cmto8cm.

•Measurediametersofdifferentobjectsusingyourverniercallipers.•Repeattheprocedureusingverniercallipersfromthelaboratoryandcompareyourresults.

ZeroErrorWhen the jaws of the vernier callipers are closed without an object betweenthem,thezeromarkofthemainscaleshouldcoincidewiththezeromarkofthevernierscale,asshowninfigure2.7.

Fig.2.7:Verniercalliperswithjawsclosed

However, sometimes thismaynotbe thecase,as illustrated in figures2.8and2.9.

Fig.2.8:Verniercalliperswithnegativezeroerror

Fig.2.9:Verniercaliperswithpositivezeroerror

The vernier calipers are then said to have a zero error. The zero error infigure2.8isnotthesameastheoneinfigure2.9.

Infigure2.8,thezeromarkofthemainscaleistotherightofthezeromarkofthevernierscale.Inthiscase,thezeroerrorisnegative,hencethemeasureddiameter is less than the actual diameter.This is corrected by adding the zeroerrortothereading.

In figure 2.9, the Zero error is positive. Therefore, the zero error issubtractedfromthemeasureddiametertogetthecorrectvalueofthediameter.

Measurements taken with such callipers are normally corrected by eitheraddingorsubtractingthezeroerror.

Exercise2.2

1.Whatarethezeroerrorsoftheverniercallipersinfigures2.8and2.9.2.Ifthecorrectdiameterofanobjectis3.65cm,whatwouldbethereadings

ofbothcallipersforthisdiameter.3.Thecallipersinfigure2.8wasusedtomeasurethediameterofacylindrical

objectofheight10cmandrecorded3.05cmwhiletheoneinfigure2.9wasused tomeasure the diameter of a spherical object and recorded 3.25 cm.Calculatethecorrectvolumesoftheseobjectsinm3.(Takeπ=3.142)

MicrometerScrewGauge

Themicrometer screw gauge is used to measure small diameters such as thediameterofathinwire.

Amicrometer screwgaugeconsistsof aU-framecarryingananvil atoneendathimblewhichcarriesacircularrotatingscaleknownasthethimblescaleand a spindle which can move forward and backwards when the thimble isrotated.

The sleevehasa linear scale inmillimetresand the thimblehasacircularscaleof50equaldivisions.

Fig.2.10:Micrometerscrewgauge

The ratchet at the end of the thimble prevents the user from exerting unduepressure on an objectwhen themicrometer screw gauge is in use. The linearscalehashalf-millimetremarks.

Thisdistancemovedbythespindleinonecompleterotationofthethimbleisknownasthepitchofthemicrometer.

In figure 2.10, since the spindle advances or retreats by 0.5 mm percompleterotationofthethimble,thepitchofmicrometeris0.5mm.

Thus,eachdivisionrepresentsaspindletravelof mm=0.01mm.

Hence, if the thimble rotates throught 1 division, the spindle advances by0.01mmor0.001cm.Somemicrometerscrewgaugeshaveapitchof1.0mmand100divisionsonthethimble.

Digitalmicrometerscrewgaugeareavailableinthemarket,seefigure2.11.

Fig.2.11:Digitalmicrometerscrewgauge

UsingaMicrometerScrewGauge

Theobjectwhosediameteristobefoundisheldbetweentheanvilandspindle(jaws).Themicrometerisclosedusingtheratchetuntiltheobjectisheldgentlybetween the anvil and the spindle. The ratchet will slip when the object isgrippedfirmlyenoughtogiveanaccuratereading.

Infigure2.12thesleevereadingis10.5,mmandthethimblescalereadingis: mm=0.24mm.

Thus,thediameterofthemarble=sleevereading+thimblereading=10.5mm+0.24mm=10.74mm

Fig.2.12:Usingamicrometerscrewgaugetomeasurethediameterofamarble

ZeroError

As in the case of vernier callipers, there occurs zero error in the micrometerscrewgauge.Itariseswhenthezeromarkofthethimblescaledoesnotcoincideexactlywith thecentre lineof thesleevescalewhen themicrometer isclosed.Theanvilisusuallyusedfortheadjustmentofthezero,sothatthemicrometerhasnozeroerror.

Themicrometer in figure2.13 (a)hasnozeroerrorwhile those in figures2.13(b)and(c)havezeroerrors.Notethepositionofthezeromarkineachcase.

Fig.2.13:Zeroerrorinmicrometerscrewgauge

Example2Whatisthereadingofthemicrometerscrewgaugeinfigure2.14?

Fig2.14

SolutionSleevescalereading=19mmThimblescalereading=0.48mmHence,micrometerreading=(19+0.48)mm=19.48mmThere are other measuring instruments which may also be used to measure

smallerdiameters,forexample,theboreofacapillarytube.Onesuchinstrumentisthetravellingmicroscope.Thisisamicroscopemountedonaheavymetallicbasewithascalealongwhichthemicroscopemovesup,downandsideways.

Exercise2.3

1.Whatisthereadingofamicrometerscrewgaugeinthefiguresbelow:

2. Compareandcontrast thethimblescalesof twomicrometerscrewgaugeswithpitchesof0.5mmand1.0mmrespectively.

3.Statethetwolimitationsofthemicrometerscrewgauge?4.Listdowntheadvantagesanddisadvantagesofthemicrometerscrewgauge

overtheverniercallipers.5.Sketchamicrometerscrewgaugescalereading:

(a)0.23mm(b)5.05mm

SignificantFigures

In measurements, some computations give answers with many digits, forexample,9403.435662.Itisadvisabletoexpresssuchanumberinsimplerform.

Thedigits1and9aresignificantwhentheyappear inanumber.Thefirstdigitfromtheleftof thenumberis thefirstsignificantfigure.Inexample,9isthe first significant figure.The number of significant figures is determined bycountingthenumberofthedigitsfromthefirstsignificantfigureontheleft.

Zero is sometimes significant and at times it is merely used as a placeholder.Whenazerooccursattheleftendofanumber,itisnotsignificant.Forexample, thezeros in0.2cm,0.002mare justplaceholders.Thus, innumber0.005734,thefirstsignificantfigureis5.

If zero occurs between non-zero digits, e.g, as in 7005, it is consideredsignificant.Also,ifzerooccursattherighthandendafterthedecimalpoint,itisalwayssignificant,e.g,1.0cm,1.00cm,1.000cm.

If a zerooccursat the righthandendof an integer, itmayormaynotbesignificant. For example, the number 750000 could be correct to 2,3,4,5 or 6significantfigures.Whenthenumberisexpressedto2significantfigures,noneof thezeros issignificant.Conversely, to6significant figuresall thezerosare

significant.On the other hand, the number 995 can bewritten as 1000 to onesignificantfigureor1000to2significantfigures.Noneofthezerosinthefirstcase is significantwhile in thesecondcase, the firstzero is significantbut thelasttwoarenot.

Example3

Find theareaofa rectangle thatmeasures4.26cmby2.77cmandwriteyouranswercorrectto:(a)Foursignificantfigures.(b)Twosignificantfigures.Solution

Area = 4.26×2.77cm2

= 11.8002cm2

= 11.80cm2tofours.f.

= 12cm2totwos.f.

StandardForm

The range of lengths in the universe is extremely large, from the diameter ofyourhair to theestimated sizeof theuniverse.Writingdown thesevery smalland very large lengths in metres is clumsy. This problem can be solved bywritingsuchlengthsinstandardform.

Apositivenumber issaid tobe instandardformwhenwrittenasA×10nwhereAissuchthat1≤A<10andindexnisaninteger.

For example, the number 9327 when written in standard form would be9.327×103.

Note that in this case, A is equivalent to 9.327 and the index n = 3 ispositivebecausethedecimalpointafterthelastdigithasbeenmovedtotheleft.

However if the number lies between 0 and 1, then the index n becomesnegative.Forexample,0.009327instandardformis9.327×10–3.

Example4Expresseachofthefollowingnumbersinstandardform:(a)201(b)2670(c)0.087

(d)0.0000009047Solution(a)201=2.01×102

(b)2670=2.670×103

(c)0.087=8.7×10-2

(d)0.0000009047=9.047×10-7

Theuse of standard form in expressing length,mass and time is a convenientmeansofwritinglargeandsmallquantities.Forexample,thedistanceofthesunfromtheearthisestimatedas1.5×108kmandthemassofanelectronisgivenas9.1×10-31kg.

TheOilDropExperiment

This is an experiment to estimate the size of a molecule in an ordinarylaboratory.

Theory

Whenanoildropiscarefullyputincontactwiththesurfaceofwater,itspreadsout to formavery thin layer,which is almost circular.This isbecause theoilbreaksthesurfacetensionofthewater,whoseparticlespullawayfromtheoil.Thethinlayercanbeapproximatedtobeonemoleculethick.

EstimatingtheVolumeofanOilDropMethod1•Filla1cm3graduatedburettewithoil.•Countthenumberofdropsastheyaresqueezedoutofthepipette.Letthenumberbey.

•Ifthenumberofdropsisy,thentheaveragevolumeoccupiedbytheonedropis cm3.

Method2•Holdadropoftheoilinfrontofafinemillimetrescaleusingafinewireandviewitthroughamagnifyingglass.Seefigure2.15(a)

•Readoffthediameteroftheoildrop.•Assumingthatitisspherical,itsvolumeisgivenby

ToobtainaMonoLayerofOilPatch•Fillaleveltraywithwateruntilitoverflows.•Cleanthesurfaceofwaterbymovingawax-coatedrodacrossit.• Placetwowaxedbeamsatthecentreofthetrayandslidethemtowardstheedgesofthetray.Theywillbeusedtoestimatethediameterofthespread.

•Sprinklelycopodiumpowderlightlyonthewatersurface.• Touchthesurfaceofthewaterwiththedropuntilitspreadsonthesurface.Theareacoveredbytheoilisseenasaclearcircularpatchofoilleftasthewaterparticlespullawayfromit,seefigure2.15(b).

•Adjusttherodsanduseametreruletoestimatethediameterofthepatch,sayd1.Repeatthesamestepsthreeorfourtimesandgettheaveragediameter.

Fig.2.15Estimatingthevolumeofanoildrop

DataAnalysisAveragediameterofthepatch,

Areaofthepatch=πr2

Volumeoftheoildrop=areaofthepatch(A)×thicknessofthepatch(h)Thus,

Therefore,

Theoildropexperimentcanbeused todetermine theextentof environmentaldamagecausedbyoilspillagefromships.Example5

In an experiment to estimate the size of amolecule of olive oil, a drop of oilvolume 0.12mm3was placed on a cleanwater surface. The oil spread into apatchofarea6.0×104mm2.Estimatethesizeofamoleculeofoliveoil.SolutionVolumeofthedrop=0.12mm3

Areaofthepatch=6.0×104mm2

Thus,thickness,t=2.0×10–9m

Thisisthethicknessofanoilmoleculeinamonolayerofoliveoil.

PrefixesusedwithSIUnits

Table2.1showsthemutiplesandsub-mutiplesusedwithSIunits,theirprefixesandsymbolsfortheseprefixes.

Table2.1:PrefixesUsedwithSIUnits

Sub-multiple/multiple Prefix Symbolforprefix

10-1 deci d

10-2 centi c

10-3 milli m

10-6 micro µ

10-9 nano n

10-12 pico p

10-15 femto f

10-18 atto a

101 deca da

102 hecto h

103 kilo k

106 mega M

109 giga G

1012 tera T

1015 peta P

1018 exa E

1µs=10-6s1s=106µs5kg=5×103g

Exercise2.4

1.Writethefollowingcorrectto:(a)Twosignificantfigures.(b)Threesignificantfigures.

(i)7321769(ii)657.65(iii)27.002(iv)0.001895(v)0.0008996

2.Thedimensionsofacuboidare6.35cm×4.75cm×0.74.Finditsvolumecorrectto:(a)Fivesignificantfigures.(b)Foursignificantfigures.(c)Threesignificantfigures.(d)Twosignificantfigures.

3.Expressthefollowingnumbersinstandardform:(a)0.00123(b)0.000000001(c)1.5(d)15(e)1595

4.Expressthefollowingnumbersinanordinaryform:(a)6.6×10–5

(b)1.257×10–2(c)7.126×10³(d)9.99×104(e)8.1234×10¹

5.Expresseachofthefollowingincm2,givingyouranswerinstandardform:(a)8.4m2

(b)40.2m2

(c)0.08m2

6.Expresseachofthefollowinginm2,givingyouranswerinstandardform:(a)7.5000cm2

(b)0.07cm2

7. Express each of the following volumes in cm³, giving your answer instandardform:(a)14mm³(b)0.002mm³

8. Express each of the following volumes in m³, giving your answer instandardform:(a)3000cm³(b)0.00039cm³

DecimalPlaces

Thenumberofdigitstotherightofthedecimalpointdeterminetheaccuracyof

thenumbergiven.Forexample,ifthenumberisgivenas6.0234,itissaidtobeaccuratetofourdecimalplaces.Thesamenumbercouldbewrittenas:6.023to3decimalplaces.6.02to2decimalplaces.6.0to1decimalplace.6tothenearestwholenumber.

Thenumber6.0234representsthehighestdegreeofaccuracy,followedby6.023.Ifwrittenas6,ithastheleastdegreeofaccuracy.

Similarly, π = 3.1415926. This is accurate to 7 decimal places. Whenexpressedas3.142,itiscorrect(to3decimalplaces).Forworkingpurposesπisgiven as 3.142 correct to 3 decimal places.While using amillimetre scale, areadingof5.4cmwouldbeaccuratelymeasured.However,areadingof5.47cmcanonlybegivenbymakinganestimateoftheseconddecimal.Therefore,thereading given by an ordinary ruler is only accurate to the first decimal place(d.p).

Example6Find the area of the rectangle measuring 3.93 cm by 5.35 cm. Express youranswercorrectto2d.p.Solution

Area = 3.93×5.35cm2

= 21.0255cm2

= 21.03cm2

RevisionExercise2

1.Expresseachofthefollowinginmillimetres,givingtheanswerinstandardform:(a)3.7m(b)0.94m(c)7.0km

2.Expressthefollowingingrams,givingtheanswerinstandardform;(a)25kg(b)0.89kg(c)3.0tones

3.Writedownthenumberofsignificantfiguresineachofthefollowing:

(a)60000(b)523.7(c)805(d)0.000030(e)0.0037(f)0.0207

4.Writeeachofthefollowingnumbersinstandardform:(a)308000000(b)0.000457(c)0.000002394

5.Writedowneachofthefollowinginstandardformto2significantfigures.(a)1000kg(b)1000000m(c)0.0000037kg(d)693(e)569000000(f)32067

6.Determinearethereadingsonthemicrometerandvernierscalesshowninfigures(a)and(b)below.

(a)

(b)

7. (a)Writedownthemicrometerscrewgaugereadingsshowninfigures(i)and(ii)below.

(b) Howmany complete revolutions of the thimble correspond to 1mmmovementin(i)and(ii)?

(i)

(ii)

8.Whatarethezeroerrorsofthemicrometerscrewgaugesinfigures(a)and(b) below? (themicrometers are closed). If themicrometerswere used to

measurethediameterofawirewhosediameteris1.00mm,whatwouldbethereadingoneach?

9.Inthefigureofthemicrometerscrewgaugebelow:(a)LabelthepartsmarkedA,B,CandD.(b)Statethereadingsshownbythescale.

10.Statethereadingsonthevernierscalebelow:

11.Ifanoildropofdiameter0.5mmspreadsonthesurfaceofwatertoformanoilpatchofdiameter0.2m,estimatethethicknessoftheoilmoleculeandexpressyouranswertotwosignificantfigures.

Chapter3

TurningEffectofaForce

MomentofaForce

Oneof the effectsof a force is that it canmakea stationarybodymove.Thistopicdwellsoncaseswhena forceappliedonanobjectmakes theobject turnaboutagivenpointknownasthepivotorfulcrum.Thisturningeffectofaforceiscalledthemomentofforce.Exampleofactivitiesinwhichaforceproducesaturningeffectareclosingopeningadoor,steeringacar,turningoffawatertap,cyclingor ridingonasee-saw, tighteninganutusingaspannerandopeningasodabottle.

Fig.3.1:See-saw

Experiment3.1:Toinvestigatetheturningeffectofaforce

Apparatus

Half-metrerule,three50gmasses,two200gmasses,string.

Fig.3.2:Turningeffectofaforce

Procedure• Tie twomassesof200g each to the stringand suspend them fromahalf-metrerule.

•Holdtherulehorizontally,asshowninfigure3.2.• Vary thepositionof themasses.Note thevariationof the turningeffectonyourhand.

• Repeat theexperimentbyhanging the twomassesataparticularpointandthenaddthreemoremassesof50goneatatime.

Observation

Theturningeffectonthehandfeelsgreaterasthemassesaremovedfarther.Thetwistingeffectofyourhand feels is the turningeffect (ormoment)of a force.Themomentofaforceaboutapointdependsonthemagnitudeoftheforceandthedistanceoftheforcefromthepointofsupport.Themaximumturningeffectat any point is obtained when the distance from the point of support isperpendicular to the line of action. At point A in figure 3.3, the maximumturning effect is obtained when the fore F is applied perpendicularly to thedistancefromthepointofsupportasin(ii)

Fig.3.3:Maximummomentofaforce

Themoment a force is its turning effect on ability tomake an object to turnabout a given point. It is the product of the force (F) and the perpendiculardistancebetweenthepointofsupportandthelineofactionoftheforce.

Thus,momentofaforce

=forceperpendiculardistance

=F×d

Themoment of a force is increasedwhen the perpendicular distance from thelineofactionoftheforcetothepivotisincreased.Thus,itiseasiertoloosenortightenanutusingaspannerwithalonghandlethanonewithashorthandle.

TheSIunitofmomentofaforceisNewton-metre(Nm).

Example1

FindthemomentoftheforceaboutOinthefollowingfigures:

Fig.3.4

Solution(a)Momentofaforce=force×perpendiculardistancefromthepivot.

ThereforemomentofforceaboutO=10N×0.3m=3Nm

(b)Notethattheperpendiculardistancebetweenthelineofactionofforceandtheturningpoint,d,isnotgiven.Sinceifformsarightanglewiththelineofaction, then,usingPythagorastheorem,52=32+d2

d2=25-9d2=16d=4mTherefore,momentofforceaboutO=6N×4m=24Nm

ThePrincipleofMoments

Experiment3.2:ToverifytheprincipleofmomentsApparatus

Metrerule,knownweights.

Fig.3.5

Procedure• Balance ametre rule (or anyother uniform rod) on a pivot at its centre, aknowndistanced.

• Place a knownweightW1 on one side of the pivot and balance it with adifferentknownweight,W2ontheotherside,seefigure3.5.

•RecordthevaluesW1d1,W2ataknowndistanced1andd2inTable3.1.

•RepeatusingdifferentvaluesofW1,d1,andW2

•CompleteW1×d1andW2×d2foreachsetoffiguresandfillintable3.1W1,d1,W2.

Table3.1

•CompareW1×d1withW2×d2foreachsetting.ObservationandConclusionThe force W1 in figure 3.5 tends to make the rule turn in an anticlockwisedirection. Themoment due toW1 is therefore referred to as an anticlockwise

moment. Similarly, the force W2 tends to make the rule turn in a clockwisedirection.It’smomentaboutthepivotisaclockwisemoment.Clockwisemoment=W2×d2=W2d2Anticlockwisemoment=W1×d1=W1d1Atequilibrium,w1d1=W2d2Theprincipleofmomentsstatesthatforasysteminequilibrium,thesumofclockwisemomentsaboutapointmustbeequaltothesumofanticlockwisemomentsaboutthesamepoint.Example2Auniformmetrerulepivotedatitscentreisbalancedbyaforceof4.8Nat20cmmarkandsomeothertwoforces.Fand2.0Natthe66cmand90cmmarksrespectively.CalculatetheforceF.

Solution

Fig.3.6

Distanceof4.8Nfromthepivot=50–20=30cm=0.3mDistanceofFfrompivot=66–50=16cm=0.16mDistance2.0Nfrompivot=(90–50)cm=0.4m

Sumofclockwisemoments=(F×0.16)+(2.0×0.4)=0.16F+0.8Sumofanticlockwisemoments=4.8×0.3=1.44Atequilibrium;Sumofclockwisemoments=sumofanticlockwisemomentsTherefore,0.16F+0.8=1.44

MomentDuetoWeightofanObject

Considerauniformmetrerulesuspendedatits50cmmarkbyaloopofcottonthreadfromspringbalance.ThereadingofthespringbalancegivestheweightW, of themetre rule.Thisweight acts at themidpoint since the rule balanceshorizontally,asinfigure3.8.

Fig.3.8

If the pivot is not placed at themidpoint of themetre rule, theweight of themetreruleactingatthemidpointwillbeconsideredalongsidetheotherforcesinthesystem.Theweightofthemetrerulethenhasamomentaboutthefulcrum.

Now consider a uniformmetre rule balanced by aweight of 1Nwith thepivotplacedatthe40cmmarkasshowninfigure3.9

Fig.3.9:Todeterminetheweightofauniformmetrerule

Since the weightW of the metre rule acts at its midpoint, the weight,W, isdeterminedasbelow:Clockwisemoments=anticlockwisemomentsW(50–40)=1(40–P)1×W=40–P

For a system of parallel forces in equilibrium, the sum of forces in eitherdirectionareequal.

