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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
.
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