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CBSEClass11Physics

NCERTSolutions

Chapter9

MECHANICALPROPERTIESOFSOLIDS

1.ASteelwireoflength4.7mandthecross-sectionalareaof stretchesby

thesameamountasaCopperwireoflength3.5mandthecross-sectionalareaof

underagivenloadofthesameamount.WhatistheratioofYoung's

modulusofsteeltothatofCopper?

Ans.LengthoftheSteelwire,

Cross-sectionareaoftheSteelwire,

LengthoftheCopperwire, =3.5m

Cross-sectionareaoftheCopperwire,

Sincechangeinlengthisgivensameforboththecases,hence

Changeinlength= = =ΔL

Forceappliedinboththecases=

Young'smodulusofthesteelwire:

…(i)

Young'smodulusofthecopperwire:

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…(ii)

Dividing(i)by(ii),weget:

Hence,theratioofYoung'smodulusforSteeltothatofCopperis1.79:1.

2.Figure9.11showsthestress-straincurveforagivenmaterial.Whatare(a)Young's

modulusand(b)approximateyieldstrengthforthismaterial?

Ans.(a)Itisclearfromthegivengraphthatthestress ,strainis0.002.

∴Young'smodulus,Y

Hence,Young'smodulusforthegivenmaterialis .

(b)Theyieldstrengthofamaterialisthemaximumstressthatthematerialcansustain

withoutcrossingtheelasticlimit.

Itisclearfromthegivengraphthattheapproximateyieldstrengthofthismaterialis

.

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3.Thestress-straingraphsformaterialsAandBareshowninFig.9.12.

Thegraphsaredrawntothesamescale.

(a)WhichofthematerialshasthegreaterYoung'smodulus?

(b)Whichofthetwoisthestrongermaterial?

Ans.(a)A(b)A

(a)Foragivenstrain,thestressformaterialAismorethanitisformaterialB,asshownin

thetwographs.

Young'smodulus

Foragivenstrain,ifthestressforamaterialismore,thenYoung'smodulusisalsogreater

forthatmaterial.Therefore,Young'smodulusformaterialAisgreaterthanitisformaterial

B.

(b)Theamountofstressrequiredforfracturingamaterial,correspondingtoitsfracture

point,givesthestrengthofthatmaterial.Fracturepointistheextremepointinastress-

straincurve.ItcanbeobservedthatmaterialAcanwithstandmorestrainthanmaterialB.

Hence,materialAisstrongerthanmaterialB.

4.Readthefollowingtwostatementsbelowcarefullyandstate,withreasons,ifitis

trueorfalse.

(a)TheYoung'smodulusofrubberisgreaterthanthatofsteel;

(b)Thestretchingofacoilisdeterminedbyitsshearmodulus.

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Ans.(a)False(b)True

(a)Foragivenstress,thestraininrubberismorethanitisinsteel.

Young'smodulus,Y=

Foraconstantstress:

Hence,Young'smodulusforrubberislessthanitisforsteel.

(b)Shearmodulusistheratiooftheappliedstresstothechangeintheshapeofabody.The

stretchingofacoilchangesitsshape.Hence,shearmodulusofelasticityisinvolvedinthis

process.

5.Twowiresofdiameter0.25cm,onemadeofsteelandtheothermadeofbrassare

loadedasshowninFig.9.13.Theunloadedlengthofsteelwireis1.5mandthatofbrass

wireis1.0m.Computetheelongationsofthesteelandthebrasswires.