Fig3.10:Parallelforces

Thus,infigure3.10(a)W2+W3=W+W1

Andinfigure3.10(b)F1+F2=F

Experiment3.3:TostudyparallelforcesApparatusTwospringbalances,knownweights,metrerule.

Fig.3.11:Balancingforces.

Procedure• SuspendthemetrerulehorizontallyfromtwospringbalancesS1andS2,asshowninfigure3.11.

• SuspendtwoknownweightsW1andW2andadjusttheirpositionsuntil themetreruleishorizontalandthespringbalancesvertical.

•ObservethereadingsonS1andS2,notethattheforcesF1andF2(onS1andS2respectively)areactingverticallyupwardswhileW1,W2andWareactingverticallydownwards.

•MeasurethedistancesofW1,W,W2,S1andS2fromanyfixedpoint.SayA.

•RepeattheexperimentbyalteringtheweightsW1andW2.

•Recordyourobservationsinatableasshownbelow.

Table3.1

ObservationThe sum of the vertically upward forces is equal to the sum of the verticallydownwards forces.Thus, ifwe take forces in one direction to be positive andthoseintheoppositedirectionnegative,thenthealgebraicsumofparallelforcesiszero.

MomentsofParallelForces

Infigure3.10,themomentsofF1andF2aboutthepointAareF1andd2andF2and d4 respectively. Forces F1 and F2 produce anticlockwise moments aboutpointAandtheirsumF1d1+F2d4

ForcesW1,WandW2produceclockwisemomentsaboutA.Theirsumis:W1d1+Wd3+W2d5Itcanbeshownthat;The sumof clockwisemoments= the sumof anticlockwisemoments, i.e., thealgebraicsumofthemomentsoftheparallelforcesiszero.Example3Figure 3.12 shows a uniformmetre rule ofweight 1.6N supported by springbalanceat the32cmmark.Themetrerule isbalancedhorizontallybya1.2Nweightsuspendedasshown.Find:(a)Thepointwherethe1.2Nissuspended.(b)Thereadingonthespringbalance.

Fig.3.12

Solution(a)Sincethemetreruleisuniform,itsweightactsthroughitscentre,i.e,the50

cm mark. Taking moments about the 32 cm mark and letting the 1.2 Nweightbedmfromthe32cmmark:Anticlockwisemoments=clockwisemoments1.2×d=1.6×(0.50–0.32)

Therefore,the1.2Nshouldbesuspendedatthe(32–24)cmmark,i.e.the8cmmark.

(b) Since the rule is balanced, upward forces =downward forces. Hence,readingonthespringbalance=1.6+1.2=2.8N

EqualParallelForcesinOppositeDirections

Whenacyclistmakesaturn,onehandappliesaninwardforcewhiletheotherappliesanoutwardforceonthehandlebars.Thesetwoforcesassisteachotherinturningthebar.Suchpairofforcesthatareequal,parallelandactinginoppositedirectionsarereferredtoasanti-parallelforcesoracouple.Otherexamplesare:(i)Forcesappliedonawheelspannertighteningoruntighteninganut.(ii)Forcesappliedwhenopeningawatertap.(iii)Forcesappliedonasteeringwheelofacarwhengoingroundabend.

Inalltheexamplesabove,apairofanti-parallelforcesproducesaturningeffect.Figure3.13showstwoequal,parallelandoppositeforcesactingonametre

rule. If the forces are d metres apart, then the turning effect is calculated asfollows:

Fig.3.13:Anti-parallelforcesactingonametrerule

Takingmomentsaboutapointp,lmfromoneoftheforces,momentduetooneforce=F×lMomentsduetotheotherforce=F×(d–l)Totalmoments=F×l+F×(d–l)=Fl+Fd–Fl=Fd

Theexpressionshowsthatthemomentofanti-parallelforcesistheproductofoneoftheforcesandtheperpendiculardistancebetweenthem.

Example5Twovertical,equalandoppositeforcesactonametreruleatthe12cmand80cmmarksrespectively.Ifeachoftheforceshasamagnitudeof3.6N,calculatetheirmomentonthemetreruleabout28cmmark.SolutionMomentduetotheupwardforce=3.6(0.28–0.12)=3.6×0.16=0.576Nm

Fig.3.14

Momentduetothedownwardforce=3.6×(0.80–0.28)=3.6×0.52=1.872NmTotalmoment=0.576+1.872=2.448Nm

Alternatively,Totalmoment=oneoftheforces×perpendiculardistancebetweenthem.

=3.6×(0.80–0.12)=3.6×0.68=2.448Nm

Example6Figure 3.15 shows two parallel forces F1 and F2 acting in opposite directionsalongthesidesADandCBofarectangularhorizontalplateABCD.

Twoequalandoppositeforcesof3NactalongthesidesCDandAB.

The platemeasures 0.8mby 0.6m.Calculate F1 andF2, given that the platedoesnotrotate.

Fig.3.15

Solution

Since the plate does not rotate, themoments of the forcesmust be equal andopposite.TakingmomentsaboutA;

=4.0NAlsotakingmomentsaboutC;F1×0.6=3×0.8

Therefore,F1=F2

ClassActivity

F1andF2bytakingmomentsaboutB,DandO.Explainwhathappens.Example7For thehammer infigure3.16,howlarge is theforceapplied to thenailwhentheforceonthehandleis320N?Assumetheforceappliedtothenailisvertical.Takethelengthsx=0.8cmandy=10cm.

Fig.3.16

Assumption:Theforce(F)appliedtothenail=Theresistance(R)thenailofferstobeingpulledout.Solution

R×0.008=320×0.1

R=F=4000N

ApplicationsofAnti-parallelForces

SteeringwheelCarsaremadetoturnroundcornersbyexertingtwoequalandoppositeforces,F,actingtangentiallytothesteeringwheelasshowninfigure3.17.

Fig.3.17:Anti-parallelforcesonasteeringwheel

The wheel is pivoted at its centre O, and rotates about O in a clockwise or

anticlockwise direction. The two forces form amoment of anti-parallel forcesaboutO.

If thediameterofawheel isD, the the turningeffectof the two forces isgiven by the product of one force and the diameter, i.e moment of the anti-parallelforces=F×DNm.

Heavy commercial vehicles have larger steering wheels to enhance theturningeffectofthevehicle.

Watertap

Awater tap isopenedor closedbyapplying twoequal andopposite forces asshown in figure3.18 (a) and (b).The two forcesproduceamoment about theaxisofrotationO.

Fig.3.18:Anti-parallelforcesonawatertap.

In figure 3.18 (a), the forces produce an anticlockwise turning effect, henceopeningthetap.Infigure3.18(b),theclockwiseturningeffectofthetwoforcesclosesthetap.

BicycleHandle-bars

Whenabicycleisturnedroundabendwithbothhandsonthehandlebars,twoequalandoppositeforcesareapplied,seefigure3.19.

Fig.3.19:Anti-parallelforcesonbicyclehandle-bars

These forces constitute anti-parallel forceswhichproduceamoment about theaxisofrotation,O.

Anti-parallel forcesand theirmomentsarealsoapplied inwatersprinklersandwheelspanners,seefigure3.20.

Fig.3.20:Wheelspanner

Revisionexercise3

1.Aboyofmass40kgsitsatpointof2.0mfromthepivotofasee-saw.Findtheweightofagirlwhocanbalancethesee-sawbysittingatadistanceof3.2mfromthepivot.(Takeg=10Nkg-1)

2.Explainwhy:(a)Itiseasiertoloosenatightnutusingaspannerwithalonghandlethan

onewithashorthandle.(b) The handle of a door is usually placed as far as possible from the

hinges.3.Thediagramcentresbelowshowsuniformbarsbalancedabouttheircanters

by different forces.Calculate the unknown distance and forcesmarked ineachcase.

(a)

(b)

(c)

4.(a)Statetheprincipleofmoments.

(b)Auniformmetreruleofmass120gispivotedatthe60cmmark.Atwhatpointonthemeterruleshouldamassof50gbesuspendedforittobalancehorizontally?

5. A uniform metal rod of length 80 cm and mass 3.2 kg is supportedhorizontallybythetwoverticalspringsbalancesCandD.BalanceCis20cm fromone endwhile balanceD is 30 cm from the other end. Find thereadingoneachbalance.

6. Auniformmetre rodof length5.0m is suspendedhorizontally from twoverticalstringPandQisattachedat0.8mfromoneendwhileQisattachedat2.0mfromtheotherend.Giventhattheweightofthemetalrodis110n;calculatethetensionineachofthestrings.

7.Auniformmetreruleofmass150gispivotedfreelyatthe0cmmark.Whatforceappliedverticallyupwardsatthe60cmmarkisneededtomaintaintherulehorizontally?

8.Auniformmetreruleisbalancedatthe30cmmarkwhenamassof50gishangingfromitszerocmmark.Calculatetheweightoftherule.

9. Ametre rule isbalancedbymassesof24gand16g suspended from itsends.Findthepositionofitspivot.

10.Ametreruleispivotedatitscentre.Aglassblockishangedfromoneendandtheruleisbalancedhorizontallybyhangingmassesof100gand50gat60cmand80cmmarksrespectively.Calculatethemassoftheglassblock.

Chapter4

EquilibriumandCentreofGravity

Ametrerulesuspendedwithastringasshowninfigure4.1isfoundtobalancewhenthepointofsuspensionisataparticularpositionontheruler.

Fig.4.1

Themetrerulecanbethoughtofasbeingmadeupofnumeroustinyparticlesofwood,eachhavingasmallgravitationalforceactingonit,seefigure4.2.

Fig.4.2:Weightsofparticlesinwood

Therulewillonlybalanceataparticularpointwheretheturningeffectsofthegravitationalforcesactingonthetinyparticlescancelout,seefigure4.3Theruleisthensaidtobeinastateofbalanceorinequilibrium.Thegravitationalforcesontheparticlesofwoodcanberepresentedbyasingledownwardforcewhichactsatthepointshowninfigure4.3.Theresultantforceistheweight(W=mg)oftherule.Thepointiscalledthecentreofgravity.

Fig.4.3

Thus,thecentreofgravityofabodyisthepointofapplicationoftheresultantforceduetotheearth’sattractiononthebody.

CentreofGravityofRegularShapes

Thecentreofgravityforauniformbodyliesatthegeometricalcentre.Thecentreofgravitycanbedeterminedbyconstruction,asshowninTable

4.1.

Table4.1:Centreofgravity(CoG)ofregularshapes

Object Diagram CentreofGravity

Squareplate

Constructthediagonals.Thepointofintersectionisthecentreofgravity.

Rectangularplate

Constructthediagonals.Thepointofintersectionisthecentreofgravity.

Triangularplate

Constructtheperpendicularbisectorsofthesides.Theirpointofintersectionthecentreofgravity.

Circularplate

Constructdiameters.Theirpointsofintersectiongivesthecentreofgravity.

Cylinder Thepointofintersection(midpointofaxis)givesthecentreofgravity.

Sphere Constructthediametersofthesphere.Thepointofintersection,whichisthecentreofthesphere,isthecentreofgravity.

Cone Constructtheperpendicularbisectorsfromthebase.Thepointofintersectionwhichis axisfromthebase,givesthecentreof

gravity.

gravity.

Ring Constructthediameters.Thepointofintersection,whichisatthecentreofthering,isthecentreofgravity.

L-Shape DividetheL-shapeintotworectanglesandconstructdiagonalsoneach.Jointhepointsofintersectionthenconstructthebisectoftheangleatthecorner.Thepointwherethebisectormeetsthelinesisthecentreofgravity.

Rectangularframe

Constructthediagonals.Thepointofintersectionofthediagonalsisthecentreofgravity.

CentreofGravityofIrregularShapes

Experiment4.1:Todeterminethecentreofgravityofanirregularshapedlamina.ApparatusPlumbline,thread,stand,cardboard

Fig4.4

Procedure•Makethreeholes,A,BandContheedgesofthecardboard.•SuspenditbyarodthroughholeAasshowninfigure4.4.•Tieaplumblineontherodbesidethecardboard.• After the cardboard and plumbline have stopped swinging, draw a verticallineonthecardboardassetbyplumbline.

•RepeattheexperimentusingtheotherholesBandC.Thelinesintersectatapoint.

• Balancethecardboardwiththetipofapencilatthepointofintersectionofthethreelines.whatdoyouobserve.

•Repeattheexperimentwithsheetsofdifferentirregularshapes.

ObservationThesuspendedobjectwillalwaysrestwithitscentreofgravityverticallybelowthepointofsupport.Theobjectbalancesonthetipofthepencilifplacedatitscentreofgravity.Note:Ifabodyisholloworifithaslegslikeatripodstand,thecentreofgravitymaybeinspaceoutsidethematerialofthebody.Example1Auniformmetalbar,100cmlong,balancesat20cmmarkwhenamassof1.5kg is attached at 0 cmmark, see figure 4.5. Calculate the weight of the bar.(Takeg=10N/kg)

Fig4.5

Solution‘Uniform’ means that the weight is evenly distributed along the bar and itthereforeactsthroughthecentreofgravity,0.5mfromeitherend.W=1.5×10=15N

Takingmomentsaboutthepivotpoint;0.2×15=0.3×W

Example2Auniformmetreruleofweight0.9NissuspendedhorizontallybytwoverticalstringsPandQ,placed20cmand30cmfromitsendsrespectively.Calculatetheforce(tension)ineachstring.

Solution

Fig4.6

Solution

Theweightoftheruleactsthroughitscentre,asshowninfigure4.6.LetF1andF2betheverticalforcesactingonthestrings.

SinceF1 andF2 are unknown, takingmoments aboutA,F1 is eliminated.Similarly,takingmomentsaboutB,F2iseliminated.Sincethesystemisinequilibrium;Sumoftheupwardforces=sumofdownwardforces,i.e.,F1+F2=0.9NTakingmomentsaboutA;F2×0.5=0.9×0.3Therefore,F2=0.54NButF1+F2=0.9HenceF1=0.36N

Activity

TakemomentsaboutBandconfirmthevaluesofF1andF2.Example3Figure 4.7 shows a drumofmass 150 kg and radius 0.5m being pulled by ahorizontalforceFagainstastep0.1mhigh.Whatinitialforceisjustsufficienttoturnthedrumsothatitrisesoverthestep?

Fig.4.7

Solution

Fig.4.8

TakingmomentsaboutA;W×d=F×0.4But,W=150×10=1500NByPythagorastheoremAnd(0.5)2=(0.4)2+d2

d2=0.09d=0.3m1500×0.3=F×0.4

Example4Calculate the force (F) required to be applied vertically to the wheelbarrowhandles in figure 4.9 to lift a load at the centre of gravity as indicated. Thecombinedmassofthewheelbarrowandloadis50kg.

Take the length a = 70 cm and b = 30 cm. Disregard the mass of thewheelbarrow.

Fig.4.9

SolutionClockwisemoments=anticlockwisemomentsLoad=50×10N=500NTakingmomentsaboutthecentreofthewheelF×(a+b)=500×bF×1.0=500×0.3F=150N

StatesofEquilibrium

Whenabodyisatrest,alltheforcesactinguponitbalanceoneanotherandthebodyissaidtobeinequilibrium.Therearethreestatesofequilibrium,namely,stable,unstableandneutral.

Tounderstand thedifferent statesof equilibrium, consider awoodenconerestingonahorizontaltableinvariouspositions.

StableEquilibrium

Theconeshowninfigure4.10(a)hasthelineofactionofitsweightL,passingthrough the centre of the surface onwhich it is resting.When it is tilted at asmallanglebyaforceF,seefigure4.10(b).Thelineofactionofitsweightisdisplacedbutstillpassingwithinthebasearea.Thustheconewhentheforceiswithdrawnthemomentof theweightmakestheconetofallbackandlie in itsoriginalposition.Theconeisinstableequilibrium.

Fig.4.10:Stableequilibrium

UnstableEquilibrium

Theconeinthisposition,seefigure4.11(a)hasaverysmallareaofthesurfaceonwhichitrestsandahighcentreofgravity.Aslightpushbytheforce,F,setsthe lineofactionof theweight,L,passoutside thesurfaceonwhich theconerests,seefigure4.11(b).

Fig.4.11:Unstableequilibrium

NeutralEquilibrium

Iftheconeislaidonitssidesasinfigure4.12andaforceappliedonit,itwillbenoticedthatnomatterhowbigtheforceofdisplacementis, thepositionofthecentre of gravity is neither lowered no raised. This condition is described asneutralequilibrium.

Fig.4.12:Neutralequilibrium.

FactorsAffectingStabilityofObjects

Thestabilityofanobjectdependsonthepositionofitscentreofgravityandtheturningeffectofitsweightaboutanaxisorpoint,inthefollowingways:

TheAreaoftheBase

If the base area is large, the line through the centre of gravity of the bodyremainswithinthebaseevenifthebodyistiltedthroughalargeangle.Hence,abodywithabroadbaseismorestablethantheonewithanarrowbase.

ThePositionoftheCentreofGravity.

Abodyismorestablewhenitscentreofgravityisaslowaspossible.Thiscanbeachievedbymakingthebaseheavier.Bodieswithhighcentresofgravityarelessstable

ApplicationsofStability

(i)Busesaremademorestablebyhavinglightmaterialsfortheupperpartsofthe body and the heavy chassis and engine as low as possible. Luggagecompartmentsaresituatedinthelowerpartssothatthecentreofgravityislowered.

(ii) A racing car has a low centre of gravity and a wide wheel base whichallows a large angle of tilt. Hence, it can negotiate sharp corners at highspeedswithouttoppling.

Fig.4.13:Aracingcar

(iii)ABunsenburnerhasawideheavybase.Theheavybaselowersthecentreofgravityandthelargebaseareaprovideslargeangleoftiltbeforetoppling.

(iv) Chairs,stools, tripods,etc.areprovidedwiththreeormorelegs.Thelegsareoftenmadeslightlyinclinedoutwardstoimprovestability.

(v) Acrobats in theirbreathtakingmanouversmustensure that theircentreofgravityremainswithinthedefinedbaseareaforstability.

Fig.4.14:Acrobats

RevisionExercise4

1.(a)Definethecentreofgravityofabody.(b)Describeanexperimenttodeterminethecentreofgravityof

(i)auniformregularobjectlikeametrerule(ii)Athinirregularshapedmetalsheet.

2.Explainwhy:(a)Itisnotsafeforadoubledeckerbustocarrypassengersstandingonthe

upperdecker.(b) Busbody-buildersbuild luggagecompartmentsunder theseats rather

thanontheroofracks.(c)Laboratoryretortstandsaremadewithwideheavybase.

3.(a)Whenisanobjectsaidtobeinstableequilibrium?(b)Describetheofequilibriumin:

(i)Amarbleatthebottomofawatchglass.(ii)Atight-ropewalker.(iii)Acylindersittingonitsbase.(iv)Asphereonaleveltabletop.(v)Abirdperchedonathinhorizontalbranchofatree.

4.(a)State:(i)Twowaysinwhichthestabilityofabodycanbeincreased.

(ii)Twopracticalapplicationsofstability.(b)Explainhowacyclistmaintainsthestabilityofamovingbicycle.

5.Explainwhyalorryloadedwithbagsofmaizepackedhighupislikelytotopplewhennegotiatingacorner.

6.Thediagrambelowshowsametalplate3mlong;1mwideandnegligiblethickness.ahorizontalforceof100NappliedatpointDjustmakestheplatetilt.Calculatetheweightoftheplate.

7.Statetheconditionsfortheequilibriumofabodywhichisacteduponbyanumberofforces.

Chapter5

ReflectionatCurvedSurfaces

Reflectionof lightbyplanemirrorswascovered inBookOne. In thischapter,considerationisgiventothetypesofcurvedsurfaces,imageformationandtheirapplications.

Curvedsurfaceshaveawide rangeofscientificandpracticalapplications,forinstance,inshavingmirrors,drivingmirrors,searchlightsandtelescopes.

TypesofCurvedSurfaces

Curved surfaces may be obtained from hollow shapes of spheres, cones orcylinders.Whenthesesurfacesarehighlypolished,theybecomereflectors.

Inahollowmetalsphereforexample,iftheinnersurfaceishighlypolished,thentheportionofthesphereisdescribedasaconcavereflector.

Conversely, if theouter surface ishighlypolished, then theportionof thesphereisdescsribedasaconvexreflector.

Fig.5.1:Shapesofreflectors

If both inner and outer surfaces of themetal spherewere highly polished, theinnersurfacebecomesconcavewhiletheouterbecomesconvex.

Curvedmirrorscanbemadebyapplyingsilverypaintonpartofahollowglasssphere,cylinderoracone,asshowninfigure5.2(a),(b)and(c).

Fig.5.2:Mirrorshapes

Thesilveringgivesrisetotwotypesofreflectingsurfaces:

Fig.5.3:Curvedmirrors

(i)Areflectingsurfacecurvinginwards,seefigure5.3(a)and(c).(ii)Areflectingsurfacebulgingoutwards,seefigure5.3(b).

Curved mirrors whose reflecting surfaces curve inwards are called concavemirrors.Seefig.5.3(a),whilethosewithreflectingsurfacesbulgingoutwardsare called convex mirrors. See fig. 5.3 (b). Mirrors made from spheres are

calledsphericalmirrors.A parabolicmirror is a specialmirror cut from a section of a cone, as in

figure5.2(c)and5.3(c).Aconcaveparabolicreflectorisobtainedbyapplyingsilverypaintontheoutsideof theconesothat thereflectingsurfacecurvesin,seefigure5.3(c)

Note:

Silveringoftheinnersurfaceofglassproducesaconvexmirror,whileahighlypolished inner surfacegives a concave reflectorwhichbehaves like a concavemirror.

