CBSE NCERT Solutions for Class 9 science Chapter 10
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CBSE NCERT Solutions for Class 9 science Chapter 10 Exercises Q.1. How does the force of gravitation between two objects change when the distance between them is reduced to half? Solution: The law of gravitation states that the gravitational force F acting between two objects is inversely proportional to the square of the distance r between them. Hence, if the distance is reduced to half, then the gravitational force would become four times. Q.2. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? Solution: Amit's friend will not agree with the weight of the gold bought. Weight of a body on the earth is given by: W=mg where, m = Mass of the body g = Acceleration due to gravity The value of g is greater at poles than at the equator. Therefore, the same mass of gold weighs lesser at the equator than at the poles. Q.3. Why will a sheet of paper fall slower than one that is crumpled into a ball? Solution: Being the same mass, the force due to gravity on the sheet and the ball will be the same. A sheet of paper has more surface area than a crumpled ball of paper. Hence, the resistance offered by air to a sheet of paper falling through it is more than the resistance offered to a falling crumpled ball of paper. This decreases the speed of the sheet of paper and hence it falls slower than the crumpled ball. Q.4. Gravitational force on the surface of the moon is only 16 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth? Solution: Weight = Mass× Acceleration Acceleration due to gravity on earth, ge=9.8 m/s2 Therefore, weight of a 10 kg object on the earth = 10×9.8=98 N As given in the question, acceleration due to gravity on the moon, gm=16ge=9.86 m/s2 Therefore, the weight of the same object on the moon = 10×9.86=16.3 N Q.5. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (Take g=9.8 m/s2) (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth. Solution: (i) v 2-u 2=2gh where, u = Initial velocity of the ball = 49 m/s (Given) v = Final velocity of the ball = 0 (At the highest point) h = Maximum height attained by the ball g = Acceleration due to gravity = -9.8 m/s2 (Ball goes up) Putting the values, we get 02-492=2×-9.8×h h=4922×9.8=122.5 m (ii) Let t be the time taken by the stone to reach the highest point. v=u+gt0=49-9.8tt=499.8=5 s Since the time of ascent is equal to the time of descent, the time taken for the stone to reach the earth's surface is, 2t=2×5=10 s Q.6. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. (Take g=9.8 m/s2) NCERT Science Grade 9 Chapter 10 Gravitation Practice more on Gravitation Page 1 www.embibe.com
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Transcript of CBSE NCERT Solutions for Class 9 science Chapter 10
CBSENCERTSolutionsforClass9scienceChapter10
Exercises
Q.1. Howdoestheforceofgravitationbetweentwoobjectschangewhenthedistancebetweenthemisreducedtohalf?
Solution: ThelawofgravitationstatesthatthegravitationalforceFactingbetweentwoobjectsisinverselyproportionaltothesquareofthedistancerbetweenthem.Hence,ifthedistanceisreducedtohalf,thenthegravitationalforcewouldbecomefourtimes.
Q.2. Amitbuysfewgramsofgoldatthepolesaspertheinstructionofoneofhisfriends.Hehandsoverthesamewhenhemeetshimattheequator.Willthefriendagreewiththeweightofgoldbought?Ifnot,why?
Solution: Amit'sfriendwillnotagreewiththeweightofthegoldbought.Weightofabodyontheearthisgivenby:W=mg
where,m=Massofthebodyg=AccelerationduetogravityThevalueofgisgreateratpolesthanattheequator.Therefore,thesamemassofgoldweighslesserattheequatorthanatthepoles.
Q.3. Whywillasheetofpaperfallslowerthanonethatiscrumpledintoaball?
Solution: Beingthesamemass,theforceduetogravityonthesheetandtheballwillbethesame.Asheetofpaperhasmoresurfaceareathanacrumpledballofpaper.Hence,theresistanceofferedbyairtoasheetofpaperfallingthroughitismorethantheresistanceofferedtoafallingcrumpledballofpaper.Thisdecreasesthespeedofthesheetofpaperandhenceitfallsslowerthanthecrumpledball.
Q.4. Gravitationalforceonthesurfaceofthemoonisonly16asstrongasgravitationalforceontheearth.Whatistheweightinnewtonsofa10kgobjectonthemoonandontheearth?
Solution: Weight=Mass×AccelerationAccelerationduetogravityonearth,ge=9.8m/s2Therefore,weightofa10kgobjectontheearth=10×9.8=98N
Asgiveninthequestion,accelerationduetogravityonthemoon,gm=16ge=9.86m/s2Therefore,theweightofthesameobjectonthemoon=10×9.86=16.3N
Q.5. Aballisthrownverticallyupwardswithavelocityof49m/s.Calculate(Takeg=9.8m/s2)(i)themaximumheighttowhichitrises,(ii)thetotaltimeittakestoreturntothesurfaceoftheearth.