Ans.Elongationofthesteelwire= m

Elongationofthebrasswire= m

Diameterofthewires,d=0.25m

Hence,theradiusofthewires, =0.125cm

Lengthofthesteelwire, =1.5m

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Lengthofthebrasswire, =1.0m

Totalforceexertedonthesteelwire:

Young'smodulusforsteel:

Where,

=Changeinthelengthofthesteelwire

=Areaofcross-sectionofthesteelwire

Young'smodulusofsteel,

Totalforceonthebrasswire:

Young'smodulusforbrass:

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Where,

=Changeinlength

=Areaofcross-sectionofthebrasswire

Elongationofthesteelwire=

Elongationofthebrasswire=

6.Theedgeofanaluminumcubeis10cmlong.Onefaceofthecubeisfirmlyfixedtoa

verticalwall.Amassof100kgisthenattachedtotheoppositefaceofthecube.The

shearmodulusofaluminiumis25GPa.Whatistheverticaldeflectionofthisface?

Ans.Edgeofthealuminiumcube,L=10cm=0.1m

Themassattachedtothecube,m=100kg

Shearmodulus( )ofaluminium=

Shearmodulus,

Where,

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F=Appliedforce=mg=100 9.8=980N

A=Areaofoneofthefacesofthecube=0.1 0.1=0.01

ΔL=Verticaldeflectionofthecube

=

Theverticaldeflectionofthisfaceofthecubeis .

7.Fouridenticalhollowcylindricalcolumnsofmildsteelsupportabigstructureof

mass50,000kg.Theinnerandouterradiiofeachcolumnare30cmand60cm

respectively.Assumingtheloaddistributiontobeuniform,calculatethecompressional

strainofeachcolumn.

Ans.Massofthebigstructure,M=50,000kg

Innerradiusofthecolumn,r=30cm=0.3m

Outerradiusofthecolumn,R=60cm=0.6m

Young'smodulusofsteel,Y=

Totalforceexerted,F=Mg=50000 9.8N

Stress=Forceexertedonasinglecolumn =122500N

Young'smodulus,Y

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Where,

Area,A=

Hence,thecompressionalstrainofeachcolumnis .

8.Apieceofcopperhavingarectangularcross-sectionof15.2mm 19.1mmispulled

intensionwith44,500Nforce,producingonlyelasticdeformation.Calculatethe

resultingstrain?

Ans.Lengthofthepieceofcopper,l=19.1mm=

Breadthofthepieceofcopper,b=15.2mm=

Areaofthecopperpiece:

A=l b

=

=

Tensionforceappliedonthepieceofcopper,F=44500N

Modulusofelasticityofcopper,

Modulusofelasticity,

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=

9.Asteelcablewitharadiusof1.5cmsupportsachairliftataskiarea.Ifthemaximum

stressisnottoexceed N ,whatisthemaximumloadthecablecansupport?

Ans.Radiusofthesteelcable,r=1.5cm=0.015m

Maximumallowablestress= N

Maximumstress=

∴Maximumforce=Maximumstress Areaofcross-section

=

=

Hence,thecablecansupportthemaximumloadof7 .

10.Arigidbarofmass15kgissupportedsymmetricallybythreewireseach2.0mlong.

Thoseateachendareofcopperandthemiddleoneisofiron.Determinetheratioof

theirdiametersifeachistohavethesametension.

Ans.Thetensionforceactingoneachwireisthesame.Thus,theextensionineachcaseis

thesame.Sincethewiresareofthesamelength,thestrainwillalsobethesame.

TherelationforYoung'smodulusisgivenas:

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……..(i)

Where,

F=Tensionforce

A=Areaofcross-section

d=Diameterofthewire

Itcanbeinferredfromequation(i)that

Young'smodulusforiron,

Diameteroftheironwire=

Young'smodulusforcopper,

Diameterofthecopperwire=

Therefore,theratiooftheirdiametersisgivenas:

11.A14.5kgmass,fastenedtotheendofasteelwireofunstretchedlength1.0m,is

whirledinaverticalcirclewithanangularvelocityof2rev/satthebottomofthe

circle.Thecross-sectionalareaofthewireis0.065 .Calculatetheelongationofthe

wirewhenthemassisatthelowestpointofitspath.