DefinitionofTerms

In order to understand some terms that are used in describing curvedmirrors,considerthemirrorsshowninfigure5.4(a)and(b).

Fig.5.4:Definitionoftermsincurvedmirrors

Theapertureofthemirroristheeffectivediameterofthelightreflectingarea.Itisthestraightlinexyinfigure5.4.

Thepoleisthecentre,P,ofthemirror.Thecentreofcurvature,C,isthecentreofthesphereofwhichthemirror

ispart.Inthecaseofaconcavemirror,thecentreofcurvatureisinfrontofthemirrorwhileinaconvexmirror,itisbehind.

The principal axis is the line drawn through the pole of the mirror and thecentreofcurvature.Itisalsocalledthemainaxisofthemirror.

Theprincipalfocus,F,foraconcavemirror,isthepointatwhichallraysparallelandclosetotheprincipalaxisconvergeafterreflection.

Foraconvexmirror,thisisthepointatwhichallraysparallelandclosetotheprincipalaxisappeartodivergefromafterreflectionbythemirror.

Theprincipalfocusofaconcavemirrorisarealfocus,thatis,reflectedraysactuallypassthroughit,whiletheprincipalfocusofaconvexmirrorisavirtualfocusbecausereflectedraysonlyappeartopassthroughit.

The focal plane is a plane perpendicular to the principal axis and passesthroughtheprincipalfocus.Parallelrayswhicharenotparalleltotheprincipalaxiswouldconvergeatapointonthefocalplane,seefigure5.5.

Theradiusofcurvature,r,istheradiusofthesphereofwhichthemirrorispart.Infigure5.4,itisthedistancePC.

Fig.5.5:Focalplane(seeexplanationofraydiagramsonpage81)

The focal length, f, is thedistance from thepoleof themirror to itsprincipalfocus.

ReflectionofLightbyCurvedMirrors

Theeffectoftheconcaveandconvexmirrorsonrayscanbestudiedusingaraybox.

Rays parallel and close to the principal axis are made incident on bothconcaveandconvexmirrors,asinfigure5.6(a)and(b).

Fig.5.6:Reflectionofparallelbeam

Fortheconcavemirror,theraysconvergeatapointF,theprincipalfocus,afterreflection.Fortheconvexmirror,theraysarereflectedsothattheyallappeartodivergefromtheprincipalfocusFbehindthemirror.If therayboxismadetoproduceabeamconvergingattheprincipalfocus,F,aparallelbeamisobtainedforboth the concaveandconvexmirror after reflection, see figure5.7 (a) and(b).

Fig.5.7:Reflectionofdivergentbeam

Comparefigures5.6and5.7.Theyshowthatthelightraysarereversible.Thisisa demonstration of the principal of reversibility of light, which states thatlightwillfollowexactlythesamepathifitsdirectionoftravelisreversed.

Experiment5.1:Todeterminethecentreofcurvatureofaconcavemirror.

ApparatusWhite screen with a hole with cross-wires, mounted concavemirror, lamp orcandle,metrerule.

Fig.5.8:Measurementofradiusofcurvatureofaconcavemirror

Procedure

•Arrangetheapparatusasshown.•Supportthewhitescreensatitscentre.•SupportaconcavemirrorMinfrontofthescreensothattheobjectcrosswireOisinlinewiththepoleofthemirror.

•Placeacandleorlampbehindthescreentoilluminatethecross-wires.•AdjustthemirrortowardsandawayfromS,untilasharpimageofOisseenbesideOonthescreen.

• Measure the distanceMI. The object cross-wires O and image I are nowpractically coincident in position. In this case, O must be at the centre ofcurvatureofthemirror.ThedistanceMIistheradiusofcurvature,Aproofgivenlaterinthischaptershowsthatthefocallength(f)ishalftheradiusofcurvaturer,thatis, .Thefocallengthofthismirrorcanalsobedeterminedfromthisrelationship.

Experiment 5.2: To determine the centre of curvature by the no-parallaxmethod

ApparatusSlidingcorkonaglassrod,concavemirror,opticalpin,clamp,stand,metrerule.

Fig.5.9:Centreofcurvatureofconcavemirrorbytheno-parallaxmethod

Procedure•Clampaglassrodwithaslidingcorkverticallyasshowninfigure5.9•Fixapintothecorksothatitishorinzontaltotheplaneofthebench.•Placeaconcavemirroronthebenchsothatthetipofthepinisinthesameverticallinewiththeaxisofthemirror.Thispinistoactastheobjectpin.

•Movethecorkupanddown,viewingitasshownuntilarealinvertedimageofthepinisseen.

• Adjust the position of the object pin carefuly until there is no parallaxbetweenitandtheimage,i.e,theimageandtheobjectcoincideandmoveinthesamedirectionastheeye.

•Measuretheverticaldistancefromthebenchtothepin.Thisgivestheradiusofcurvature,ofthemirror.

•Repeattheexperimenttoobtainmorevaluesofr.Calculatetheaverageofthevalues.

Experiment5.3:Toestimatethefocallengthofaconcavemirror

Apparatus

Metrerule,distantobject,concavemirror,whitescreen.

Fig.5.10:Estimationoffocallength

Procedure•HoldthemirrorMsothatitsreflectingsurfacefacesanobject,e.g,awindow.•MovethewhitescreenSinfrontofthemirroruntilaninvertedsharpimageofthewindowframeisformedonit,seefigure5.10.

•MeasurethedistancefromMtoS.Parallelraysfromadistantwindowarereflected,convergingatthefocalplaneofthemirror.ThedistanceMSisapproximatelyequaltothefocallength,f,ofthemirror.

LawsofReflectionandCurvedMirrors

ConsideranincidentrayQM,parallelandclosetotheprincipalaxis,asshownin figure 5.11 (a).AtM, the ray is reflected throughF.A line drawn fromC

throughthepointofincidentMisnormaltothesurfaceofthemirror.

Fig.5.11:Lawofreflectionandcurvedmirrors

Fromgeometry,itfollowsthat;

∠QMC=∠MCF(alternateangles).ItwasmentionedearlierthatPC=2PF

Therefore,PF=FC

WhenMisveryclosetoP,then;

PF=MF

Therefore,∠MCF=∠CMF

Since∠QMC=∠MCF,itfollowsthat;

∠QMC=∠CMF

But∠QMCistheangleofincidence,i,and∠CMFtheangleofreflection,r.

Therefore,i=r.

Itfollowsfromthisrelationshipthatreflectionbycurvedmirrorsobeysthelawsofreflection.Thiscanalsobeprovedfortheconvexmirror,seefigure5.11(b).

RayDiagrams

Raydiagramscanbeusedtoexplainhowimagesareformedbycurvedmirrorsandthecharacteristicsoftheseimages.Thereflectingsurfaceisrepresentedbyastraightlineandasmallcurveisusedtoshowthetypeofmirror.

The object is represented by a straight line perpendicular to the principalaxis,withanarrowtoshowitstip,seefigure5.12(a)and(b).

Thepoleofthemirroristakenastheintersectionofprincipalaxisandthemirrorline.

Fig.5.12:Representationofconcaveandconvexmirrors

Amongthemanyrayswhichcouldbedrawninconstructingaraydiagram,fourspecialonesareconsidered.Theseare:

(i)AraythroughCorappearingtopassthroughC

Fig.5.13:AraythroughC

Thisrayisreflectedalongthesamepath,seefigure5.13

(ii)AraycloseandparalleltothePrincipalAxis.Incaseofaconcavemirror,thisrayisreflectedthroughtheprincipalfocus,whileforaconvexmirror,therayisreflectedinsuchawaythatitappearstoemergefromtheprincipalfocus,seefigure5.14(b).

Fig.5.14:Araycloseandparalleltotheprincipalaxis

(iii) A ray through Principal Focus of ConcaveMirror or appearing to bedirectedtoPrincipalFocusofConvexMirrorThisrayisreflectedparalleltoprincipalaxis,seefigure5.15(a)and(b).

Fig.5.15:AraythroughF

(iv)ArayatanangletothePrincipalAxisandincidenttothePoleThis ray is reflected in such a way that the angle of incidence, i, equalsangleofreflection,r,seefigure5.16(a)and(b).

Fig.5.16:Arayatanangletotheprincipalaxis

ImageFormationandCharacteristicsConcaveMirrors

The nature, size and position of the image of an object formed by a concavemirrordependsonthepositionoftheobjectfromthemirror.

Experiment5.4:Tolocatetheimageformedbyaconcavemirror

ApparatusConcavemirroronaholder,screenS,objectslit,lamporcandle.

Fig.5.17:Imageformationbyaconcavemirror

Theobjectinthisexperimentisaslitonacard‘O’,informofanarrow.

Procedure• Position the card andplace a lampor a candlebehind the slit, as in figure

5.17.•SupporttheconcavemirrorMinfrontoftheobjectsothatthedistanceOMisgreaterthantheradiusofcurvatureofthemirror.

• Move the white screen S between the object and themirror until a sharpimage is formedon it.Compare the size and appearance of the imagewiththatoftheobject.

• Repeat the experiment with the object at different positions towards themirror,untilitisveryclosetoit.IfnoimageisformedonS,trytolocateitbehindO.

ObservationWhen the object is far away from themirror, the image is real, inverted andsmallerthantheobject.

Whentheobjectismovednearertothemirror,butstillfurtherthanF, theimageremainsinverted,realandbecomesbigger.

When the object is at the centre of curvature of themirror, the image isinverted,realandsamesizeastheobject.

When theobject isbetweencenterof curvatureandprincipal focusof themirror,theimageisinverted,largerthantheobjectandbeyondC.

The imagebecomes larger as theobject approaches theprincipal focusofthe mirror. When the object is moved from the principal focus towards themirror,noimageisobtainedonthescreen.

Thefollowingraydiagramsillustratetheobservations

ObjectatInfinity

Fig.5.18:Objectatinfinity

Theimageformedis:(i)atF.(ii)real.(iii)inverted.(iv)smallerthantheobject.

ObjectbeyondC

Fig.5.19:ObjectbeyondC

Theimageformedis:(i)betweenCandF.(ii)real.(iii)inverted.

(iv)smallerthantheobject.

ObjectatC

Fig.5.20:ObjectatC

Theimageformedis:(i)atC(ii)real(iii)inverted(iv)samesizeastheobject.

ObjectbetweenCandF

Fig.5.21:ObjectbetweenCandF

Theimageformedis:(i)beyondC.

(ii)real.(iii)inverted.(iv)Magnified(largerthantheobject).

ObjectatFThe image is formedat infinity, see figure5.22.This isbecause the light raysemergeparallelafterreflection.Comparethiswithobjectatinfinity.

Fig.5.22:ObjectatF

ObjectbetweenFandP

Fig.5.23:ObjectbetweenFandP

Theimageformedis:(i)behindthemirror.(ii)virtual.(iii)erect(upright).

(iv)largerthantheobject.By convention, full lines represent real rays andobjectswhiledotted linesrepresent virtual rays and images. In a ray, an arrow is drawn to show thedirectioninwhichthelightistravelling.A real image can be focused on a screenwhile a virtual image is formed byapparent intersectionofraysandcannot thereforebeformedonascreen.Notealsothatarealimageisformedbyaconvergentreflectedbeamwhileavirtualimageisformedbyadivergentreflectedbeam.

ConvexMirrors

Whereas concavemirrors form either real or virtual images depending on theposition of the object, images formed by convex mirrors are always upright,smallerthantheobjectandbetweenPandF.Figure5.24showstheraydiagramfortypicalimageformedbyaconvexmirror.

Fig.5.24:Imageformationbyaconvexmirror

GraphicalConstructionofRayDiagrams

The images obtained from a curved mirror can be drawn to scale in a raydiagram.Theconstructionofaraydiagramisbestdoneonagraphpaperusingsuitablescale.Note:Whenconstructingaraydiagram:(i)Drawahorizontalstraightline,sayAB,torepresenttheprincipalaxis.(ii)Drawaline,sayQR,atrightanglestoABatpolePtorepresentthemirror.

Example1Anobjectofheight10mmisplaced50mminfrontofaconcavemirroroffocallength30mm.Byscaledrawing,determinethe:

(a)Positionoftheimage.(b)Sizeoftheimage.(c)Natureoftheimageformed.Solution

Fig.5.25

A suitable scale is 1 mm on the paper to represent 1mm of actual length onverticalscaleand1mmtorepresent2mmonhorizontalscale.•Marktheprincipalfocus30mmfromP.•DrawalineOO,10mmlong,perpendiculartoABand50mmfromP.Thisrepresentstheobject.

•DrawtwoincidentraysfromO′.(i)ArayfromO′paralleltotheprincipalaxisCP,isreflectedthroughF.(ii)ArayfromO′throughC,isreflectedalongthesamepath.ArayfromO′

throughFcanalsobeusedhere.These two intersectafter reflection togivearealimageatI.

•CompletetheimagebydrawingMIperpendiculartotheaxisPC.

Fromthecompletediagram,andaccording to thescaleused, itcanbeseenthat:(a)Theimageisformed75mmfromP.(b)Theimageis15mmtall,hencemagnifiedandinverted.(c)Theimageisreal,becauseitisformedbyactualintersectionofrays.

Example2

Aconvexmirroroffocallength9cmproducesanimageonitsaxis6cmawayfromthemirror.Iftheimageis3cmhigh,determinebyscaledrawing:(a)Theobjectdistancefromthemirror.(b)Thesizeoftheobject.

Solution

Asuitablescaleinthiscaseis1cmtorepresent3cmofactuallengthonverticalscaleand1cmtorepresent4cmonhorizontalscale,seefigure5.26.

Fig.5.26

• Usingthescale,marktheprincipalfocusFandthecentreofcurvatureC,9cmand18cmrespectivelyfromPbehindthemirrorasshowninthegraph.

• DrawabrokenlineIM,3cmlong,perpendiculartoABand6cmfromthepole,P,ofthemirror.Thisrepresentstheimage.Tworayswillbeusefulinconstructionoftheraydiagram:(i)Arayfromthetopoftheobjectandparalleltotheprincipalaxis.(ii)ArayfromthetopoftheobjectthroughC.

• Draw a dotted line FS and extend it to S′ as a continuous line. SS′ is thereflectedrayandSthereforeisthepointofincidence.Fromthispoint,drawalineQSparalleltotheprincipalaxis.Thisistheincidentrayandthetopoftheobjectliessomewhereonthisray.

•DrawadottedlinefromCthroughItothemirror.Sincethislineisnormaltothemirror,arayfromthe topof theobjectalongthis line isreflectedalongthesamepath.ExtendthislineasacontinuouslinetointersectwithQSatO′.O′isthetopoftheobject.

• Draw a perpendicular line fromO′ tomeetAB atO.O is the foot of theobject.

• Usingyourscale,determinethedistancePO(objectdistance)andheightofobjectOO′.Fromthecompletediagram:(a)Objectdistance,PQ,is18cm.(b)ObjectheightOO′is9cm.

LinearMagnification

Imagesformedbycurvedmirrorsvaryinsize.Itisalwaysimportanttocomparethesizeoftheobjectwiththatoftheimageformed.

The numerical comparison of the image size with object size is calledmagnification. In this chapter, linear or transverse magnification, i.e,magnificationofonedimension,isconsidered.

Linearortransversemagnification,m,isgivenbytheformula:

Consider figure5.27 inwhich the imagehasbeenconstructedusinga rayfromOthroughCandanotherrayfromOtoP.

Inthefigure,AOistheheightoftheobjectandBItheheightoftheimage.

Fig.5.27:Linearmagnification

Fromthedefinitionoflinearmagnificationm;

Theright-angledtrianglesBIPandAOParesimilar.Therefore,m

Ifvistheimagedistanceandutheobjectdistance,thenlinearmagnificationcanalsobegivenby:

Linearmagnificationhasnounitssinceitisaratiooftwolengths.

Example3

Fig.5.28

A concavemirror of focal length 20 cm forms a sharply focused image of asmallobjectonascreenplacedatadistance80cmfromthemirror.Graphicallydetermine:(a)Thepositionoftheobject.(b)Thelinearmagnificationoftheimage.

Solution

Asuitablescaleinthiscaseis1cmtorepresent10cmoftheactuallength.•usingthescale,marktheprincipalfocus20cmfromthemirror.

MarkthecentreofcurvatureC,40cmfromthemirror(r=2f)• Chooseasuitableimageheightsuchas20cm(allrealimagesareinverted,forconcavemirror.

•Drawtheimage80cmfromthemirror.

Arayfromthe topof theobject throughF is reflectedparallel to theprincipalaxis. Therefore, a ray from the top of the image parallel to principal axis isreflected through F(principle of reversibility of light). The top of the objectthereforeliessomewhereonthisray.•DrawarayfromtopoftheimagethroughC.Thisrayisreflectedbackalong

the same path. The intersection of the two rays between C and F aboveprincipal axis is the top of the object. Complete the object by drawing aperpendicularlinefromthispointtotheprincipalaxis.

Fromthegraph:(a)Theobjectis26cmfromthemirror.

Alternatively,

Example4A concave mirror of focal length 20 cm produces an upright image ofmagnification2.Graphicallydeterminetheobjectdistance.

Solution

Scale:1cmrepresents10cmSince(i)magnificationis2;

objectheight:imageheight=1:2(ii)theimageiserect,itisvirtualandbehindthemirror.Asuitablescale in thiscase is1cmonpaper to represent10cmof theactuallength,seefigure5.29.•DrawtwolinesKHandEGparalleltotheprincipalaxisatheightsabovetheaxisintheratio1:2,soastomeetthemirroratHandGrespectively.

•JoinFHandproduceittomeetEGproducedatI.Iisthetopoftheimage.• CompletetheimagebydrawingaperpendicularlinefromItotheprincipalaxisatB.

• Draw a perpendicular line from O the principal axis at A. The object isthereforeOA.Fromthegraph,theobjectdistancePAis8.5cm.

Fig.5.29

Example5

Fig.5.30

Aconcavemirroroffocallength10cmformsarealimagefourtimesthesizeoftheobject.Iftheobjectheightis5cm,determinegraphically:(a)Theobjectdistance.

(b)Theimagedistance.

SolutionAsuitablescaleinthiscaseis1cmtorepresent10cmofactuallength.•Usingthescale,drawlineEGparalleltotheprincipalaxisataheight0.5cm(heightofobject)fromit.

•DrawanotherlineKH,2cmbelowandparalleltotheprincipalaxis.Notethatthe ratio of distances of EG and KH from the principal axis is 1: 4 sincemagnificationis4.

•DrawalinefromGthroughFandproduceittocutKHatI.Iisthetopoftheimage.

•Completetheimagebydrawingaperpendicularlinetotheprincipalaxisat1.•DrawalinefromIthroughCandproduceittomeetEGatOsothatOisthetopoftheobject.

•Completetheobjectbydrawingaperpendicularlinetotheprincipalaxis.Theobjectis,therefore,OA.

Fromthegraphs:(a)Theobjectdistanceis12.5cm.(b)Theimagedistanceis50cm.

Relationshipbetweenfandr

Wehave seen that focal length is half the radius of curvature, that is, .This relationship holds for spherical mirrors and can be proved theoreticallyusinggeometry.

Consider a single ray AB above and parallel to the principal axis andincidenttothemirroratB.TherayisreflectedatBthroughF,seefigure5.31.

Fig.5.31:RelationshipbetweenFandr

Bygeometry,CBisnormaltothetangenttothemirroratB.Itthereforefollowsfromthelawsofreflectionthat:

∠ABC=∠CBFBut∠ABC=∠BCF(alternateangles)Therefore,∠CBF=∠BCFTherefore,BF=FC.

WhenBisclosetoP,thatis,ABisincidenttothemirrorandveryclosetoP,then

BF=PF.

Therefore,PF=FC.Hence,FismidwaybetweenPandC.

Thus,FP=

But,FP=fandCP=r

Therefore,

TheMirrorFormula

Ifanobjectisatadistanceufromacurvedmirroroffocallengthf,itsimageisformedatadistancevfromthemirror.Therelationshipbetweenu,vandfcanbedeterminedexperimentally.

Experiment 5.5: To investigate the relationship between u, v and f for aconcavemirror

ApparatusTwoopticalpins,ruler,mountedconcavemirror.

Fig.5.32:Relationshipbetweenu,vandf

Procedure•Placeanobjectpinatadistanceu=15cminfrontoftheconcavemirror,seefigure5.32.

• With the eye positioned as shown, locate the position of the imagewith asearchpinusingthe‘noparallax’method.

•RecordtheimagedistancevinTable5.1•Repeattheexperimentwithobjectpinatu=17,19,21and23cm.

Recordthecorrespondingvaluesofvonthetable.•Completethetableandcalculatethemeanvalueof

•Comparethemeanvalueof with ,thereciprocaloffocallengthofthe,withmirrorused.

Table5.1

Itisobservedthattheobjectdistanceu,imagedistancevandthefocallengthfofthemirrorarerelatedbytheformula;

Thisiscalledthemirrorformularandappliestoallsphericalmirrors.Fromtheformular,

SignConvention

Inorder todetermine thepositionandnatureof the imageformedbyacurvedmirror,asignconventionisnormallyused.

Thesignconventionusedinthisbookisreal-is-positivesignconvention.