Solution: (i)v2-u2=2ghwhere,u=Initialvelocityoftheball=49m/s(Given)v=Finalvelocityoftheball=0(Atthehighestpoint)h=Maximumheightattainedbytheballg=Accelerationduetogravity=-9.8m/s2(Ballgoesup)Puttingthevalues,weget02-492=2×-9.8×hh=4922×9.8=122.5m
(ii)Lettbethetimetakenbythestonetoreachthehighestpoint.v=u+gt0=49-9.8tt=499.8=5sSincethetimeofascentisequaltothetimeofdescent,thetimetakenforthestonetoreachtheearth'ssurfaceis,2t=2×5=10s
Q.6. Astoneisreleasedfromthetopofatowerofheight19.6m.Calculateitsfinalvelocityjustbeforetouchingtheground.(Takeg=9.8m/s2)
NCERTScienceGrade9 Chapter10Gravitation
PracticemoreonGravitation Page1 www.embibe.com
Solution: Given,Theinitialvelocityofthestone,u=0Finalvelocityofthestone,v=?Heightofthetower,h=19.6mAccelerationduetogravity,g=9.8m/s2Usingtheequationofmotion,v2-u2=2ghv2-0=2×9.8×19.6v2=19.62v=19.6m/sHence,thevelocityofthestonejustbeforetouchingthegroundis19.6m/s.
Q.7. Astoneisthrownverticallyupwardwithaninitialvelocityof40m/s.Takingg=10m/s2,findthemaximumheightreachedbythestone.Whatarethenetdisplacementandthetotaldistancecoveredbythestone?
Solution: Theinitialvelocityofthestone,u=40m/s(Given)Thefinalvelocityofthestone,v=0(Atthehighestpoint)Themaximumheightreachedbythestone,h=?Accelerationduetogravity,g=-10ms-2(Stonegoesup)Usingtheequationofmotion:v2-u2=2gh
Puttingthevalues,weget,0-402=2×10×h
h=4022×10=80m
Therefore,totaldistancecoveredbythestoneduringitsupwardanddownwardjourney=80+80=160mThenetdisplacementofthestoneduringitsupwardanddownwardjourney=0(sincefinalpositioncoincideswiththeinitialposition).
Q.8. CalculatetheforceofgravitationbetweentheearthandtheSun,giventhatthemassoftheearth=6×1024kgandoftheSun=2×1030kg.Theaveragedistancebetweenthetwois1.5×1011mandthegravitationalconstantisG=6.7×10-11Nm2kg-2.
Solution: TheforceofattractionbetweentheearthandtheSunisgivenby:F=GMsunMearthr2Here,MSun=MassoftheSun=2×1030kgMEarth=MassoftheEarth=6×1024kgr=AveragedistancebetweentheearthandtheSun=1.5×1011mG=Universalgravitationalconstant=6.7×10-11Nm2kg-2Onsubstitutingthevalues,wegetF=6.7×10-11×2×1030×6×10241.5×10112N=3.57×1022N
Q.9. Astoneisallowedtofallfromthetopofatower100mhighandatthesametime,anotherstoneisprojectedverticallyupwardsfromthegroundwithavelocityof25m/s.Calculatewhenandwherethetwostoneswillmeet.(Useg=10m/s2)
NCERTScienceGrade9 Chapter10Gravitation
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Solution: Letthetwostonesmeetafteratimetandatheighthabovetheground.
Forthestonedroppedfromthetopofthetower:Initialvelocity,u1=0Thedisplacementofthestoneintimetis100–h.Accelerationduetogravity,g=10ms-2usingtheequationofmotion,s=ut+12gt2100-h=0+12×10×t2-----------(1)
Forthestonethrownupwards:Initialvelocity,u2=25ms-1Thedisplacementofthestonefromthegroundintimetbeh.Accelerationduetogravity,g=-10ms-2Usingtheequationofmotion,s=ut+12gt2h=25t-12×10×t2---------(2)
Addingequation(1)andequation(2),weget100=25tt=10025s=4sTocalculatetheheightofthestonein4s,useequation(2).h=25×4-12×10×42=20m
Q.10. Aballthrownupverticallyreturnstothethrowerafter6s.(Useg=10m/s2)Find(a)thevelocitywithwhichitwasthrownup,(b)themaximumheightitreaches,and(c)itspositionafter4s.