Ans.Mass,m=14.5kg

Lengthofthesteelwire,l=1.0m

Angularvelocity, =2rev/s=2 2πrad/s=12.56rad/s

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Cross-sectionalareaofthewire,

LetΔlbetheelongationofthewirewhenthemassisatthelowestpointofitspath.

Whenthemassisplacedatthepositionoftheverticalcircle,thetotalforceonthemassis:

F=mg+

=

=2429.53N

Young’smodulus

Young'smodulusforsteel=

Hence,theelongationofthewireis

12.Computethebulkmodulusofwaterfromthefollowingdata:Initialvolume=100.0

litre,Pressureincrease=100.0atm(1atm= ),Finalvolume=100.5litre.

Comparethebulkmodulusofwaterwiththatofair(atconstanttemperature).Explain

insimpletermswhytheratioissolarge.

Ans.Initialvolume,

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Finalvolume,

Increaseinvolume,

Increaseinpressure,Δp=100.0atm=100 1.013 105Pa

Bulkmodulus=

Bulkmodulusofair

Thisratioisveryhighbecauseairismorecompressiblethanwater.

13.Whatisthedensityofwateratadepthwherepressureis80.0atm,giventhatits

densityatthesurfaceis ?

Ans.Letthegivendepthbeh.

Pressureatthegivendepth,p=80.0atm=

Densityofwateratthesurface,

Let bethedensityofwateratthedepthh.

Let bethevolumeofwaterofmassmatthesurface.

Let bethevolumeofwaterofmassmatthedepthh.

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LetΔVbethechangeinvolume.

Volumetricstrain=

…………(i)

Bulkmodulus,

Compressibilityofwater

………..(ii)

Forequations(i)and(ii),weget:

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Therefore,thedensityofwateratthegivendepth(h)is .

14.Computethefractionalchangeinvolumeofaglassslab,whensubjectedtoa

hydraulicpressureof10atm.

Ans.Hydraulicpressureexertedontheglassslab,p=10atm=

Bulkmodulusofglass,B=

Bulkmodulus,

Where, =Fractionalchangeinvolume

Hence,thefractionalchangeinthevolumeoftheglassslabis .

15.Determinethevolumecontractionofasolidcoppercube,10cmonanedge,when

subjectedtoahydraulicpressureof Pa.

Ans.Lengthofanedgeofthesolidcoppercube,l=10cm=0.1m

Hydraulicpressure,p= Pa

Bulkmodulusofcopper,B= Pa

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Bulkmodulus,

Where, =Volumetricstrain

ΔV=Changeinvolume

V=Originalvolume.

Originalvolumeofthecube,V=

Therefore,thevolumecontractionofthesolidcoppercubeis .

16.Howmuchshouldthepressureonalitreofwaterbechangedtocompressitby

0.10%?

Ans.Volumeofwater,V=1L

Itisgiventhatwateristobecompressedby0.10%.

Fractionalchange,

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Bulkmodulus,

Bulkmodulusofwater,B=

Therefore,thepressureonwatershouldbe .

17.Anvilsmadeofsinglecrystalsofdiamond,withtheshapeasshowninFig.9.14,are

usedtoinvestigatebehaviourofmaterialsunderveryhighpressures.Flatfacesatthe

narrowendoftheanvilhaveadiameterof0.50mm,andthewideendsaresubjectedto

acompressionalforceof50,000N.Whatisthepressureatthetipoftheanvil?

Ans.Diameteroftheconesatthenarrowends,d=0.50mm= m

Radius,r=

Compressionforce,F=50000N

Pressureatthetipoftheanvil:

Therefore,thepressureatthetipoftheanvilis2.55 Pa.

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18.Arodoflength1.05mhavingnegligiblemassissupportedatitsendsbytwowiresof

steel(wireA)andaluminium(wireB)ofequallengthsasshowninFig.9.15.Thecross-

sectionalareasofwiresAandBare1.0mm2and2.0mm2,respectively.Atwhatpoint

alongtherodshouldamassmbesuspendedinordertoproduce(a)equalstressesand

(b)equalstrainsinbothsteelandaluminiumwires.