Real-is-PositiveSignConvetionWhenapplyingthisconvention:(i)Alldistancesaremeasuredfromthemirrorastheorigin.(ii)Distancesofrealobjectsandimagesareconsideredpositive(+).(iii)Distancesofvirtualobjectsandimagesareconsiderednegative(–).

Intheconvention,aconcavemirrorhasarealprincipalfocusandthereforepositive focal length,while a convexmirrorwith a virtual principal focus hasnegativefocallength.

Example6Anobjectisplaced30cmfromaconcavemirroroffocallength20cmCalculate:(a)Theimageposition.(b)Themagnification.Solution(a)Themirrorisconcave,hence;

Theimageis60cmfromthemirrorandsincevispositive,itisreal.

Example7An object is placed (a) 18 cm (b) 6 cm in front of a concavemirror of focallength 12 cm.Determine the position and nature of the image formed in eachcase.Solution

Theimageisformed36cmfromthemirror,vispositive,sotheimageisreal.(b)u=+6cm

Substitutingintheformular;

The image is formed12cmfromthemirror,v isnegative,hence the image isvirtual.

Example8Aconvexmirroroffocallength9cmproducesanimageonitsaxis6cmfromthemirror.Determinethepositionoftheobject.Solution

f=–9cm(convexmirror)

v=–6cm

Theplussignshowsthattheobjectisrealand18cminfrontofthemirror.

GraphicalAnalysisoftheMirrorFormular

(i)FromTable5.1ifagraphof= against= isplotted,astraightlinewithanegativegradientisobtained.

Fig.5.33:Graphof against=

The x-intercept or the y-intercept gives . The gradient is negative,implyingthattheimageisinvertedrelativetotheobject.

(ii)Agraphofuvagainst(u+v)isastraightlinepassingthroughtheorigin.

Fig.5.34:Graphofuvagainstu+v

Thegradientgivesthefocallengthf.Since ,multiplyingthroughbyvgives;

(iii)Agraphofmagainstvisastraightlinewithgradient .They-interceptis–1.

Fig.5.35:Graphofmagainstv

ApplicationofCurvedMirrors

ConcaveMirrors

Concavemirrorsareused:(i)Asshavingmirrors(ii)Bydentists,whenexaminingteeth.

Ineachcase,theobjectisplacedwithinthefocallengthofthemirrorsothatamagnifiederectimageisobtained,seefigure5.36.

Fig.5.36:Useofconcavemirrorasashavingmirror

(iii)AsReflectorbehindaProjectorLampThelampisplacedatthecentreofcurvatureoftheconcavemirrortoreflectlighttravellingawayfromtheprojector,henceincreasingtheilluminationoftheslide,seefigure5.37

Fig.5.37:Concavemirrorwithprojectorlamp

(iv)InTelescopes,forAstronomicalObservationsWhenanobject suchasastar isvery far fromthemirror (at infinity), theraysfromanypointonitappearstooriginatefromaparticularpointandarethereforeparallel.Theimageisthusformedatthefocalpoint.

(v)SolarConcentratorsTheheatandlightenergyfromthesuncanbebroughttofocusbyaconcavemirror.Thisfactisemployedinsolarcookers,whereasmallovenisplacedatthefocalpointofalargemirror.

ConvexMirrors

Theyareused:(i)AsdrivingMirrors(ii)InSupermarkets,sothattheattendantscanmonitorlargefloorarea

Thisisbecause:•Theyformanuprightimage,regardlessoftheobjectdistance.• They provide awide field view, so that the overtaking traffic can beeasilyseen,seefigure5.38.

Fig.5.38:Narrowandwidefieldofview

Theonlydisadvantageofusingaconvexmirrorasadrivingmirroristhatitforms a diminished image, giving the impression that the vehicles behindarefartherawaythantheyactuallyare.Thisisdangerousandthedriverhastolearntojudgedistancesaccuratelywhenusingthemirror.

DefectsofSphericalMirrors

Inthischapter,wehaveusednarrowparallelbeamsoflightclosetoandparallelto the principal axis in studying reflection of light by curved mirrors. Forconcavemirrorinthiscase,allthereflectedraysconvergeatonedefinitepoint,the principal focus.However, ifwide parallel beam is incident on a sphericalmirror of wide aperture, then parallel rays are not brought to a focus at theprincipalfocusFafterreflection.

Rayswellawayfromtheprincipalaxisarebroughttoadifferentpoint,suchasF1,afterreflection,seefigure5.39(a).Theparallelbeam,therefore,produces

ablurredfocus.Thisiscalledsphericalaberration.The reflected rays intersect to forma surface calledacaustic curve, a curvedshapecentredabouttheprincipalfocusF.

Fig.5.39:Causticcurves

Abrightcausticcurveisoftenseenonthesurfaceofteainacup.Thisisformedwhenlightfromadistancesourceisreflectedfromtheinsideofthecup,whichactsasacurvedmirroroflargeaperture.

From theprincipleof reversibilityof light, it follows thatpartof the lightfromasmalllampplacedattheprincipalfocusFwillbereflectedasadivergentbeamfromtheouterpartsofthemirror.Sincethereflectedlightenergyspreadsout from themirror, it becomesweaker as the distance increases.Due to this,sphericalmirrorsareusedinsearchlightsorheadlampsofcars.

Aparabolicmirrorovercomesthisdefectoffocusinasphericalmirror.Theshape of a parabolic mirror is shown in figure 5.39 (b). All parallel rays arebroughttoasinglefocusattheprincipalfocusF.

By the principle of reversibility of light, a light source placed at theprincipal focus of the parabolic mirror will produce a parallel beam after

reflection.Parabolic concavemirrors are used therefore for the propagationofparallel beams in search lights, car headlights or hand torches. Parabolicreflectorsinalltheseproduceabeamofhighintensity.

RevisionExercise5

1.Definethefollowingtermsasappliedtoconcaveandconvexmirrors:(a)Principalfocus.(b)Centreofcurvature.(c)Focalplane.

2. Youaregiven twocurved reflectors, a sphericalconcaveandaparabolicconcavereflector.Whichonewouldyouchooseforatorch.Explainwhy.

3. Stateandexplainoneadvantageandonedisadvantageofusingaconvexmirrorasadrivingmirror.

4. Adentisthas a choiceof three smallmirrors, a convex, a concaveandaplaneonetoexaminethebackofyourteeth.Statewhichoneheshouldusetogivethebestview.Givereasonsforyourchoice.

5.(Insertartwork)

Diagrams(a)and(b)showaconcaveandaconvexmirrorrespectively.Tworaysof lightareshown ineachdiagram.Trace thediagramsonpaperanddrawtheapproximatepathofeachrayafterreflection.

6.Aconcavemirrorisusedtoformanimageofanobjectpin.Wheremustthe

objectpinbeplacedtoobtain:(a)Anupright,magnifiedimage?(b)Aninverted,diminishedimage.(c)Animagethesamesizeastheobject?

7. A concave mirror with radius of curvature 20 cm produces an invertedimagetwotimesthesizeofanobjectplacedinfrontofitandperpendicularto theprincipleaxis.Byanaccuratescaledrawing,determine thepositionof:(a)Theobject.(b)Theimage.

8.Aconcavemirroroffocallength10cmformsavirtualimage5cmhighand30cmfromthemirror.Byanaccuratescaledrawing,determine:(a)Thepositionoftheobject.(b)Theheightoftheobject.(c)Magnificationoftheimage.

9.Aconcavemirroroffocallength20cmformsavirtualimagefourtimesthesizeoftheobject.(a)Useascalediagramtodeterminetheobjectdistance.(b)Usethemirrorformulatodeterminetheobjectdistance.

10.Anobjectissetup20cminfrontofamirrorA.Thedetailsoftheimagearenotedasshowninthetablebelow.TheprocessisrepeatedwithadifferentmirrorB.

MirrorA Real,inverted,magnifiedandagreatdistance.

MirrorB Real,inverted,samesizeastheobject.

(a)Statethetypeof:(i)MirrorA.(ii)MirrorB.

(b)Determinethefocallengthsofthetwomirrors11.Thedistancebetweenanobjectanditsmagnifiedrealimageproducedbya

concavemirroris20cmwhentheobjectisplaced10cmfromthepoleofthemirror.Determinethe:(a)Linearmagnificationoftheimage.(b)Thefocallengthofthemirror.

12.Thedistancebetweenanerectimageandtheobjectis30cm.theimageistwiceastallastheobject.(a)Whatistheobjectdistance?

(b)Determinetheradiusofcurvature.(c)Iftheimageisvirtual,whattypeofmirrorisused?

13.Aconcavemirrorofradius180cmformsarealimagehalfitsheight.Whereshouldtheobjectpositionbe?

14. A Magnified erect image and a magnified inverted image can both beformedbyaconcavemirror.Drawraydiagramstoshowthis.

15.Showhowparabolicreflectorspropagateparallelbeamsoflight.16. Describe a method that can be used to determine the focal length of a

convexmirrorexperimentally.17.Aconcavemirrorandilluminatedobjectareusedtoproduceasharpimage

oftheobjectonascreen.Theobjectdistancesandimagedistancesaregivenbelow:

Objectdistance(cm) Imagedistance(cm)80 20

26.67 40

22.4 56

20.57 72

19.55 88

(a)(i)Plotagraphof against(ii)Determinethevalueof fromthegraph.

(b)Plotagraphofmagnificationmagainstv.(c)Usethegraphtofind:

(i)Theobjectdistancewhenm=1.0.(ii)Theimagedistancewhenm=1.0.(iii)Thefocallengthofthemirror.

(d)(i)Plotagraphofuvagainstu+v.(ii)Usethegraphtofindf,thefocallength.

Chapter6

MagneticEffectofanElectricCurrent

Professorofphysics,HansOersted,discoveredintheyear1820thataconductorcarryingcurrenthasamagneticfieldaroundit.Thishasledtotheunderstandingthataflowofchargegeneratesamagneticfield.

The magnetic effect of an electric current has been explored and it is nowproviding mankind with a wide range of devices like electric motors,loudspeakersandelectricbells.

Thischapterlooksintothemagneticfieldproducedbyelectriccurrentflow,anditsapplications.

Experiment 6.1: To investigate the magnetic effect of a current flowingthroughaconductor.

Fig.6.1:Magneticeffectofanelectriccurrent

Apparatus

Direct current power source, connectingwires, switch, rheostat, twomagneticcompassesandanammeter.

Procedure

•Setupthecircuitasshowninfigure6.1.•HoldthewireoverbutinlinewiththecompassneedleAandinlinewithbutbelowcompassneedleB.

•Closetheswitchandobservethecompassneedles.•Varythestrengthofthecurrentusingtherheostatandnotetheeffectontheneedles.

•Opentheswitchandnotetheeffectonthecompassneedles.• Repeattheexperimentwiththebatterypolaritiesreversed,i.e.,thedirectionofcurrentreversed.

Observation(i)Whentheswitchisclose,itisobservedthatthenorthpoleofthecompass

needles deflect, as shown in figures 6.2 (b) to (e), with respect to thedirectionofcurrentflow.

Fig.6.2

(ii) Theextent towhich theneedlesdeflect increaseswith thestrengthof thecurrentflowing.

(iii)Reversingthedirectionofcurrentreversesthedirectionofdeflection.(iv) The compass needles settle back into their original positions when the

switchwhichisopen,asshowninfigure6.2(a)

ExplanationA flow of current is basically a flow of charge.When charges flow they

haveanassociatedmagneticfieldwhichinteractswiththefieldofthemagneticcompass,causingdeflectionoftheneedle.

Thestrengthofthefieldincreaseswiththestrengthofthecurrentandhencegreaterdeflectionofthecompassneedle.

Amperedevisedarule,Ampere’sswimmingrule,topredictthedirectionofdeflectionofthecompassneedle.

If one imagines to be swimming along a wire in the direction of thecurrentandfacingthecompassneedle,thenthenorthpoleoftheneedlewillbedeflectedtowardstheswimmer’slefthand.

MagneticField-PatternofaStraightCurrent-carryingConductor.

Experiment6.2:Toinvestigatethemagneticfieldpatternduetoastraightconductorcarryingcurrent.

UsingIronFilings.

ApparatusIron filings a straight conductor, high current power source, four magneticcompassesandacard.

Fig.6.3:Ironfillingsshowingmagneticfieldpattern

Procedure•Makeasmallholethroughthecard.•Clampthecardinahorizontalposition.• Pass theconductor throughtheholeonthecardandconnect itsends to thepowersupplythroughaswitch.

•Sprinkleironfilingsevenlyonthecard•Closetheswitchandtapcardgently.•Drawthepatternassumedbytheironfilings.

ObservationsTheironfilingssettleinconcentriccircles,aroundtheconductor,becominglessthesignificantasthedistancefromthecentreincreases.

UsingMagneticCompasses

Fig.6.4

Procedure•Settheapparatusasshowninfigure6.4.•Placethecompasesequidistantfromthestraightconductor.•Closetheswitchandobservecompassneedles.•Repeattheexperimentwiththecompassesplacedanotherequidistancefromthestraightconductor.

Observation

Thecompassneedlesarealigned inacircle,pointing inaclockwisedirection,seefigure6.4.Whateverthedistancefromthestraightconductor.

Conclusion

The magnetic field produced by a straight conductor forms a pattern ofconcentriccirclesaroundtheconductor.

The direction of the magnetic field produced by a straight conductor can bepredictedusingMaxwell’sRight-handgripruleorMaxwell’scorkscrewrule.

Maxwell’sRight-handgrip rule fora straightconductorcarryingcurrent statesthat if a conductor carrying current is grasped in the righthandwith thethumbpointingalongthewireinthedirectionoftheconventionalcurrent,thefingerswillencircletheconductorinthedirectionofthemagneticfield.

Thisisillustratedinfigure6.5(a)and(b).

Fig.6.5(a):Right-handgriprule

Fig.6.5(b):Right-handgriprule

Maxwell’scorkscrewrulestatesthatifaright-handedscrewisdrivenforwardinthe direction of the conventional current, then the direction of rotation of thescrewisthedirectionofthefieldlines.

Fig.6.6

MagneticFieldPatternofaCircularCurrent-carryingLoop

Experiment 6.3: To investigate magnetic field pattern due to a circularcurrent-carryingloop

Apparatus

Acard,thickwire,ironfilings,magneticcompass,powersourceandaclamp.

Fig.6.7:Alignmentofironfilingsduetoaloop

Procedure•Drilltwoholesinthecardandpassthewirethroughtheholestomakealoop.•Settheapparatusasshowninfigure6.7.•Gentlysprinkletheironfilingsonthecard.• Switch on the current and tap the card gently.Note the effect on the ironfilings.

•Switchoffthecurrentandremovetheironfilings.• Switch on the current and, using the magnetic compass needle, trace themagneticfieldlines.

ObservationThepatternformedbytheironfilingsissimilartothatofasmallmagnet.Themagneticcompasstracesthefieldpattenshowninfigure6.8.

Fig6.8:Magneticfieldpatternduetoasinglecoil

Theright-handgripruleforaloopcarryingacurrentassertsthatifthefingersof the right hand encircle the current loop such that they point in thedirectionofcurrent,thethumbpointsinthedirectionofthemagneticfieldformedthroughtheinsideoftheloop,seefigure6.9.

Fig.6.9:Right-handgripruleforacurrent-carryingloop

MagneticFieldPatternofaSolenoidCarryingCurrent

Themagneticfieldaroungawirecarryingcurrentcanbeintesifiedbywindingthewireintoalongcylindricalcoilwithmanyconnectedloopsotherwisecalledsolenoid.Usinganumberofcompassesorironfilings,itcanbeshownthatthefieldsetupbyasolenoidissimilartothatofapermanentmagnet,seefigure6.10.WhenacompassneedleisplacednearendX,itsnorthpoleisrepelledindicatingthatanorthpoleiscreatedatendX.Theotherendofthesolenoidisasouthpole,seefigure 6.10. When a compass needle is placed near end X, its north pole isrepelled.

Fig.6.10:Magneticfieldpatternduetoasolenoid

When the field inside and outside the solenoid is explored, the followingpropertiesemerge:(i)Thefieldresemblesthatofamagnet.(ii) The field near the ends is non-uniform compared to the field inside the

solenoid.(iii) The field near the end of the solenoid is weaker than that inside the

solenoid.(iv) The field outside the solenoid is oppositely directed to that inside the

solenoid(iv)Thefieldoutsidethesolenoidislessthanthatinsidethesolenoid.

Thus,thesolenoidcarryingcurrentbehaveslikeabarmagnet.Itisreferredtoasan electromagnet since itsmagnetism arises from the flow of electric current.Therulecanbeusedtopredictthepolaritiesoftheelectromagnetformed.Therulestatesthatifthedirectionofcurrentinthecoilasobservedfromoneendisclockwise,thisendisthesouthpoleandifthecurrentisanticlockwise,theendbecomesthenorthpole.

Fig.6.11:Clockrule

The right-hand grip rule can also be used to predict the north pole of theelectromagnetasfollows;

Ifacoilcarryingcurrentisheldintherighthandsuchthatthefingersencircle the loops while pointing in the direction of the current flow, thethumbpointsinthedirectionofthenorthpole.Seefigure6.12.

Fig.6.12:Righthandgripruleforacoilcarryingcurrent

Practical electromagnets require that the coilsbewoundona soft iron core toincreasemagneticpower,

Thefactorsaffectingthestrengthofanelectromagnetinclude:(i)thesizeofcurrentinthesolenoid.(ii)thenumberofturnsofwireinthesolenoid.(iii)theshapeofthecore.(iv)thelengthofthesolenoid.

Theexperimentsoutlinedbelowexplorehowthesizeofthecurrent,thenumberofturnsandtheshapeofthecoreaffectthestrengthoftheelectromagnets.

Experiment 6.4: To investigate how the size of current flowing in thesolenoidaffectsthestrengthofanelectromagnet

Apparatus

Longinsulatedcopperwire,rheostat,ammeter,powersource,preferrablya6.0Vlead-acidaccumulator,switch,connectingwires,adishcontainingpaperpins,balanceandironrod.

Procedure•Wind20turnsofinsulatedcopperwiretightlyroundtheironrod.•Connectittothebatteryasshowninfigure6.13.•Switchonthecurrentandadjusttherheostattogiveacurrentof0.5A.Findwhatmassofthepaperpinstheelectomagnetcansupport.

•Repeattheexperimentwiththecurrentof1.5Aand2.0A.

Fig.6.13:Effectofsizeofcurentonthestrengthofelectromagnet

Observation

As the current is increased, the mass of the pins supported (representing thestrengthofthemagnet)increases.Beyondaparticularvalueofcurrent,however,thestrengthoftheelectomagnetremainsconstant.

Experiment 6.5: To investigate how the strength of an electromagnet isaffectedbythenumberofturns

Apparatus

Ironcore,powersource,ammeter,rheostat,steelpins,balance,switchandlonginsulatedwire.Procedure•Windfiveturnsofthecoilontheironcore.•Connectthecoiltothebattery,asshowninfigure6.14.•Setthecurrenttoaconstantvalue,say1.5A.

Fig.6.14:Effectofnumberofturnsonstrengthofmagnet

•Recordthemaximummassoftheclipssupportedbytheelectromagnet.• Now increase thenumberof turns in stepsof fiveand ineachcase set thecurrent to the same value as before (1.5 A) with the turns occupying thelengthL.

• Sketch the graphof themass of the clips supported against the number ofturns.

Observation

Increasing the number of turns increases the maximum mass of the pinssupported.Ingeneral,themagneticstrengthofanelectromagnet:(i)increaseswithincreaseincurrent.(ii)increaseswithincreaseinthenumberofturnsofthesolenoid.(iii)decreaseswiththelengthofthesolenoid.

OtherexperimentsshowthattheU-shapedcoreproducesastrongermagnetthanthatofastraightcore.

Experiment 6.6: To investigate how the shape of the core affects thestrengthofanelectromagnet

Apparatus

U-shaped iron core, straight iron core, light spring with a pointer, iron bar,insulatedcopperwire,ammeter,metrerule,currentsourceandrheostat.

Procedure•Wind20turnsofwireontheU-shapedcore.• Set the apparatus as shown in figure 6.13.Note the position of the springpointerwhentheswitchisopen.

Fig.6.13:Magneticpowerdependsontheshapeofthecore

•Switchonthecircuitandadjustthecurrentto0.5A.Recordtheextensionofthespring.

•Repeattheexperimentwithcurrents1.0A,1.5Aand2.0Aflowingthroughthecoil.

• Replace theU-shaped ironcorewith thestraight ironcorehaving thesamenumberofturnswoundonitandrepeattheexperiment.

•Onthesameaxes,plotgraphsofextensionagainstcurrentforU-shapedandstraightironcore.

Observation

Figure6.14showspossiblegraphresultingfromtheexperiment.Theextensionproducedrepresentsthemagneticforceoftheelectromagnet.

The U-shaped core produces more extension for a given current that thestraightcoresinceitattractstheironbarwithtwopoles.Theironbarprovidesapath to complete the magnetic field lines, making the U-shaped core more

efficient than the straight iron core that uses only one pole for the attraction.WhenwindingonaU-shapedcore,caremustbe taken todo is in suchawaythatoneendof thecoreproducesanorthpolewhile theotherendproducesasouthpole.

Fig.6.14:ComparisonofU-shapedandstraightcoreelectromagnet

Ingeneral,themagneticstrengthofanelectromagnet:(i)increaseswithincreaseincurrent.(ii)increaseswiththeincreaseinthenumberofturnsofthesolenoid.(iii)decreaseswiththelengthofthesolenoid.(iv)dependsontheshapeofthecore.