Solution: (a)Timetogoupanddown=6sLettheinitialvelocityoftheballbeu.Displacement,s=0Accelerationduetogravity,g=-10m/s2Usingtheequationofmotion,s=ut+12gt20=u×6-12×10×62u=30m/s
(b)Letthemaximumheightattainedbytheballbeh.Initialvelocity,u=30m/sFinalvelocity,v=0Accelerationduetogravity,g=-10m/s2Usingtheequationofmotion,v2=u2-2gh0=302-2×10×hh=3022×10m=45m
(c)findthepositionafter4s:u=30m/st=4sg=-10m/s2Usingtheequationofmotions=ut+12gt2s=30×4-12×10×42=40m
Q.11. Inwhatdirectiondoesthebuoyantforceonanobjectimmersedinaliquidact?
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Solution: Thebuoyantforceonanobjectimmersedinaliquidactsinanupwarddirection.Buoyancyistheupwardforceexertedbyafluidthatopposestheweightofanimmersedobject.
Q.12. Gravitationalforceactsonallobjectsinproportiontotheirmasses.Whythen,aheavyobjectdoesnotfallfasterthanalightobject?
Solution: Thespeedofthefallingobjectdependsontheaccelerationofthebodyandisindependentofthemassoftheobject.Hence,theaccelerationduetogravityisthesameandalltheobjectswillfallwiththesamespeed.
Q.13. Whydoesablockofplasticreleasedunderwatercomeuptothesurfaceofthewater?
Solution: Incaseofablockofplastic,theupwardbuoyantforceisgreaterthantheweightoftheobject.Thelargebuoyantforceontheblockisduetoitsdensitybeingsmallerthanthatofwater.Duetothelargerbuoyantforce,theblockofplasticcomesupwhenreleasedunderwater.
Q.14. Thevolumeof50gofasubstanceis20cm3.Ifthedensityofwateris1gcm-3willthesubstancefloatorsink?
Solution: Thedensityofthesubstanceisgivenby:density=massvolume=50g20cm3=2.5gcm-3Sincethedensityofthesubstanceisgreaterthanwater,thesubstancewillsink.
Q.15. Thevolumeofa500gsealedpacketis350cm3.Willthepacketfloatorsinkinwaterifthedensityofwateris1gcm-3?Whatwillbethemassofthewaterdisplacedbythispacket?
Solution: Thedensityofthesealedpacketiscalculatedas:Density=massvolume=500350gcm-3=1.43gcm-3
Thedensityofthesealedpacket(1.43gcm-3)ismorethanthedensityofwater(1gcm-3).Hence,itwillsinkinwater.
Massofthewaterdisplacedbypacket:=volumeofthepacket×densityofthewater=350×1=350g
Q.16. Whatisthemagnitudeofthegravitationalforcebetweentheearthanda 1kgobjectonitssurface?(Massoftheearthis6×1024kg,radiusoftheearthis6.4×106mandthegravitationalconstantisG=6.7×10-11Nm2kg-2).
Solution: Asperthelawsofgravitation,thegravitationalforcebetweentwoobjectsofmassesMandmatadistancerfromeachotherisgivenby:F=GMmr2
ThemassofearthberepresentedbyM=6×1024kgThemassoftheobjectberepresentedbym=1kgUniversalgravitationalconstant,G=6.7×10-11Nm2kg-2SincetheobjectisonthesurfaceoftheEarth,r=radiusoftheEarth=(R)r=R=6.4×106mF=GMmr2=6.7×10-11×6×1024×16.4×1062=9.8N
Q.17. Theearthandthemoonareattractedtoeachotherbygravitationalforce.Doestheearthattractthemoonwithaforcethatisgreaterorsmallerorthesameastheforcewithwhichthemoonattractstheearth?Why?
Solution: TheearthattractsthemoonwithaforcewhichissameastheforcewithwhichthemoonattractstheearthbecauseaccordingtoNewton'sthirdlawofmotion,theforceofactionandreactionarealwaysequalandopposite.Therefore,theforceofattractionoftheearthonthemoonisequalandoppositetotheforceofattractionofthemoonontheearth.
Q.18. Ifthemoonattractstheearth,whydoestheearthnotmovetowardsthemoon?
Solution: Themotionofanobjectdependsuponitsacceleration,notonitsforce.Itistruethattheforceonbothisthesame;theaccelerationwillbedifferent.Asthemassoftheearthisverylarge,itsaccelerationisverysmall,andhence,theearthdoesnotmovetowardsthemoon.
Q.19. Whathappenstotheforcebetweentwoobjects,ifthemassofoneobjectisdoubled?