Ans.(a)0.7mfromthesteel-wireend

(b)0.432mfromthesteel-wireend

Cross-sectionalareaofwireA,

Cross-sectionalareaofwireB,

Young'smodulusforsteel,

Young'smodulusforaluminium,

(a)LetasmallmassmbesuspendedtotherodatadistanceyfromtheendwherewireAis

attached.

Stressinthewire=

Ifthetwowireshaveequalstresses,then:

Where, =Forceexertedonthesteelwire

=Forceexertedonthealuminumwire

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…………….(i)

Thesituationisshowninthefollowingfigure.

Takingtorqueaboutthepointofsuspension,wehave:

…..(ii)

Usingequations(i)and(ii),wecanwrite:

Inordertoproduceanequalstressinthetwowires,themassshouldbesuspendedata

distanceof0.7mfromtheendwherewireAisattached.

(b)Young’smodulus=

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Ifthestraininthetwowiresisequal,then:

………..(iii)

Takingtorqueaboutthepointwheremassm,issuspendedatadistancey1fromtheside

wherewireAattached,weget:

….(iii)

Usingequations(iii)and(iv),weget:

Inordertoproduceanequalstraininthetwowires,themassshouldbesuspendedata

distanceof0.432mfromtheendwherewireAisattached.

19.Amildsteelwireoflength1.0mandcross-sectionalarea is

stretched,wellwithinitselasticlimit,horizontallybetweentwopillars.Amassof100g

issuspendedfromthemid-pointofthewire.Calculatethedepressionatthemidpoint.

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Ans.

Lengthofthesteelwire=1.0m

Areaofcross-section,A=

Amass100gissuspendedfromitsmidpoint.

m=100g=0.1kg

Hence,thewiredips,asshowninthegivenfigure.

Originallength=XZ

Depression=l

Thelengthaftermassm,isattachedtothewire=XO+OZ

Increaseinthelengthofthewire:

Δl=(XO+OZ)–XZ

Where,XO=OZ=

Expandinganneglectinghigherterms,weget:

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Strain=

LetTbethetensioninthewire.

∴mg=2Tcosθ

Usingthefigure,itcanbewrittenas:

Expandingtheexpressionandeliminatingthehigherterms:

Stress=

Young’smodulus=

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Young'smodulusofsteel,Y=

Hence,thedepressionatthemidpointis0.0106m.

20.Twostripsofmetalarerivetedtogetherattheirendsbyfourrivets,eachof

diameter6.0mm.Whatisthemaximumtensionthatcanbeexertedbytherivetedstrip

iftheshearingstressontherivetisnottoexceed Pa?Assumethateachrivet

istocarryonequarteroftheload.

Ans.Diameterofthemetalstrip,d=6.0mm= m

Radius,r=

Maximumshearingstress

Maximumstress=

Maximumforce=Maximumstress Area

=

=

=1949.94N

Eachrivetcarriesonequarteroftheload.

∴Maximumtensiononeachrivet=4 1949.94=7799.76N

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21.TheMarinatrenchislocatedinthePacificOcean,andatoneplaceitisnearly

elevenkmbeneaththesurfaceofwater.Thewaterpressureatthebottomofthetrench

isabout .Asteelballofinitialvolume0.32 bisdroppedintotheocean

andfallstothebottomofthetrench.Whatisthechangeinthevolumeoftheballwhen

itreachestothebottom?

Ans.Waterpressureatthebottom,p=

Initialvolumeofthesteelball,V=0.32

Bulkmodulusofsteel,B=

TheballfallsatthebottomofthePacificOcean,whichis11kmbeneaththesurface.

LetthechangeinthevolumeoftheballonreachingthebottomofthetrenchbeΔV.

Bulkmodulus,B=

Therefore,thechangeinvolumeoftheballonreachingthebottomofthetrenchis

.