ForceonCurrent-carryingConductorinaMagneticField

The effect of a magnetic field on a current-carrying conductor can bedemonstratedintheexperimentbelow.

Experiment 6.7: To investigate how magnetic field affects a conductorcarryingcurrent

Apparatus

U-shapedmagnet, two straight brass rods, insulating support, rheostat, switchandbattery.

Fig.6.15:Conductorinamagneticfield

Procedure•MounttwostiffstraightbrassrodsXandYparalleltoeachotheronaplastic(insulator)supportasshowninfigure6.15(a).

•Clampthesupportsothatthebrassrodsareonthesamehorizontalplane.•PlaceanotherbrassrodABacrossXandY.• Postitionastronghorse-shoemagnetsothatbeamABisbetweenthepolesandperpendiculartothemagneticfield.

•ConnectabatteryandrheostatinserieswithrodsXandY.•SwitchonthecurrentandobservethebehaviouroftherodAB.•ReversethedirectionofthecurrentandnotethebehaviourchangeofrodAB.• Reverse thedirectionof themagnetic fieldso that thenorthpole isupandobservethebehaviouroftherodwhenthecurrentflowsinit.

•Increasethestrengthofthecurrentandrepeattheexperiment.•NowturnthemagnetsothatthefieldisparalleltothelengthoftherodAB,asinfigure6.15(b).

•Observewhathappenswhenthecurrentflowsintherod.

Observation(i)WhenthecurrentflowsalongAB,therodrollsalongthebrassrodsXand

Ytowardstheplasticsupport.(ii) When either the direction of the current or that of themagnetic field is

reversed,thedirectionofmovementofABalsochanges.

(iii)Whencurrentisincreased,therodmovesfaster.(iv) When themagnet is turned so that themagnetic field is parallel to the

lengthAB,therodremainsstationary.Aforceactsonacurrent-carryingconductorwhenitisplacedinamagnetic

field. The magnitude of this force increases with increase in current andtherefore field strength. The force is maximum when the angle between theconductorandthefieldis90°.Theforceisminimum(zero)whentheconductorisparalleltothemagneticfield.

It can also be shown that the force on the conductor can be increased byincreasingthelengthoftheconductorinthemagneticfield.

Explanation

Figure6.16(a)showsthefieldaroundtheconductorABasviewedfromA.Thisfieldinteractswiththefieldduetopermanentmagnetshowninfigure6.16(b)toform the field pattern shown in figure 6.16 (c). Note that the field tends toconcentratemoreononesideoftheconductorthantheother.Theweakfieldontheoppositesideisaresultofthetwofieldsopposingeachother.

Considerthemagneticlinesofforceactinglikeplasticbands,concentrationof the lines on the other side of the conductor produces a catapult effect thatpushes theconductor in the opposite direction to rid the plastic of the strain.Whenthedirectionofthecurrentormagneticfieldisreversed,thedirectionoftheforceontheconductoralsoreverses.

Fig.6.16:Forceonaconductorcarryingcurrentinamagneticfield

For a conductor carrying current in amagnetic field, the direction of theforce acting on it can be predicted using the Fleming’s left-hand rule (motorrule),statedbelow;

Ifthelefthandisheldwiththethumb,thefirstfingerandthesecondfingers mutually at right angles so that the first finger points in thedirectionofthemagneticfieldandthesecondfingerinthedirectionofthecurrent,thenthethumbpointsinthedirectionofmotion,seefigure6.17.

Thedirectionofcurrentinthisruleistheconventionaldirection.Itshouldbe noted that the rule applies only if the magnetic field and current areperpendicular to each other. When the field and current are parallel to each

other,thereisnoforceontheconductor.

Fig.6.17:Fleming’sleft-handrule

ForceonaCurrent-carryingCoilinaMagneticField

Figure6.18 (a) showsa rectangularcoilABCDput inamagnetic field.Whenthe current flows through the coil in the direction DCBA, the resultant fieldpatternisshowninfigure6.18(b).Thecatapult forceactingon thesidesof thecoilcauses it to turn inclockwisedirection.ApplicationofFleming’sleft-handrulemakesiteasier topredict thedirectionofmotionthandrawingthefieldpattern.

Fig.6.18:Magneticfieldpatternofacoilinamagneticfield

ForceonaChargedParticleMovinginaMagneticField

Earlier in this chapter, it was stated that moving charges produce a magneticfield.Anelectronmoving throughamagnetic fieldwill thereforeexperienceaforce,seefigure6.19(a).

Fig.6.19(a):Electronmovementinamagneticfield

Fig.6.19(b):Electronmovementinamagneticfield

Consideringthatthedirectionofmovementofanelectronisoppositetotheflowofcoventionalcurrent,thedirectionoftheforceonthechargecanbepredictedusingFleming’sleft-handruletobedownward,seefigure6.19(b).

ForceBetweenParallelConductorsCarryingCurrent

Experiment 6.8: To investigate the force experienced by two parallelconductorscarryingcurrent

Apparatus

Two stands with clamps, two aluminium foils 30 cm each, connecting wires,powersourceandarheostat.

Procedure•Settheapparatusasshowninfigure6.20(a),sothatcurrentflowsinthesamedirection.

•Closetheswitch.Observewhathappenstothemetalfoils.

•Increasethemagnitudeofcurrentflowinginthecircuitandobservetheeffectonthemetalfoils.

•Repeatwiththecurrentflowingintheoppositedirection,asshowninfigure6.20(b).

Fig.6.20

Observation

The aluminium foil strips attract each otherwhen the current flowing throughthemisinthesamedirectionandrepelwhenthecurrentisflowinginoppositedirection.

Explanation

Fig.6.21showsthemagneticfieldpatternfor thestripscarryingcurrent in thesamedirection.Thefieldbetweenthestripscanceleachother,leavingaregionof zero resultantmagnetic field called the neutral point (X). The field on theoutside part of each strip generates catapult force that pushes the two strips

towardseachother.Infigure6.21(a),thefieldsbetweenthestripscanceleachother,leavinga

regionofzeroresultantmagneticfieldcalledtheneutralpoint(X).Thefieldonthe outside part of each strip act as catapult forces, pushing the two stripstowardseachother.

Fig.6.21:Magneticfieldpatternduetoparallelconductors

Infigure6.21(b),showthepatternforcurrentflowinginoppositedirection,themagneticfieldsbetweenthetwostripscomeveryclosetoeachother.Sincefieldlines mutually repel one another, the two strips are pushed away from eachother.Fleming’sleft-handrulecanbeappliedtothetwosituations,asshowninfigure6.22.

Fig.6.22:Fleming’sleft-handruletoparallelconductors

If stripA is taken to produce the field, Fleming’s left-hand rule applied at BindicatesthattheforceonBistowardsA.IfstripBistakentoproducethefield,theruleindicatesthattheforceonAistowardsB.

Thesametreatmentcanbeappliedtothesituationwherethecurrentsflowinoppositedirectionstoconfirmtherepulsionbetweenthestrips.

MagneticFieldPatternofaConductorCarryingCurrentintheEarth’sMagneticField

Theearth’smagneticfieldlinesaretakentobeparallel,exceptatthepoles.Theinteractionof theearth is fieldand the fielddue to theconductorproduces thepatternshowninfigure6.23.TheconductorthusexperiencesaforceFshown.

Fig.6.23

UsesofElectromagnets

Electromagnets are used in various domestic instruments or devices, industrialandeveninmedicine.Someofthecommonapplicationsareillustratedbelow.

ElectricBellAnelectricbellconsistsofaU-shapedelectromagnetwhosewindingononearmisoppositetothatontheother,seefigure6.24.Acontactscrewpressesontoa

softironstrip,whichactsasaspring.TheelectriccircuitiscompletedthroughabatteryandaswitchS.

Fig.6.24:Electricbell

WhenS is closed, current flows through the circuit and the corebecomesmagnetised. The electromagnet induces magnetism in the soft iron strip(armature), which is then attracted to the poles of the electromagnet. Thehammerattachedtothearmaturethusstrikesthegong.

Theattractionofthesoftironarmatureseparatesthecontacts,breakingthecircuit.Themagnetismin thecore thereforediesoffand thespringreturns thearmature to its original position. Contact is made again and the process isrepeated. So long as the switch is closed, the hammer strikes the gongrepeatedly,makingcontinousringingsound.Thesteelspringandscrewcontactwhere the current is automatically switchedon andoff constitute amake-and-breakmechanism. The frequency at which themake and break takes place iscontrolledbythescrewwiththeadventofelectricsounderssuchbellsarebeingreplaced.

TelephoneReceiver(Earpiece)

The telephone receiver has a U-shaped magnet formed by placing a shortpermanentbarmagnetacrosstheendsoftwosoftironbars.Thearrangementismade in such a way that a steady current flowing through the coils of theelectromagnetcauses it toexertapullon thespringymagnetalloydiaphragm,seefigure6.25.

Fig.6.25:Telephonereceiver

When a person speaks into themicrophone at the other endof the line, avarying electric current representing the speech is set up. When this electriccurrentiscausedtopassthroughthesolenoidintheearpiece.Thestrengthofthemagneticfieldinthediaphragmisalteredcorrespondingly.Thediaphragmthusvibratesandreproducesthesoundmadebythepersonthroughthemicrophone.

ElectromagneticRelay

Thisisanelectricallyoperatedswitchthatmakesuseofanelectromagnet.Itisasystemwherebysmallcurrentflowinginonecircuitproducesmechanicaleffectthatcontrolsotherusuallyheavycurrentcircuit(s).

AsimplerelayconsistsofanelectromagnetEandasoftironarmatureleverApivotedatO,seefigure6.26.Ithastwocompletelyseparatecircuits,PandQ.When thecircuitQ iscompleted,asmallcurrent flows through thesolenoidEwhichactsasanelectromagnetandinturnattactsthesoftironarmatureA.ThisclosesthecontactincircuitPbypushingthespringmetalstrip.WhencurrentincircuitQisswitchedoff,ElosesitsmagnetismandAgoesbacktoitsoriginalposition,thusswitchingoffthecurrentincircuitP.Thesamearmaturelevercanoperateseveralsetsofcontacts.

Thus,inamagneticrelay,currentflowinginonecircuitisusedtoactivateanothercircuitwithoutanydirectelectricalconnectionbetweenthetwocircuits.Incaseswherearelay isused tocontrolcircuitscarryinghighcurrents,onlyasmallcurrentisneededinthesolenoidtoenergisesthesoftironbarthatswitcheson a high voltage or high current supply. Although relays have wide uses intelephone circuits, controlling power supply to sockets and lights in big

industries,theyarebeingreplacedbythemoreefficientelectronicsystems.

Fig.6.26:Magneticrelay

Moving-CoilLoudspeakerThemoving-coilloudspeakerisusedtoconvertelectricalsignalsfromanaudioamplifiertosoundinradios,televisionsetsandotheraudiosystems.Itworksbythe fact that a current-carrying conductor experiences a force when put in astrongmagneticfield.Figure6.27showsthemainfeaturesofaloudspeaker.

Fig.6.27:Moving-coilloudspeaker

Thepowerfulmagnetshownproducesaradialmagneticfield.Aspeechcoilwound on a cylindrical former and positioned in the narrow gap between thepolesofamagnetisfreetomovebackandforthinthemagneticfield.

Avaryingelectriccurrentwhosefrequequencycorrespondstothesoundtobe produced is passed through the speech coil. Since the radial field of themagnetcutstheturnatrightangles,amaximumforceactsonthecoil,movingitinaccordancewithFleming’sleft-handrule.

Sincethecurrentinthespeechcoilisvarying,thecoilexperiencesaforcevarying in magnitude and direction and at the frequency of the speech. The

diaphragm (cone) attached to the coil vibrates at the same frequency as thecurrent in the speech coil vibrates at the same frequency as the current in thespeechcoil.Thelargemassofairincontactwiththediaphragmisthussetintovibration,reproducingthesound.Figure6.28(a)and(b)showthecross-sectionofthemovingcoilloudspeakerandtheradialfieldproducedbythemagnet.

Fig.6.28:Cross-sectionandmagnetofamovingcoilloudspeaker

Moving-CoilMeterFigure6.29 showsamovingcoilmeter.The instrument is adapted tomeasurethestrengthofcurrentfloworvoltage.Ithasthefollowingcomponents:(i)Acoilofinsulatedcopperwirewoundonsoftironcore.(ii)Twohair-springs(iii)Twojewelledpivots(iv)Curvedmagneticpoles(v)Pointerandscale.

Fig.6.29:Amovingcoilmeter

When an electric current flows through the coil via the hair-springs, themagnetic fiels set up by the coil interactswith the field due to the permanentmagnet, producing a force. This force is experienced by the sides of the coil,causingittoturnaboutthelow-frictionjewelledpivots.Thepolesofthemagnetare curved and togetherwith the soft iron core produce an intense radial fieldthatcuts thecoilat rightangleswhatever thepositionof thecoil.Thisensuresthatamaximumforcethatisdirectlyproportionaltothecurrentactsonthecoil.

As thecoil turns, thespringsarewoundup,producingamechanical forcethatopposestheturningofthecoil.Theturningstopswhentherestoringforcebythespringsisequaltotheforceduetothecurrent.Thespringsunwindwhenthecurrentstopsflowinginthecoil,returningthecoiltoitsoriginalposition.

Figure 6.30 shows the coil with respect to the radial field generated by themagnet.

Fig.6.30:Coilinaradialfield

It is worth noting that moving-coil meters are being replaced by moreaccurateelectronicones.

CircuitBreakersIn modern domestic wiring, circuit breakers, and not fuses, often protectelectricalcomponentsfromexcessiveflowofcurrent,seefigure6.31.

Whenexcesscurrentflowsthroughthecircuit,increasedmagneticpowerofthe electromagnet opens the switch, stopping electric current flow. Once theproblem causing the excessive current flow has been corrected, the switch isclosedbymechanicalmeans.

Fig.6.31:Circuitbreaker

MagneticTapePick-up(Head)

Themagneticpick-upofa tape recorderhasacoilwoundona soft ironcore.Themusic or speech to be recorded is converted to a corresponding electricalcurrent, which flows through the coils of the electromagnet producing amagneticfieldofvaryingpower.Amagneticaudiomovingpastthegaphasitsmagnetic coating magnetised corresponding by the intense field of theelectromagnet,seefigure6.32.

Fig.6.32:Magnetictapepick-up

ElectricMotorAnelectricmotorisadevicethatconvertselectricalenergytorotationskineticenergy.AsimpledirectcurrentelectricmotorconsistsofacoilofinsulatedwireABCD,whichcanturnaboutafixedaxis,withinmagneticfieldprovidedbyastrongcurvedpermanentmagnet,seefigure6.33.

Fig.6.33:Electricmotor

current enters and leaves the coil through a split copper ring P and Q calledcommutator.Thetwohalvesareinsulatedfromeachotherbrushespresslightlyagainstthecommutatorandareconnectedbybatteryterminals.

Suppose the coil is in the horizontal position shown. When current isswitchedon,itflowsthroughthecoilinthedirectionshown.ByFleming’sleft-handrule,sideABofthecoilexperiencesanupwardforceandCDadownwardforce.Sincethecurrentinbothsidesissame,theforcesareequalandopposite.These forces cause the coil to rotate in clockwise direction until it reaches itsverticalpositionwithsideABupandCDdown.

Inthisposition,thebrushestouchthespacebetweenthetwohalvesofthesplitrings,cuttingoffcurrentflowinginthecoil.Consenquently,noforceactson thesidesABandCD.Since thecoil is in rotation, itsmomentumcarries itpast this position and the two split rings exchange brushes. The direction ofcurrentthroughthecoilisreversedandconsenquentlythedirectionofforceoneach sideof the coil changes.This process is called commutation.SideAB isnow on the right hand side and side CD is on the left hand side. Side ABexperiences a downward force and sideCDan upward force.The coilABCDwillcontinuerotatingintheclockwisedirectionsolongasthecurrentisflowingthrough it.The speed of rotation of the coil increaseswith the increase in thestrengthofthecurrentflowingthroughthecoil.Iftheterminalsofthebatteryareinter-changed, thedirectionof current reverses and thedirectionof rotationofthecoilisalsoreversed.

SidesADandBCdonotexperienceanyforcebecausecurrentinthesesides

isparalleltothedirectionofthemagneticfield.

The simple electric d.c.motor described above is not powerful. It can beimprovedby:(i)windingthecoilonasoftironcore.Thesoftironcorebecomesmagnetised

andconcentrates itsmagnetic field in thecoil.This increases the forceonthecoil.

(ii)increasingthenumberofturnsoftherotatingcoil.(iii)usingastrongermagnet.(iv)multiplyingthenumberofcoilsandcommutatorsegments.

Inpracticalmotors,thepermanentmagnetisreplacedbyanelectromagnet.

RevisionExercise6

1.Drawamagneticfieldpatternaroundacircularcoil.2.Drawmagneticfieldpatternsaroundastraight,current-carryingconductor.3.Stateandexplaintheeffectonanelectromagnetof:

(a)removingthecore.(b)replacingtheironcorewithasteelcore.

4.(a)Drawalabelleddiagramofasimpled.c.electricmotor.(b)Stateandexplainthreewaysinwhichtheforcesonthecoil,andhence

thespeedofthemotor,canbeincreased.5.Asmallelectromagnet,usedforliftingandthenreleasingasmallsteelball,

ismadeinthelaboratoryasshowninthefigurebelow.

(a)Explainwhysoftironisabettermaterialthansteeltouseforthecore.(b)Inordertoliftaslightlylargerball,itisnecessarytomakeastronger

electromagnet. State two ways in which the electromagnet could bemadepowerful.

6. ThefigurebelowshowsarectangulaplanecoilABCDofseveralturnsofwirelocatedinamagneticfieldduetotwopolesnorthandsouth.ThecoilisfreetorotateontheverticalaxisXY.

Whena current is passed through the coil in thedirectionABCD the coilstarts to turn, and eventually comes to rest. With the aid of a diagram,explain:(a)whythecoilbeginstoturn.(b)inwhichdirectionitbeginstoturn.(c)whyitcomestorest.(d)thepositioninwhichitcomestorest.

Chapter7

Hooke’sLaw

Amaterial isselectedforaparticularusedependingon itsability towithstandthe forces it may be subjected to. The following characteristics are used todescribematerials:

Strength

This is the ability of a material to resist breakage when under a stretching,compressingorshearingforce.Astrongmaterial isonewhichcanwithstandalargeforcewithoutbreaking.

Stiffness

Thisistheresistanceamaterialofferstoforceswhichtendtochangeitsshapeorsize,orboth.Stiffmaterialsarenotflexibleandresistbending.

Ductility

This is the quality of amaterialwhich leads to permanent change of size andshape. Materials which elongate considerably under stretching forces andundergoplasticdeformationuntiltheybreakareknownasductilematerials.eglead,copper,wroughtironandplasticine.

Ductilematerialscanberolledintosheets,drawnintowiresorworkedintootheruseful shapeswithout breaking. They are used inmaking such implements asstaples,rivetsandpaperclips.

Brittleness

This is the quality of amaterialwhich leads to breakage just after the elasticlimitisreached.Brittlematerialsarefragileanddonotundergoanynoticeableextensiononstretchingbutsnapsuddenlywithoutwarning.

Blackchalk,bricks,castironboard,glassanddrybiscuitsareexamplesofbrittlematerials.

Elasticity

Elasticity

This is the abilityof amaterial to recover its original shape and size after theforcecausingdeformationisremoved.Amaterialwhichdoesnotrecoverbutisdeformedpermanently,likeplasticine,issaidtobeelastic.

StretchingofMaterialsTheforcesbetweenthemoleculesinasolidaccountforitscharacteristicelasticor stretching properties. When a solid is stretched, the spaces between themolecules increase slightly. The tension felt in a stretched rubber band, forexample,isduetoalltheforcesofattractionbetweenthemoleculesinit.

Experiment7.1:Toinvestigatethestretchingofaspiralspring

Apparatus

A spiral spring with pointer attached, a metre rule, retort stand, two sets ofclampsandbosses,20gmasses.

Fig.7.1Stretchingaspring

Procedure•Arrangetheapparatusasshowninfigure7.1.Notethepositionofthepointerwhenthespringisunstretched,ornotloaded.

• Suspend amass at the end of the spring and note the new position of thepointer.

• Increase the load in stepsof20gand record the readingof thepointer for

each load in table 7.1 (care should ne taken not to use too heavy weightswhichwouldoverstretchthespring).

•Unloadthespringinstepsof20gandagainrecordthepointerreadings.•Plotagraphofstretchingforce(F)againstextension,e,

Observation

Providedweightsusedarenottooheavy,thespringalwaysreturnstoitsoriginallengthonunloading.Theratioofstretchingforcetoextensionisconstant.

ThegraphofstretchingforceFagainstextensioneisastraightlinethroughtheorigin,seefigure7.2.

Table7.1

Fig.7.2Graphofstretchingforceagainstextensionforaspring

Conclusion

Theextension,e,ofaspringisdirectlyproportionaltothestretchingforceF,i.e,ifthestretchingforceisdoubled,theextensionisalsodoubled.

Thesamekindofresultisobtainedifastraightsteelwireisstretched

Ifthestretchingforceisincreasedbeyondacertainvalue,permanentstretchingoccurs.Thegraphof extensionagainst stretching force is shown in figure7.3.

OP represents the permanent stretching or permanent extension of the spring.PointEiscalledtheelasticlimitofthespring.Beyondthispoint,asinpointB,furtherextensioncausespermanentextension.

Fig.7.3:Asteelspringstretchedbeyonditselasticlimit

Hooke’sLawRobertHookeinvestigatedandformulatedalawrelatingthestretchingforceandextension.Thelawstatesthatforahelicalspringorotherelasticmaterial,theextension is directly proportional to the stretching force, provided theelasticlimitisnotexceeded.Therelationshipcanbeexpressedas;Feintermsofforce(F)andextensioneHence,F=ke,wherek is the constant ofproportionalitywhichdependson thematerialofthespring.Theconstantisreferredtoasthespringconstant.

Thisisthespringconstant,whoseunitsaregivenasNm1orN/m.

Fig.7.4:Thespringconstant

The area under a force versus extension graph is equal to the work done instretchingthespring,seefigure7.5.

Fig.7.5:Workdoneinstretchingaspring.

Areaunderthegraph= Fe,whereFistheforceappliedandetheextensionattained

ButF=ke(wherekisthespringconstant)

Hence,workdone= ke2

Example1Amass of 100 g is suspended from the lower end of a spring. If the springextendsby100mmandtheelastic limitof thespring isnotexceeded,what isthespringconstant?

Solution

Sincetheelasticlimitisnotexceeded,Hooke’slawapplies,i.e.,

F=ke.

Hence,k=

ForceF=100×10–3kg×10Nkg-1

=1N

Extension,e=100×10-3m

Therefore,springconstant

Example2Ametalcubesuspendedfreelyfromtheendofaspringcauses it tostretchby5.0cm.A500gmasssuspendedfromthesamespringstretchesitby2.0.Iftheelasticlimitisnotexceeded:(a)Findtheweightofthemetalcube.(b)Bywhatlengthwillthespringstretchifamassof1.5kgisattachedtoits

end?

Solution(a)SincethespringobeysHooke’sLaw;

F=keSo,K=

ButF=mg

Sincethesamespringisused:weightofthecube=250×5×10-2

=12.5N(b)Lettheextensionproducedbythe1.5kgmassbee.

Force=1.5×10=15NFromF=ke

Example3Aspiralspringisfittedwithascalepanasshowninfigure7.6.Thepointerisatthe 40 cmmark on the scale.When a packet of salt is placed in the pan, thepointermoves to the 20 cmmark.When a 20 gmass is placed on top of thepacketofsalt,thepointerindicates10cm.

Fig.7.6

Findthe:(a)extensionproducedbythepacketofsalt.(b)extensionproducedbythe20gmass.(c)themassofthesalt.

Solution(a) Assuming that the spring does not exceed its elastic limit; extension

producedbythepacketofsalt=40cm–20cm=20cm=0.2m

(b)Extensionproducedbythe20gmass=20cm–10cm=10cm=0.1m

(c)Usingtheextensionproducedbythe20gmass;

Example4TwoverylightidenticalspringsPandQarearrangedasshowninfigure7.7(a)and(b)

Aweightof4.8Nissupportedbyeacharrangement.

Fig.7.7:

Given that each spring has a spring constant of 10N/cm, determine the totalextensionforeacharrangement.

Solution

SincethespringsPandQareidentical,

Forfigure7.7(a)bothspringsPandQexperiencealoadof4.8Nforce.

Thus,extensionineachspringis

Totalextension=2×0.48=0.96cm

Infigure7.7.(b),thetwospringssharetheload

orForceperspringis

Example5Thelengthofaspringis160cm.itslengthbecomes20cmwhensupportingaweightof5.0N.Calculatethelengthofthespringwhensupportingaweightof:(a)2.5N(b)6.0N(c)200N

Commentontheanswertopart(c).

Solution

Lengthofloadedspring

=extension+lengthofunstretchedspring.

Extension

=lengthofloadedspring–lengthoftheunstretchedspring.

5.0Nproducesanextensionof20–16=4cm

Therefore,1.0Nproducesanextensionof =0.8cm

(a)2.5Nproducesanextensionof2.5×0.8=2.0cm.Therefore,lengthofspring=2.0+16=18cm

(b)6.0Nproducesanextensionof6.0×0.8=4.8cmLengthofthespring=4.8+16=20.8cm

(c)200Nproducesanextensionof200×0.8=160cm.Lengthofspring=160+16=176cm

Inpart (c) theextension is too largeand thespringwouldget straightenedoutandpossiblysnap.

Exercise7.1

1.Aspringstretchesby8.00mmwhensupportingaloadof2.0N.(a)Byhowmuchwoulditstretchwhensupportingaloadof5.0N?(b)Whatloadwouldmakethespringextendby2.5cm?

2.Asinglespringextendsby3.6cmwhensupportingaloadof2.0kg.Whatistheextensionineachofthearrangementsshownbelow?Assumethatallthespringsareidenticalandofneglegibleweight.

CompressingaSpring

When the two ends of a spring are squeezed together, it shortens. There ischangeinlengththatisreferredtoascompression.Thespringonitspartexertsacounterforcewhichresiststhecompression.

Figure 7.8 shows the variation of length against compression of a springwhichobeysHooke’sLaw.BeyondthepointE, the turnsofcoilsarevirtuallypressing onto one another and further increase in the force achieves nonoticeabledecreaseinlength.

Someoftheapplicationsofcompressedspringaretop-panbalance,springshockabsorbersinmotorvehicles.

Fig.7.8Compressingaspring

Fig.7.9:Toppanbalance

Fig.7.10:Shockabsorber

Hooke’sLawAppliedtoLoadingofBeamsExperiment7.2:Toinvestigatethesaggingofaloadedbeam

Apparatus

Woodenbeamoflengthapproximately50cmwithapointer,six100gmasses,G-clamp,metrerule,retortstand,string(about30cmlong),weightholder.

Fig.7.11:loadingabeam

Procedure•Arrangetheapparatusasshowninfigure7.10.•Recordthepointerreadingwhennoloadissuspendedonthebeam,inTable7.2.

•Suspendone100gmassandrecordthenewpointerreading.• Continue adding the load in 100 g steps, each time recording the pointerposition.Ensurethatthebeamisnotoverloaded.

•Determinetheamountofsagging,x,andfillinthetable.•PlotagraphofforceFagainstamountofsaggingx.

Table7.2

ObservationA graph of load against amount of sagging, x, is a straight line through theorigin.ThisshowsthatthesaggingisinaccordancewithHooke’slaw.

Some materials regain their original shapes after being stretched, eventhough they do not extend according to Hooke’s law. Rubber is one suchmaterial.Figure7.9showsthevariationofforcewithextensionforrubberandmaterialswithsimilarbehaviour.

Initially, rubber stretches by a large amount for a small increase instretchingforcebutbeyondacertainpoint,tendstostiffenupshowingverylittleextensionwithincreaseinforce.Whenthestretchingforceiswithdrawn,thereisnopermanentextension.Thisshowsthatthematerialiselastic.

Bristlematerialslikeconcreteandglassexhibitelasticitybutsuddenlysnapwithout becoming plastic. Materials like polythene and metal wires displayelasticity,butgothroughplasticitybeforesnapping.

Fig.7.12:Stretchingrubber

MakingaSimpleSpringBalance

Apparatus

String, firm cardboard cylinder or cuboid, strong helical spring with pointerattached, wooden strip with hook, wooden rod, 100 g weights, strip of graphpaper,smallwoodenblocksorstones.

Procedure•Preparethecuboidsothatthespringwiththepointerandwoodenstripwiththehookhandfreely,asshowninthefigure7.12.

•Fixthestripofgraphpaperononesidealongtheslot.Thespringbalanceisnowreadyforcalibration.

•Withnomassonthehook,markthepointerpositionasthezeroonthescale.•Suspenda100gmassonthehookandmarkthenewpointerpositionas1N.• Continue adding the masses in steps of 100 g while marking thecorrespondingpointerpositions.

Thisisyourspringbalance

Fig.7.13:ASimplespringbalance

ActivityDeterminetheweightsofobjectsprovidedtothenearestonenewton.

Revisionexercise7

1 (a) Youaregivena scalewhich isgraduated innewtons.Youarealsoprovidedwithmassesof20g,50g,100g,200gandaspringbalancecalibratedingrams,explainhowyouwouldcalibratethespringbalanceinnewtons.

(b)Aspringstretchesby1.2cmwhen600gmassissuspendedonitWhatisthespringconstant?

2.Thefigurebelowshowsaspringwhenunloaded,whensupportingamassof5g and when supporting a stone. Study the diagrams and use them todeterminethemassofthestone.

3. Thefollowingreadingswereobtainedinanexperiment toverifyHooke’sLawusingaspring.

(a)Foreachreading,calculate:(i)Thevalueofforceapplied.(ii)Theextensioninmm.

(b)Plotagraphofextensionagainstforce.DoesthespringobeyHooke’slaw?

(c)Fromthegraphdeterminethe:(i)elasticlimit.(ii)springconstant.(iii) weight of a bottle of ink hung from the spring, if the reading

obtainedis12cm.(iv)extensioninmillimetres,whenaforceof0.3Nisapplied.(v)scalereading,incentimetresforamassof0.02kg

4.Apieceofwireoflength12misstretchedthrough2.5cmbyamassof5kg.AssumingthatthewireobeysHooke’slaw:(a)Throughwhatlengthwillamassof12.5kgstretchit?(b)Whatforcewillstretchitthrough4.0cm.

5.Thefigurebelowshowsthevariationofforcewithextensionforasteelcoilspring.Inthesamediagram,sketchthevariationofforcewithextensionforawire fromwhich the spring ismade.Explain thedifferencebetween thesketches.

6.Thedatabelowrepresentsthetotallengthofaspringastheloadsuspendedonitisincreased:

(a)Plotagraphoftotallength(y-axis)againstweight:(b)Usethegraphtodeterminethe.

(i)lengthofthespring.(ii)springconstant.

7.Thefigurebelowshowsasimpleapparatusforstudyingthebehaviourofaspringwhensubjectedtoforcesofcompression.

(a) Describe how the apparatus may be used to obtain readings ofcompressionforceandcorrespondinglengthofspring.

(b)Inasimilarexperiment,thefollowingreadingswereobtained:

Plotagraphof:(i)Compressionforceversuslengthofthespring.(ii)Compressionforceversuscompressionofspring.Fromthegraph

plotted, determine the minimum force that will make the springcoilstojustcomeintocontact.

8.Thefigurebelowillustratessystemsofidenticalsprings.Equalmassesaresuspendedonthespringstostudythevariationofextensionwithforce.(a) On the same axes, sketch the variation of extension with stretching

forceforeachofthesystems.(b)Explainthedifferencesbetweenthesketches.

9. A60gmass is suspended froma spring.When15gmore is added, thespringincreases1.2cm.GiventhatthespringobeysHooke’sLaw,find:(a)Springconstant.(b)Totalextensionofthespring.

Chapter8

Waves

A pebble thrown into a pond sends out ripples in the water. The disturbancespreadsoutonthewatersurfaceinaformofconcentriccirclesaroundthepointoforigin.Atoyboatinthepathofthedisturbancewillbeobservedtobobonthewaterastheripples(waves)moveoutwards.Theboatishowevernotmovedtothe side to the pond. The disturbance sets upwaves in thewater,which onlytransfertheenergywithoutdraggingthewaterfromoneendtotheother.

Waveshavemanyuses indaily life.Radio televisiontransmission,mobilecommunication, remote control systems and heat energy radiation are allapplicationsofwaves.

A view of thewave profile forwater andwaves appears as illustrated infigure8.1.Thewatersurfaceisdistortedintheformofcrestsandtroughs.

Fig.8.1:Wavesinwater

Awavethereforeisthetransmissionofadisturbance.Wavescanbeclassifiedaselectromagneticinnatureormechanical.

Electromagneticwaves

Electromagnetic waves do not require material medium for transmission, e.g,radiowaves,radiantheat,lightandmicrowaves.

MechanicalWaves

MechanicalWaves

Mechanicalwavesrequirematerialmediumfortransmission.Thistransmissioniseffectedbythevibrationof theparticles in themedium.Examplesarewaterwaves and sound waves. Mechanical waves can be either transverse orlongitudinal.

TransverseWaves

In transverse waves, the vibration of the particles is at right angles to thedirection ofwave travel.Waterwaves,waves on a string and electromagneticwaves(light,radio,microwaves,etc)areexamplesoftransversewaves.

To illustrate the formation of transversewaves, a slinky spring or a ropemaybeused.Thespringorropeisstretchedalongasmoothfloororbenchtop.Oneend isattached toa rigid supportwhile theotherend isheld in thehand.Theendheldinthehandisswungupanddownatrightanglestothespringorrope,asinfigure8.2.Suchawavetravelsasaseriesofcrestsandtroughs.

Fig.8.2Transversewavesonaslinkyspring.

Figure 8.3 shows the displacement of an individual particle in relation to thedirectionofwavemotion.

Fig.8.3:Particledisplacementinatransversewave

LongitudinalWaves

LongitudinalWaves

In longitudinalwaves, thevibrationof theparticles is inadirectionparallel tothe direction of the wave travel. Examples of longitudinal waves are soundwaves.

To illustrate the formationof a longitudinalwave, a slinky springmaybeused.Thespringisstretchedalongasmoothfloororbenchtop.Oneendisfixedtoarigidsupportandtheotherheldinthehand.Thisendisvibratedinato-and-fromovementcontinuallyalongitslengthasshowninfigure8.4.

Fig.8.4:Longitudinalwavesonaslinkyspring

The continuous to and fro movements at one end result in the formation ofsections of compression alternating with rarefactions along the length of thespring.Figure8.5illustratesthedisplacementofaparticleinalongitudinalwaveinrelationtothedirectionofwavemotion.

Fig.8.5:Particledisplacementinalongitudinalwave

Notethatindividualparticlesintheslinkyspringaresetintoperiodicvibrationsinlinewiththedirectionofthewavemotion.

Thewavemotion affects the inter-particle spacing.Particles in the sections ofcompression are pushed closer together while those in the sections ofrarefactionsarepulledslightlyfartherapart.Variationininter-particleseparationisaccompaniedbyvariationinpressure,sothatsectionsundercompressionareat a higher pressure while those under rarefaction are at low pressure. Thispressurevariationcausesthewavemotion.

ProgressiveWaves

These are waves that move continually away from the source. They can betransverseorlongitudinal.Ifalongslinkyspringiscontinuouslyvibratedatoneend, the wavesmove forward, carrying the energy of the vibrations along itslength.Similarly,ifastoneisdroppedontoawatersurface,theresultingwater

wavesmoveoutwards,carryingtheenergyoftheimpactawayfromthesourceasinfigure8.1.Asthewavemovesawayfromthesource,theenergyisspreadoveranincreasinglylargearea.Thiscausesgradualdecreaseinitsamplitude.

Pulses

Apulseisgeneratedwhenasinglevibrationissentthroughamedium.Itcanbetransverseor longitudinal innature.Figure8.6 (a)and (b) representspulsesoftransverseandlongitudinalnaturerespectively.

Fig.8.6:Transverseandlongitudinalpulses

Wavetrainsaregeneratedasaresultofcontinuousvibrationsataconstantrateinamedium.Themediumisdistortedintorepeatedpatternsofcrestsalternatingwithtroughsforatransversewave(Figure8.2),whileforthelongitudinalwavetrain, the medium is set into repeated patterns of sections of compressionalternatingwiththoseofrarefaction(figure8.4).

CharacteristicsofWaveMotionThe characteristics of wave motion can be explained with reference to theoscillatorymotionofamassattachedtoaspringorthatofthebobofaswingingpendulum,seefigure8.7(a)and(b).

Fig.8.7:Oscillatingspringandpendulum

Considerthemassatrestattheendofaspiralspring(positionM).IfthemassisdepressedslightlytopositionNandreleased,itoscillatesupanddownaboutthemeanpositionM.OnecompleteoscillationoccurswhenthemassmovesthroughpositionsN-M-L-M-N, i.e.,when it has returned to its starting position and ismovinginthesamedirection.

If,forexample,themassstartsatM,thenM-N-Misnotacompleteoscillation.This is because although the mass has returned to its starting position, it ismovingintheoppositedirection.

For the pendulum, the bobmakes a complete oscillationwhen, after an initialdisplacement to, say,positionX, it swings throughX-Y-Z-Y-X, see figure8.7(b).

Ifthemassinfigure8.7(a)takestwosecondstomakeacompleteoscillation,asketchofdisplacement-timeforthemotionwillappearasinfigure8.8.Similargraphwouldbeobtainedfortheswingingpendulum.

Fig.8.8:Displacementtimegraphforamassoscillatingonaspring

Thedisplacement-timegraphisasinecurvesimilartotheprofileoftransversewaves (compare with figure 8.2). With reference to figure 8.7 and 8.8, thefollowingtermsassociatedwithwavesaredefined:

Amplitude

Theamplitude(A)ofawaveisthemaximumdisplacementoneithersideofthemean position. Its S1 unit is themetre. In figure 8.7 (a), the amplitude is thedistanceLMorMNwhilein8.7(b),theamplitudeisdistanceXYorYZ.

Frequency

Thefrequency(f)ofawave is thenumberofcompleteoscillationsmadebyaparticle in one second. The unit of frequency is the hertz (Hz) or cycles persecond.

Period

The period T of oscillation is the time taken by a particle to complete oneoscillation. The S1 unit of period is the second (s). In figure 8.8, the particletakes 2 s to go through 1 complete oscillation and its period is therefore 2seconds.

Itfollowsthatf= ,thus,thefrequencyfortheoscillationshownis;

PhaseandPhaseDifference

Wavescanbeofthesameamplitudebutdifferentfrequencyorsamefrequencybutdifferentamplitude.Figures8.9and8.10illustratethetwocases.

Fig.8.9:Wavesofthesameamplitudebutdifferentfrequency

Thewavewithhigherfrequencyhasasmallerperiod.

Fig.8.10:Wavesofthesamefrequencybutdifferentamplitude

ConsidertwoidenticalpendulumswithbobsPandQ,seefigure8.11.Ifthetwomasses are both given some displacement towards the left and then releasedsimultaneously, they will be set in oscillation. The masses pass through restposition Y at the same time as they move in the same direction. They attainamplitudedisplacementtogetheratZandswingbacktocompletetheoscillationtogether at X. At any one time, the two masses will be moving in the samedirectionandatthesamelevelofdisplacementintheiroscillations.Themassesaresaidtobeoscillatinginphase.

Fig.8.11:Massesoscillatinginphase

Particles in a wave motion which happen to be oscillating in the samedirectionandatthesamelevelofdisplacementintheiroscillationaresaidtobein phase. The displacement-time graphs for such particles are identical, seefigure8.12.

Fig.8.12:Particlesoscillatinginphase

Particles in wave motion can be in phase even though they have differentamplitude. Thus, for masses P and Q above, if P is initially given a largerdisplacement than Q, the two will still oscillate in phase even though P willalways be at larger magnitude of displacement than Q. The resultingdisplacement-timegraphsareshowninfigure8.13.

Fig.8.13:Oscillationsofdifferentamplitudebutinphaseforparticles

SupposemassesPandQarenowdisplacedone topositionXandtheother topositionZ,seefigure8.14.

Fig.8.14:Oscillationsthatare180°outofphase

When released simultaneously, theypass through the rest position at the sametime, but moving in opposite directions. They reach positions of maximumdisplacement,butonoppositesidesoftherestposition.Thus,theyarealwaysonopposite levels of displacement in their oscillations and moving in oppositedirections.Theyaresaidtobeoppositeinphase(180°phasedifference).Figure8.15isasketchofthedisplacement-timegraphforthetwooscillations.

Fig.8.15:Oscillationsthatare180°outofphase

Theoscillationscanbeatotherlevelsofphasedifference.Thus,forthemassesPandQabove,iftheyarebothdisplacedtopositionXandPisreleasedfirstthenQasPcrossespositionY,theresultingoscillationswillbe90°outofphase.Thedisplacement-timegraphisasshowninfigure8.16.

Fig.8.16:Oscillationsthatare90°outofphase

Wavelength

A transverse wave train, for example, waves on water viewed from the sidewouldgiveadisplacement-positiongraphasinfigure8.17.

Fig.8.17:Displacement-positiongraph

Thewavelength,λ,isthedistancebetweentwopointsonawavetrainwhichareinphase.Itmayalsobedefinedasthedistancebetweentwosuccessivecrestsortroughsinatransversewaveorthedistancebetweentwosuccessiverarefactionsorcompressionsinalongitudinalwave.TheS1unitforwavelengthisthemetre.In figure 8.17, distance AC, and BX and EF are all equivalent to onewavelength.

Speed

The speed v is the distance covered by a wave in one second. Its S1 unit ismetrespersecond.

RelationshipBetweenSpeed,WavelengthandFrequencySupposetheperiodofawaveisT.Then,thedistancecoveredintimeTisλ.

Thissimplifiestov=fλ

Itshouldbenoted thatwhile the rateofvibrationof thesourcedetermines thefrequency of the waves, the speed in a given medium is constant. From therelationshipv=fλ,anincreaseinfrequencyresultsinadecreaseinwavelength,seefigure8.18.

Fig.8.18:Variationoffrequencywithwavelength

Example1Wavesonaspringareproducedattherate20wavelengthevery5s.(a)Findthefrequencyofthewavemotion.(b)Ifthewavelengthofthewavesis0.01m,findthespeedofthewaves.(c)Findtheperiodofthewaves.

Solutions

(b)Usingv=fλ,v=4×0.01

=0.04ms–1

Example2A water wave travels 12 m in 4 s. If the frequency of the waves is 2 Hz,calculate:(a)Thespeedofthewave.(b)Thewavelengthofthewave.

Solution

(b)Speed=frequency×wavelength(v=fλ)

Therefore,3=2×λHence,λ=

Thewavelengthis1.5m.

Example3

Figure8.19showsawaveforminastring.Thenumbersinthediagramshowsthescaleincentimeters.Thespeedofthewaveis10.0ms-1.

Fig.8.19

Withthereferencetothiswavemotion,determinethe:(a)Wavelength

(b)Amplitude(c)Frequency(d)Periodoftheoscillation

Solution(a)Wavelength=40cm(b)Amplitude=5cm(c)Fromf=fλ;

RevisionExercise8

1.(a)Explainthefollowingterms;progressivewave,wavelength,frequencyandamptitude.

(b)Statetheequationrelatingspeed,frequencyandwavelengthofawave.2.Giveoneexampleof:

(a)Atransversewave.(b)Alongitudinalwave.Whatisthemaindifferencebetweenthesetwotypesofwaves?

3.Thefigurebelowshowsthedisplacement-timegraphforawave:

(a)Howmanycompletecyclesareshown?(b)Whatisthefrequencyofthewaveformshown?

(c)Drawadiagramtoshowthewaveformofawavewhosefrequencyistwiceandwhoseamplitudeishalftheoneshowninthediagram.

(d)Aradiowavehasafrequencyof3MHzandtravelswithavelocityof3×108ms-1.Whatisitswavelength?(1MHz=106Hz).

4. What ripplesarecaused to travel across the surfaceofa shallow tankbymeansofastraightvibrator.Thedistancebetweensuccessivecrests is3.0cmandthewavestravel25.2cmin1.25seconds.Calculatethewavelengthandthevelocityofthewaves.Findalsothefrequencyofthevibrator.

5. Wavesenteraharbouratarateof30crestsperminute.Amanwatchesaparticular wave crest passing two buoys which are 12 m apart along thedirectionoftravelofthewaves.Thetimethewavetakestomovefromonebuoytotheotheris2.0seconds.Calculate:(a)Thefrequencyofthewavemotion(b)Thewavelengthofthewaves

6.Avibratorissendingouteightripplespersecondacrossawatertank.Theripplesareobservedtobe4cmapart.Calculatethevelocityoftheripples.

7.Thediagrambelowshowsadisplacement-positiongraphforaslinkyspringasitiscontinuallyvibratedatoneend:

(a)Whattypeofwavesaregeneratedintheslinky?(b)Whatisthe:

(i)Amplitudeofdisplacement?(ii)Wavelengthofthewaves?

(c)Inthesamediagram,showthewaveformwhen:(i)Therateofvibrationisdoubled.(ii)Theamplitudeishalved.

(d) On the same axes, draw awaveformwhose vibration is opposite inphaseandtheamplitudehalftheoneshown.

Chapter9

SoundI

Sound originates from vibrating bodies. The nature of these vibrationsdeterminesthetypeofsoundproduced.Insomecases,thevibrationsmaybefeltorseen,forexample,whenastringofaguitarispluckedorwhenatuningforkisstruck.Inothercases,thevibratingoriginofthesoundmaynotbesoobvious,forexample,windrushingthroughasmallopeningor thenoiseproducedbyathunderstorm.

SomeSourcesofSound

VibratingWoodenStrip

Whena thin strip clampedat one end in avice is struck sharply, the free endvibratesto-and-fro,producingsound,seefigure9.1.

Fig.9.1:Vibratingstrip

VibratingWire

Awire is tightly tied on two pegs that have been fixed on awooden box, asshowninfigure9.2.

Fig.9.2:VibratingWire

Whenpluckedatitsmiddlepoint,thewirevibratesasshown,producingsound.Whenalightfeatherisplacedonthewire,thefeatherjumpsoff,confirmingthatthewireisvibrating.

VibratingDrum

Whenadrumisstruck,itvibrates,producingsound.

Fig.9.3:Drums

If a few grains of dry sand are placed on a vibrating drum, the grains aredisplacedupanddown,showingthattheskinvibrates.

TuningFork

Theprongsofatuningforkaremadetovibratebystrikingthemagainstahardsurface. The sound produced can be heardmore clearlywhen the stem of theforkisfixedtightlyonahollowboxasshowninfigure9.4.Pressingthestemoftheforkonabenchproducessimilarresults.

Fig.9.4:Photoofvibratingtuningfork

Atuningforkofhigherfrequencywhenstruckvibratesfasterthanoneoflowerfrequency,producingsoundofahighernote.

When the prongs are made to touch the water surface, the water splashes,confirmingthattheprongsarevibrating.

VibratingAirColumns

Ifairisblownacrossthemouthofatest-tube,soundisheard.Thisisbecausethecolumnofairinthetubeissetinvibration.Soundsofdifferentfrequenciesareproduceddependingontheoftheaircolumn.

Air Siren in form of a disc with a ring of equally spaced holes which areequidistantfromthecentreisusedinthiscase.Whenitisrotatedataconstantrate and air blown through theholes as shown in figure9.6, soundof agivennoteisheard.Thediscmayhaveconcentricsetsofholesaboutitsaxistovarythefrequency.

Fig.9.5:Asiren

CogWheelandCard

Acogwheel is rotatedwith a stiff cardpressing lightly against the teeth.Theteethoftherotatingwheelstrikethecard,producingsound.

Fig.9.6Cogwheel

If the teeth are equally spaced, thewheelswith fewer teeth produce sound oflowerfrequency.

VoiceBox(Larynx)

The human voice box contains vocal cords which vibrate to produce sound.Otherexamplesofvibratingbodiesareloudspeakersandcellphoneortelephonemembranes.

PropagationofSoundEnergy

Vibratingprongsofatuningforkproducecompressions(areasofhighpressure)and rarefactions (areas of lowpressure) of airmolecules.This is illustrated infigure9.8.

Fig.9.7:Propagationofsoundwaves

As the prong of the tuning fork moves from A to B, it compresses the airmolecules, transferring energy to the molecules in the direction in which thecompressionoccurs.Ahighpressureregionisthuscreated.Thisleavesaregionoflowpressure(ararefaction)ontheleftofA.TheprongmovesbacktoAandthentoCandtheprocessisrepeated.Aseriesofcompressionsandrarefactionsarethusproduced,transferringenergytotheairparticles(molecules)totheleft

and right. The energy transfer alternates in direction just as themotion of theprong.Wemaydescribeaprogressivesoundwaveinairasatravellingpressurewave,asshowninfigure9.9.

Fig.9.8:Propagationofsoundwaves

Soundenergymovesforwardinthemediumwithoutnetforwardmovementofthe medium. The direction of vibrations of the particles is parallel to thedirectionofthemovementofsoundenergy.Hence,soundwaveisalongitudinalwave.

Experiment9.1:Toshowthatsoundrequiresamediumforpropagation

Apparatus

Anelectricbell,switch,bell-jar,vacuumpump,rubberbung,flexibleconnectingwires,cells,glassplate,rubbercord.

Fig.9.9:Soundrequiresamaterialmediumforpropagation

Procedure•Settheapparatusasshowninfigure9.9.•Switchonthecurrenttomakethebellringcontinuouslyasairispumpedoutslowlyusingthevacuumpump.Observewhathappens.

Observation

Theintensityofsounddiminishesastheairinthejarbecomesless.

ExplanationThe sound grows faint because the jar is deprived of air.A vacuum does nottransmit sound and the little sound that reaches us does so only through theconnectingwires,rubberandthewallsofthejar.

FactorsAffectingVelocityofSound

Inair thespeedofsoundisabout330m/s.Thisspeedisdependentonvariousfactors,namely;•Temperatureofair.•Humidityintheair.•Winddirection.

Sound travels faster in air at high temperature. It also happens that when themoisturecontent in theair ishigh, soundsaregenerally transmittedat a fasterrate.Wind affects the velocity of sound in that when the two, i.e., wind andsound happens to be moving in the same direction, the velocity of sound isincreased.

Generally, solids transmit sound at a speed of 6 000 m/s. This velocityvaries from solid to solid, dependingondensityof thematerial.Denser solidstransmitsoundfaster.

In addition to gases and solids, liquids also propagate sound energy. Aswimmereasilyhearssoundofwaterwaveswhenunderneaththewaterandfishsimilarlyrespondtosoundsproducedinwater.Liquidstransmitsoundenergyatdifferentspeedsdependingontheirdensities.

Thevelocity of sound in freshwater is 1400m/s and in saltywater 1500m/s.Gasestransmitsoundslowest,whilesolidstransmitsoundfastest.

ReflectionofSound

Whenasharpsoundfallsonanobstacle,itisreflected.Reflectedsoundiscalledanecho.

Reflectedsoundismorepronouncedfromhardsurfacesuchaswood,stonewallsandmetalsurfaces.Reflectionfromliquidsurfacesisconsiderablyweaker.

Insomehalls,soundwavesarereflectedfromthewalls, floorandceiling.Since the echo time is short, the echo overlaps with the original sound. Theoriginalsoundthusseemstobeprolonged,aneffectcalledreverberation.

Surfacesofmaterialssuchascottonwoolandfoamrubberabsorbmostoftheenergyofincidentsoundwaves.Becauseofthisproperty,suchmaterialsareused inplaceswhere echoeffects arenotdesirable.Thewallsofbroadcastingstudiosandconcerthallsarethusmadeofabsorbentmaterials.

Experiement9.3:Todemonstratereflectionofsound

Apparatus

Twoplastictubes,aticklingclock,smoothhardwall.

Fig.9.11:Reflectionofsound

Procedure•Placetheclockneartheendofoneofthetubesasshowninfigure9.11•Pointtheopenendofthetubetowardsahardwallatanangleofincident,i•Withtheearclosetotheendofthesecondopentube,listentothereflectionof the sound from thewall at different angles of reflection, r, and note theangleatwhichthereflectedsoundisloudest.

Observation

Itisobservedthatmaximumloudnessofthereflectedsoundoccurswhen:

(i)Theangleofreflection,r,isequaltotheangleofincidence,i.(ii)Bothtubesandthenormaltothereflectingsurfacelieinthesameplane.

Conclusion

Soundwavesobeythelawsofreflection.

ApplicationsofReflectionofSound

DeterminationoftheSpeedofSound

Experiment9.4:Todeterminethespeedofsoundinairbyecho

ApparatusHardsmoothwall,twopiecesofwood,stop-watch.

Fig.9.12:Determinationofvelocityofsoundbyecho

Procedure•Standsomedistanceawayfromanddirectlyinfrontofalargehighwall.•Tapthetwopiecesofwoodtogetherandlistentotheecho,seefigure9.12.•Nowtaprepeatedlyandtrytoarrangethenexttaptocoincidewiththeechoheard.

• Timeanumberof successive tapsanddetermine theaverage timebetweensuccessivetaps.

• Measuretheperpendiculardistancebetweentheobserverandthewallasinthefigure.

Since the echo fromone tap coincideswith the sound fromnext tap, itmeansthatthetimetakentomakeatapaftertheprecedingoneequalsthetimetakenbysoundtotravelfromtheobservertothewallandback.Ifthedistancebetweentheobserverandthewallisdmetres,thenumberoftapintervalsnandthetimetseconds,thesoundtravels2dmetresin seconds.

Pulse-EchoTechnique

Thepulse-echo technique involvesmeasuringdistancesbyproducingsoundofknown speed and measuring the time taken to receive the echo. Sound offrequencyofover20kHz(ultrasound)isused,becauseitpenetratesdeepestandcanbereflectedeasilybytinygrains.

The distance of the reflecting obstacle from the source of sound is thencalculatedusingtheformula;

distance(d)

Thistechniqueisusedinchipstodeterminethedepthofthesea,seefigure9.14.

Fig.9.13:Usingpulse-echotechnique

Thetechniqueisalsoused:(i)Inunder-waterexplorationforgasandoil.(ii)Infishingboatswithpulse-echoequipmenttolocateshoalsoffish.(iii)Inspecialtypesofspectaclesusedbyblindpeopletotellhowfarobjects

areaheadofthem.Thespectacleshavetransmittersthatemitultrasoundandreceiversthatcollecttheechoandconvertthemintoaudiblesound.

Batsuseechoestodetectthepresenceofobstaclesintheirflightpath.

Example1

Adiscsirenwith100holesisrotatedatconstantspeedmaking0.20revolutionspersecond.Ifairisblowntowardstheholes,calculate:(a)Thefrequencyofthesoundproduced(b) Thewavelengthof thesoundproduced, ifvelocityof sound inair is340

m/s.

Solution(a)Numberofholesreceivingairpersecond

=

=20holesFrequencyofthesound

Example2

A cog wheel rotating uniformly produces sound of wavelength 1.65m. If itmakes10revolutionspersecond,findthenumberof teethonthewheel,giventhatthevelocityofsoundinairis330m/s.

Solution

Example3

Two boys stand 200m from awall. One bangs two pieces of wood togetherwhilethesecondstartsastop-watchandstopsitwhenhehearstheecho.Ifthetimeshownonthestop-watchis1.2seconds,calculatethespeedofsound.

Solution

Example4

Thespeedofsoundinairis340ms–1.AloudspeakerplacedbetweentwowallsAandB,butnearertowallAthanwallB,issendingoutconstantsoundpulses.How far is the speaker fromwall B if it is 200m fromwall A and the timebetweenthetwoechoesreceivedis0.176seconds?

Solution

Fig.9.15

Speedofsoundis340ms–1.LetthetimetakentohearechofromwallsAandBbet1andt2respectively.Then,t2–t1=0.176secondsIfdistancebetweentheloudspeakerandwallBisxmetres,then;

Example5

Theshipsendsoutanultrasoundwhoseechoisreceivedafter10seconds.Ifthewavelength of the ultrasound in water is 0.05 m and the frequency of thetransmitteris50kHz,calculatethedepthoftheocean.

Solution

Velocityofwaves

=frequency×wavelength

=50000×0.05

=2500ms–1

Depthd=(2500× ×10)m

=12500m

RevisionExercise9

1.(a)Howissoundproduced?(b) State the law of reflection of sound and explain how an echo is

produced.(c)Whyistheinsideofsomeconcerthallscoveredwithsoftmaterial?

2.(a)Howdoesasoundwavedifferfromthewavesproducedwhenawatersurfaceisdisturbed?

(b) Therangeofaudiblefrequenciesvariesfrom20Hzto20kHz.If thespeed of sound is 340 ms–1, what is the corresponding range ofwavelength?

3. Aman standing in a gorge between two large cliffs claps his hands at a

steady rate andhears twoechoes.The first comesafter2 secondsand theotherafter3 seconds. If the speedof sound is340ms–1, find thedistancebetweenthetwocliffs.

4. (a) Agirlstands160mawayfromahighwallandclapsherhandsatasteadyratesothateachclapcoincideswiththeechooftheonebefore.Ifshemakes60clapsinoneminute,calculatethespeedofsound.

(b)Ifshemoves40cmclosertothewall,shefindsthattheclappingratehastobe80perminute.Calculatethespeedofsound.

5.(a)Howissoundpropagated?(b)Whatevidenceistheretoshowthatsoundisawavemotion?(c)Describeexperiments,oneineachcase,toshowthat:

(i)Thesourceofsoundisavibratingbody.(ii)Amaterialmediumisnecessarytotransmitsound.

6.Statetwodifferencesbetweensoundandlightwaves.7. Apersonstanding49.5mfromthefootofa tallcliffclapshishandsand

hearsanecho0.3secondlater.Calculatethevelocityofsoundinair.8.Twopeoplestandfacingeachother200mapartononesideofahighwall

andatthesameperpendiculardistancefromit.Whenonefiresapistol,theotherhearsareport0.6secaftertheflashandasecondsound0.25seclater.Explainthisandcalculate:(a)Velocityofsoundinair(b)Perpendiculardistanceofthepeoplefromthewall.

9. (a) What is the relationship connecting the frequency,wavelength andvelocityofsoundinair?(b) An echo sounder in a ship produces a sound pulse and an echo is

receivedfromtheseabedafter0.2sec.Ifthevelocityofsoundinwateris1400m/s,calculatethedepthofthesea.

(c) If the echo sounder hadproduced continuouswavesof frequency25kHz,whatwouldbetheirwavelengthinthewater?

10.Indeterminingthedepthofanocean,anechosounderproducesultrasonicsound.Givereasonswhythissoundispreferred.

11. The figurebelowshowsavibrating tuning fork.The time interval for theprongtogofromAtoBis0.005sec.

Findthe:(a)Frequencyofthefork.(b)Wavelengthofthevibrations,ifthevelocityofsoundinairis340ms–1.

12.Thediagrambelowisarepresentationofsoundwavespassingthroughair.Studyitandanswerthequestionsthatfollow.

(a)Labelthefollowing:(i)Compression(ii)Rarefaction(iii)Wavelength

(b)Thewavefronttakes0.1sectotravelfromAtoB,findthe:(i)Thefrequency.(ii)Thewavelength,ifvelocityofsoundinairis330ms–1.

13. Awoodenstrippressesonacogwheelasitrotatesat300revolutionspersecond.Ifthewheelhas100teeth,findthefrequencyandthewavelengthofsound,giventhatitsvelocityinairis330ms–1.

14.Airisblownintotheholesofadiscsirenasit isrotatedat10revolutionspersecondsothatitproducessoundoffrequency5kHz.Findthenumberofholesonit.

Chapter10

FluidFlow

The term ‘fluid’ refers to both gases and liquid. A fluid flows as a result ofpressuredifference.Examplesof fluid flows include the flowofwateralongariverbed, flowofhotwater incentralheatingsystemsandconvectioncurrents.Thestudyofmovingfluidsisveryimportantbecauseoftheirvariedapplicationsin day, to, day life, for example, the dynamic lift on an aeroplane wing, theworkingofapaintspray-gunandthebunsenburner.Aflowingfluidexperiencesinternalfriction,whichisknownasviscosity,betweenitslayers.Inthischapter,itisassumedthattheviscosityofanyfluidunderdiscussionisnegligible.

TypesofflowTheflowofafluidcanbeclassifiedasstreamline(steady),laminarorturbulent.

Streamline(steady)flow

Ifalltheparticlesofafluidpassingthroughanygivenpointinthefluidhavethesamevelocity,thentheflowissaidtobestreamlineorsteady.Thepathtracedbytheparticlesiscalledthelineofflow,whichisrepresentedbyalineandanarrow,seefigure10.1(a).Astreamlineisacurvewhosetangentatagivenpointisalongthedirectionofthedisplacementofthefluidparticlesatthatpoint,seefigure10.1 (b).When thestreamlineand the lineof flowcoincide, the flow issteadyorstreamline.

Fig.10.1:Lineofflowandstreamline

LaminarFlow

Thistermistakentomeansteadyflow.Amovingfluidhasmanystreamlinesorlayers.Theflowislaminariftheparticlesinagivenstreamlineorlayerhavethesame velocity. Which may be different from other particles in the adjacentparallellayers,seefigure10.2(a)and(b).

Fig.10.2:Fluidflowasseriesofconcentriccylindersinapipe

Turbulentflow

Turbulent fluid flowoccurswhen thespeedanddirectionof the fluidparticlespassingthroughapointvarywithtime.

Experiment 10.1: To investigate the effect an object e.g. a ruler onstreamlines.

Apparatus

Ruler,abasinofwater

Procedure• Move the ruler with its sharp edge cutting through water and note thefollowing:

•Theforcerequiredtomovetheruler.•Theripplesonthewatercausedbyitsmovement.•Repeattheexperimentwiththebluntorflatsideoftheruler.

Observation

Thestreamlinesobtainedareasshowninfigure10.3(a)and(b).Figure10.3(a)shows streamlines around the edge of the ruler while 10.3 (b) shows whathappenstothestreamlineswhentheexperimentisrepeatedwiththebluntorflatside.Itisobservedthatwiththeflatsideoftheruler,moreeffortisrequiredtomove the ruler thanwhen it ismovedwith its sharp edge cutting through thewater.Ripplesaresetupinthewater,whichtendtofollowtherulerasitmoves.

Fig.10.3:Effectofaruleronstreamlines

This breaking of streamline into ripples (disorderly flow) is referred to asturbulent flow.The ripples or eddies have a drag effect on the objectmovingthroughthefluid.

Experiment10.2:Toinvestigatetheeffectsofvariousshapesonstreamlines

Apparatus

Basinofwater,piecesofwoodofdifferentshapes,strings,hooks

Procedure•Fixastringattheendofeachshapesasshowninfigure10.4(a),(b)and(c):• Pull it through a pool or a large container ofwater. In each case, note theeffortrequiredtopulltheobjectthroughthewaterandtheripplesoreddiesas

theobjectsailsthrough.

ObservationThestreamlinesobtainedareasshowninfigures10.5(a),(b)and(c).

Fig.10.4:Effectofshapesonstreamlines

Shape (a) requires littleeffort tomove,andhasnoeddiesbehind it.Shape (b)requiresmore effort than (a) andgives rise tomore eddies than (c).Shape (c)requiresmoreeffortthan(a),butlesseffortthan(b).

ConclusionObjects with sharper or pointed ends require relatively less force to movethrough the fluid as they cause less turbulence (drag effect). Objects withstreamlinedshapescauselessturbulencetoafluidflow.

ShapesDesignedforStreamlineFlow

Shapessuitedtostreamlineflowaredesignedinsuchawaythattheyeasilycut

through fluids and reduce the formation of eddies behind them. This reducesresistancetotheirmotion.

Fig.10.5:Aquaticandflyinganimals

In nature some animals have bodies that are shaped tomake their movementeasy.Examplesincludeaquaticandflyinganimals.

Fig.10.6:Bodiesdesignedforstreamlineflow

EffectofSpeedofFlowonStreamlines

Thearrangementinfigure10.7canbeusedtostudytheeffectofspeedofflowonstreamlines.

Fig.10.7:Effectofspeedonstreamlines

Reservoir A contains potassium permanganate solution, which is released incontrolledamountsintothewaterflowingthroughacylindricalglassjacketCbya fine jetB.The speedofwater throughC is varied by the clipD. If a smallamount of water is allowed to flow out through D, a fine coloured stream isobservedalongthetubeC,indicatingasteadyflow.Butwhenalargeamountofwaterisallowedtoflowout,thevelocityofwaterinthetubeCincreasesrapidlyand this breaks the coloured stream, indicating that turbulence has set in.Turbulence sets inwhen the fluid flow is beyond a certain velocity known ascriticalvelocity.

TheEquationofContinuity

Inderivingthisequation,theassumptionsmadearethatthefluidis:(i)Flowingsteadily.(ii)Incompressible,i.e.changesinpressureproduceinsignificantchangeinits

density.(iii)Non-viscous.

Someusefuldefinitions

Volumeflux(flowrate)Consider a tube of flowwith regionW having cross-section area ofAW andregionXwithcross-sectionAreaAX.Thevolume flux is thevolumeofa fluidpassing throughagivensectionofatubeofflowperunittime,seefigure10.8.

Fig.10.8:Volumeflux

Let the velocity of the fluid through region X be vx and the average cross-sectionareabeAx.Ifthedistancecoveredbythefluidperunittimeisdxthenthevolumefluxthroughthatregionisgivenby:

Volumeflux=areaofcross-section×distance

Hence,velocityvx=dxsincet=1sec.Thus,volumeflux=AxvxIfAxisinm2andvxisinm/s,thentheunitofvolumefluxism2×m/s=m3/s(m3s–1).

Massflux

Mass=density×volumeHence,massflux=density×volumeflux

=ρ×AxVx

=Axvxρ

Ifvolumefluxisinm3s–1anddensityisinkgm–3,thentheunitofmassfluxiskg/s(kgs–1).

Massfluxisthereforethemassofthefluidthatflowsthroughagivensectionperunittime.

Sincethefluidisincompressible,themassofthefluidenteringregionXisequaltothemassofthefluidleavingregionWwithinthesameperiod,i.e.,massisconserved.

MassfluxatW=massfluxatXAwvwρ=AxvxρHence,Awvw=AxvxThisistheequationofcontinuity.AwvworAxvxisalsocalledtheflowrateandisconstant,i.e;Area×velocity=constantAV=kTherefore, it can be deduced that for a non-viscous steady flow, the area ofcross-sectionofthefluidisinverselyproportionaltothevelocityofthefluid.

The speed of the fluid increases when it flows from a pipe of big cross-sectionareatoasmallerone.Theconverseisalsotrue,seefigure10.9.

Fig.10.9:Velocitychangewitharea

Example1

Alawnsprinklerhas40holes,eachofcross-sectionarea2.0×10–2cm2.Itisconnectedtoahose-pipeofcross-sectionarea1.6cm2.Ifspeedofthewaterinthehose-pipeis1.2ms–1,calculatethe:(a)Theflowrateinthehose-pipe(b)Thespeedatwhichwateremergesfromtheholes

Solution(a)Flowrate=cross-sectionareaxspeed

=1.6×10–4m2×1.2ms–1

=1.92×10–4m3s–1

(b)Theflowrateisconstant.Volumeefflux(volumeofwatercomingout)=v×2.0102×10–4×40=8×10–5vm3s–1

Volumeinflux=Volumeefflux8×10–5v=1.92×10–4

v=2.4ms–1

Example2

Waterflowsalongahorizontalpipeofcross-sectionarea40cm2whichalsohasa constrictionof cross-section area 5 cm2. If the speed at the constriction is 4ms–1,calculatethe(a)Speedinthewidesection.(b)Massflux(takethedensityofwatertobe1×103kgm–3).

Solution(a)Volumefluxinthewidersection

=40×10–4×vVolumefluxintheconstriction=5×10–4×4Butvolumeflux=constant40×10–4v=5×10–4×4

Massflux=densityxvolumeflux=1×103×2×10–3

=2Kgs–1

Example3

Itisnotedthat250cm3offluidflowsoutofatube,whoseinnerdiameteris7mm,inatimeof41s.Whatistheaveragevelocityofthefluidinthetube?

Solution

AreaA=πr2=3.142×(3.5×10–3)2

=3.85×10–5m2

Sincevolumeflux=Av

Therefore6.098×10–6=3.85×10–5×v

Exercise10.1

1.Distinguishbetweenthefollowingterms:(a)Streamlineflowandturbulentflow.(b)Steadyflowandlaminarflow.

2.Waterflowsthroughahorizontalpipeatarateof1.00m3/min.Determinethevelocityofthewateratapointwherethediameterofthepipeis:(a)1.00cm.(b)4.00cm.

3.Inthefigurebelow,thetubeABCisfilledwithaliquid.ThepistonmovesfromAtoBin1second.

Calculate:(a)thevolumeoftheliquidinpartAB.(b)thevelocityoftheliquidbetweenAandB.(c)thevelocityoftheliquidthroughBC.

4. A garden sprinkler has small holes, each 2.00mm2 in area. If water issuppliedattherateof3.0×10–3m3s–1andtheaveragevelocityofthesprayis10ms–1,calculatethenumberoftheholes.

5.Oilflowsthrougha6cminternaldiameterpipeatanaveragevelocityof5m/s.Findtheflowrateinm3/sandcm3/s.

6.Thevelocityofglycerineina5cminternaldiameterpipeis1.00m/s.Findthevelocityina3cminternaldiameterpipethatconnectswithit,bothpipesflowingfull.

Bernoulli’sprincipleThe pressure of a fluid at rest in a uniform horizontal tube is the same at allpointsinthetube.However,ifthefluidflows,thepressurewillvaryfrompointtopoint,seefigure10.10.

Fig.10.10:Pressurevariation

Apressuregradientisneededtomakealiquidflowthroughapipe.Thecauseofthe pressure difference is the friction between the liquid and the walls of thepipe.

Experiment 10.3: To investigate the relationship between speed of waterandthepressureitexerts

Apparatus

Glasstubewithvaryingcross-sectionareaandfittedwithverticaltubesatX,YandZ,sourceofrunningwater.

Fig.10.11:Investigatingspeedandpressureofmovingwater

Procedure•Filltheglasstubewithwater.•Filltheverticaltubes(manometers)withwateruptothesameheight.•Opentheoutletandsupplytheglasstubewithwatersuchthattheamountofwaterenteringtheglass tubeis thesameas theamountflowingout.Note iftherearechangesintheheightsofthewaterintheverticaltubes.

Observation

ItisobservedthatthelevelinmanometerBislowerthanthelevelintheothertubes,seefigure10.12.

Explanation

RecallthatpressureinstationaryfluidsisgivenbyP=hρg(whereρ=density).Hence,thepressurebeingexertedbythefluidinthenarrowconstrictionislowerthanthatofXandZ.ItisalsoslightlyloweratZthanatX.Recallalsothatthevelocityof the fluid at thenarrowconstriction is higher than that at thewidersections.Thus,thehigherthespeedofthefluid,thelowerthepressureitexerts.ThisrelationisknownasBernoulli’seffect,whichcanbestatedasfollows:

Provideda fluid isnon-viscous, incompressibleand its flowstreamline,anincreaseinitsvelocityproducesacorrespondingdecreaseinthepressureitexerts.(a)Somebooksarearrangedonatableandapieceofpaperplacedonthem,as

showninfigure10.12(a).

Fig.10.12:Bernoulli’seffect

Whenairisblownintothechannelmadebythebooks,thepressureunderthepaperdecreasesandtheatmosphericpressureactingontopofthepaperpressesitdown.Thepaperthuscurvesinasshowninfigure10.13(b).Thepressure in the channel decreases because the velocity of the air in thechannelincreases.

(b)Ifalightpaperisheldinfrontofthemouthandairblownhorizontallyoverthepaper,itwillbeobservedthatthepapergetsliftedup.Initially, part of the paper is suspended because its weight and theatmosphericpressureactingonthetwosurfacesbalance.Whenairisblownover thepaper, its velocitygetshigher than at the initial statewhenair isstationary. Increase in velocity causes a corresponding decrease in thepressure being exerted on the top side of the paper. The atmosphericpressure acting underneath, therefore, becomes higher and produces theforcethatliftsupthepaper,seeFigure10.13(a).

(c)Iftwopiecesofpaperareplacedclosetoeachotherandairblownbetweenthem as shown in figure 10.14 (a), the two papers close in towards eachother.Themovingairbetween thepapers lowers thepressure it exertsontheir innersurfaces.Thehigheratmosphericpressureactingontheoutsidesurfacescausesthepaperstomoveclosertoeachother.Thesameeffectisobserved when air is blown between two suspended pithballs, see figure

10.13(b).

Fig.10.13:Bernoulli’seffect

(d)TheSpinningBallWhena tennisball of negligibleweight ismoving through still airwith aconstantspeed,thestreamlinesarounditareuniformlyspread,asshowninfigure 10.14 (a). The direction of the streamlines is the direction of therelativemotionbetweentheballandtheair.

Fig.10.14:Spinningball

Iftheballisnowmadetospinasitmoves,itisobservedtocurveoutofitsinitialpath.Astheballspins, itdragsairalongwith it,whichopposes therelative motion on one side of the ball. This causes a reduction in the

relative speedand the streamlinesare spread, see figure10.14 (b).On theopposite side, the dragged air is in the direction of the relative motion,resulting inan increase inspeedandconsequentialdecrease inpressure inaccordancewithBernoulli’seffect.Thepressuredifferenceonthetwosidesoftheballproducesaresultantforcethatcausestheballtocurveoutofitsinitialpath.

(e)LiftingalightballusingaFunnelThestreamlinesasair isblowndown thenarrowsectionof the funnelareveryclosetoeachother,signifyinghighvelocityandthereforelowpressure,see figure 10.16. However, when the streamlines emerge into the widersection, they spread, signifying reduced velocity and therefore highpressure.Thehighpressurebelow theball (atmosphericpressure) lifts theballuptotheneckofthefunnel.

Fig.10.15:Liftingalightball

Bernoulli’sEquation

Bernoulli’s principle can also be expressed as an equation.Consider liquid ofmassmflowingthroughapipewithvelocityv.Letthepressureatagivenpointbep.Then,kineticenergyperunitvolume

Hence,kineticenergyperunitvolume

Similarly,potentialenergyperunit

Bernoulli found that if the liquid is incompressible, non-viscous and its flowstreamline;

That is, the sum of pressure, kinetic energy per unit volume and thepotentialenergyperunitvolumeisaconstant.ThisisBernoulli’sprinciple.

ApplicationsofBernoulli’sPrinciple

TheAerofoilThis is a structure constructed in such a way that the fluid flowing above itmoveswithahigherspeedthanthatflowingbelow,seefigure10.16.

Fig.10.16:Aerofoil

Aircraftwingsandhelicopterrotorbladesareexamplesofaerofoils.Becausethefluidflowingabovetheaerofoilhastotravelalongerdistancethanthatflowingbelow, it has to travel at a higher speed (low pressure) compared to the lowspeed(highpressure)underneath.ThepressureP1 is thusgreater thanpressureP2.Thepressuredifference(P1–P2)givesrisetotheliftoftheaerofoil,calledthedynamiclift.TheforceoftheliftisgivenbyF=(P1–P2)A,whereAistheareaoftheaerofoil.

BunsenBurner

When gas is made to flow into the bunsen burner from the gas cylinder, itsvelocity is increased when it passes through the nozzle. This decreases thepressure above the nozzle. Because of the atmospheric pressure outside thebarrel,airisthendrawninasshowninthediagram.Theairandgasthenmixastheyriseupandwhenignited,aflameisproduced.

Fig.10.17:Bunsenburner

SpraygunFigure10.19showsahandspraygun.Whenthepistonismovedforward,airismade to flow through the barrel, some of it going down tube A and theremainder blowing past the mouth of tube B, where it causes low pressure.Becauseofincreasedpressureonthesurfaceoftheliquid,andreducedpressureatthemouthofB,theliquidiscompelledtomoveuptubeBandblowntothenozzle by the air from the barrel. The velocity of the liquid is increased as itpassesthroughthenozzlebecauseofthereducedcross-sectionarea.Theliquidthusemergesasafinespray.

Fig.10.18:Spraygun

TheCarburettor

Duetotheactionoftheenginepistons,airisdrawnintotheventuri,seefigure

10.20.The fast-moving air causes low pressure above the petrol pipe. Petrol is

drawn into the venturi due to low pressure in the venturi and atmosphericpressure in the floatchamber.Themixtureofairandpetrol is thusdrawn intothecylindersforcombustion.

Fig.10.19:Carburettor

FlowMeters

VenturimeterThisdeviceisusedinmeasuringthevolumefluxofafluid.

Fig.10.20:Venturimeter

LetP1andP2bepressureandv1andv2 thevelocitiesof the fluidatXandYrespectively.SinceP+ pv2=constant(horizontaltube);

Applyingtheequationofcontinuity;

A1v1=A2v2

Fromthis,v1andhenceA1v1canbecalculated.

Pitot-tube

This device is used for measuring the velocity of moving fluid. The staticpressure ismeasuredby the static tubewhile the pitot tubemeasures the totalpressure.

Fig.10.21:Pitottube

PressureatCisgivenby:

TotalpressureatBequalPc

PressureatAisgivenby;PA=P

WherePBisthetotalpressureandPAthestaticpressure.ButPB–PA=(hB–hA)pg

HazardsofBernoulli’seffect

BlowingoffofRoof-topsTheairflowingoveraroof-tophasahighvelocitycomparedtotheoneflowingunderneath,seefigure10.24.Consequently,thepressureactingontherooffromtheunderneath,P2,will be higher than that acting fromabove,P1.Hence, theroofmaybeblownoff.

Fig.10.22:Winddrivenairflowoverabuildingroof

RoadAccidents

Asmall car travellingat averyhigh speed is likely tobedragged into a longtrucktravelingintheoppositedirection,alsoatahighspeed.Thisisbecausetheair in between them moves with a very high speed, reducing the pressurebetween them.Theatmosphericpressureacting from thesidesof twovehicleswillpushthemclosertogether,increasingchancesofanaccident.

Example4

Waterflowssteadilythroughahorizontalpipeofvaryingcross-sectionarea.AtapointAofcross-sectionarea10cm2,thevelocityis0.2ms–1.Calculatethe:(a)VelocityatapointB,ofcross-sectionarea2.5cm2

(b) Pressure difference betweenA andB, given that the density ofwater is

1,000Kgm–3.

Solution(a)Fromtheequationofcontinuity;

A1v1=A2v210×10–4×0.2=vB×2.5×10–4

vB=0.8ms–1

Example5

Airflowsovertheuppersurfacesofthewingsofanaeroplaneataspeedof100ms–1andpastthelowersurfaceofthewingsat90ms–1.Calculatethe:(a)Pressuredifferenceonthewings.(b)Liftforceontheaeroplane,ifithasatotalareaof10m2(takethedensityof

airas1.3Kgm-3).

Solution(a)Pressuredifference

=1235Nm–2

(b)Liftforce=pressuredifference×area=1235×10=1.235×104N

Example6

Aliquidofdensity8.75×103 gm is flowing throughahorizontalpipewhosecross-sectionareachangesfrom40cm2,to25cm2,withtwoopenverticaltubes

atthetwosections.Theverticaltubescontainthesameliquid.Ifthereisaliquidleveldifferenceof1.25cmbetweenthetwotubes,calculatetherateofflowinmassperunittime.

Fig.10.23

Fromtheequationofcontinuity,A1v1=A2v240×10-4×v1=25×10-4v2

So,v1=0.625v2ms-1

Thechange inpressure is responsible for the change in thekinetic energyperunitvolumeofthefluid.Thatis;

ButP=hpgThereforeΔP=1.25×10-2×8.75×10=109.4Nm-2

Massflux=volumeflux×density=AV×density=20×10-4×0.64×8.75×102

=1.4kgs-1

Example7

Water flows steadily along a uniform flow tube of cross-section area 20 cm4.The total pressure is 3 × 104 Pascal and the static pressure 2.5 × 104 Pa.Calculatethe:(a)Velocityofthefluid.(b)Massofwaterflowingpastasectionintwoseconds(takedensityofwater

as103kgm–3)

Solution(a)V2= (PT–PS),wherePTisthetotalpressureandPSstaticpressure

Intwoseconds,themassofwaterflowingpastagivensectionis;2×6.324=12.64kg.

Example8

Thedepthofwaterinatankoflargecross-sectionareaismaintainedat40cm.Wateremergesinacontinuousstreamoutofahole2mminadiameterat thebase(a)Calculate:

(i)Speedatwhichthewateremergesfromthehole.

(ii)Massflux(densityofwater=1000kgm–3)(b)Whatassumptionshaveyoumadeinpart(a)?

Solution

Fig.10.24

AtpointX,P1=atmosphericpressurePh1=40cmVelocityv1=0AtpointY,P2=atmosphericpressurePh2=0velocity=v2

UsingBernoulli’sprinciple;

P+hpg+ pv2=constant

P1+h1pg+ pv12=P2+h2pg+ pv22

Substituting,P+(40×10-2×103×10)

Thereforev2=2.828ms–1

Massflux=volumefluxxdensity

=3.142×10-6×2.828×103

=8.89×10–3kgs–1

(b)Assumptions:(i)Theareaoftheholeisnegligiblecomparedtothatofthetank.(ii)Theflowisstreamline.

RevisionExercise10

1.Thewindowanddoorofahousearebothleftopenduringawindyday.Thedoor, which opens inwards, tends to slum shut when it is slightly ajar.Explain.

2.Explainyourobservationwhenairisblown:(a)Throughthetunnelshownindiagram(i)below.(b)Betweenlightballssuspendedonlightstringsasshownindiagram(ii).

3.Explainthefollowing:(a)Acartravellingathighspeedappearslighter.(b)Itisdangeroustostandclosetoarailwaylineonwhichafastmoving

trainispassing.(c)Anaeroplaneislikelytotakeoffmuchearlierthanexpectedwhenthe

speedofthewindblowingintheoppositedirectiontoitsmotionontherunwaysuddenlyincreases.

4.Explainusingadiagramwhyaspinningbaseballtravelsinaslightlycurvedpath.

5. State Bernoulli’s principle and write its equation. Define the physicalquantitieswhichappearinitandtheconditionsrequiredforitsvalidity.

6. Draw a cross-section of an airplane’s wing and clearly indicate the twocomponentsoftheresultantforceoftheairmovingovertheairplane.Whichcomponentisusefulandwhichoneactsasaretardingforce?

7. A ship’s captain, in anattempt towarnoff another ship steeredhis ship,parallelandclosetotheintrudership.Thetwoshipsfinallycollided.Whatcouldhavecausedtheaccident?

8.Whenastrongwindblewoveraprimaryschoolwithcrowdedbuildings,theroofswereblownoffandthewallsthatwereclosecollapsedtowardseachother. Explain these observations. What physical modifications could bemadeonthebuildingstominimisesuchhazards?

9. Redraw thediagrambelow, showing thenewmercury levelswhen air is

movingveryfastthroughthetube.

10. A venture tube is used to measure the speed v of a fluid in a pipe bycomparingthepressureinthewideandnarrowsections,as inthediagrambelow.Finvifthedifferencebetweenthemercurylevelinh=25mm.Thedensityofmercuryis13600kgm–3

11. Water flows steadily through ahorizontal pipewithvarying cross-sectionarea. At one point, the pressure is 130 kPa and the velocity is 1.20m/s.Determinethepressureatapointinthepipewherethespeedis12.0m/s.

12.Alargetankcontainswatertoadepthof2m.Wateremergesfromasmallholeonthesideofthetank50cmabovethebottomofthetankasbelow.

Calculatethe:(a)Speed(v)atwhichthewateremergesfromthehole.(b)Distancemarked(x).

13.(a)Explainwhythelevelsoftheliquidintheverticaltubesbelowarenotthesame.

(c)Usingadiagram,explainhowafilterpumpworks.14.Waterflowssteadilyalongahorizontalpipeatavolumerateof8×10–3m3

s–1.Iftheareaofcross-sectionofthepipeis20cm2:(a)Calculatethevelocityofthefluid.(b)Findthetotalpressureinthepipeifthestaticpressureinthehorizontal

pipeis1.0×104Pa,assumingthatwaterisincompressible,non-viscousanditsdensityis103kgm-3.

(c)Whatisthenewflowvelocityifthetotalpressureis2×104Pa?15.Airflowsovertheuppersurfacesofthewingsofanaeroplaneataspeedof

12.0ms–1. If the lift forceon theaeroplane is2.97×104N, calculate thespeedoftheairpastthelowersurfacesofthewing(takethetotalwingareatobe20.0m2andthedensityofairas1.29kgm–3).