NCERTScienceGrade9 Chapter10Gravitation
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Solution: Asperthelawofgravitation,theforceofgravitationbetweentwoobjectsisgivenby:
F=Gm1m2r2
Fisdirectlyproportionaltotheproductofmassesoftheobjectsandisinverselyproportionaltothesquareoftheseparationdistancebetweenthetwoobjects.Sincethemassofoneobjectisdoubled,thenthegravitationalforcewillalsogetdoubled.
Q.20. Whathappenstotheforcebetweentwoobjects,ifthedistancebetweentheobjectsisdoubledandtripled?
Solution: Asperthelawsofgravitation,theforceofgravitationbetweentwoobjectsisgivenby:
F=Gm1m2r2
Fisinverselyproportionaltothesquareofthedistancebetweentheobjects.Ifthedistanceisdoubled,thenthegravitationalforcebecomesone-fourthofitsoriginalvalue.Similarly,ifthedistanceistripled,thenthegravitationalforcebecomesone-ninthofitsoriginalvalue.
Q.21. Whathappenstotheforcebetweentwoobjects,ifthemassesofbothobjectsaredoubled?
Solution: Asperthelawsofgravitation,theforceofgravitationbetweentwoobjectsisgivenby:
F=Gm1m2r2
Fisdirectlyproportionaltotheproductofmassesoftheobjects.Ifthemassesofboththeobjectsaredoubled,theproductofmassesbecomesfourtimesandhencethegravitationalforcewillalsobecomefourtimestheoriginalvalue.
Q.22. Whatistheimportanceofuniversallawofgravitation?
Solution: Importanceofuniversallawsofgravitation:ItexplainsthemotionofplanetsaroundtheSun,themotionofthemoonandotherartificialsatellitesaroundtheearth,tidesduetothemoonandtheSunandmanyotherphenomena.
Q.23. Whatistheaccelerationoffreefall?
Solution: Anobjectisinastateoffreefallifitismovingundertheinfluenceofonlygravitationalforce.Neartheearth,thisforceproducesauniformaccelerationintheobjects,whichiscalledaccelerationoffreefalloraccelerationduetogravity.Itsvalueis9.8m/s2.
Q.24. Whatdowecallthegravitationalforcebetweentheearthandanobject?
Solution: Thegravitationalforcebetweentheearthandanobjectiscalledtheweightoftheobject.Weightisalsoequaltoproductofmassandaccelerationduetogravityofanobject.
NCERTScienceGrade9 Chapter10Gravitation
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Questions
Q.1. Statetheuniversallawofgravitation.
Solution: TheuniversallawofgravitationwasformulatedbySirIsaacNewton.Theuniversallawofgravitationstatesthateveryobjectintheuniverseattractseveryotherobjectwithaforcethatisproportionaltotheproductoftheirmassesandinverselyproportionaltothesquareofthedistancebetweenthem.Theforceactsalongthelinejoiningthecentresoftwoobjects.
Q.2. Whatdoyoumeanbyfreefall?
Solution: Whenanobjectisfallingundertheinfluenceofgravity,itiscalledfreefall.Inthecaseoffreefall,theonlyforceactingontheobjectshouldbethegravityoftheearth.Whenanobjectisreleasedfromaheight,itfallstowardsthesurfaceoftheEarthundertheinfluenceofgravitationalforcealoneandthismotionoftheobjectiscalledfreefall.
Q.3. Whatarethedifferencesbetweenthemassofanobjectanditsweight?
Solution: Mass WeightItisthequantityofmattercontainedinabody
Itistheforceexertedbythegravityonthebody
Massisconstant Weightisnotconstantanddiffersatdifferentplaces
Mass WeightIthasonlymagnitude IthasmagnitudeaswellasdirectionItsSIunitiskilogram(kg) ItsSIunitisthesameasforce,i.enewton(N)
Q.4. Whyisitdifficulttoholdaschoolbaghavingastrapmadeofathinandstrongstring?
Solution: Sincepressureisdefinedastheforceexertedbyanobjectperunitarea.Forathinstrap,theareaissmallduetowhichthepressureexertedwillbelarge.Sincethepressureislarge,itwillbedifficulttoholdthebag.
Q.5. Youfindyourmasstobe42kgonaweighingmachine.Isyourmassmoreorlessthan42kg?
Solution: Weighingmachineweighsourweight.Duetothebuoyancyforceofair,ourweightwillberecordedslightlyless.Hence,theactualmasswillbemorethan42kg.
Note:Massistheamountofmatterpresentinthebodyandweightofthebodyistheproductofmassandgravitationalacceleration
Q.6. Writetheformulatofindthemagnitudeofthegravitationalforcebetweentheearthandanobjectonthesurfaceoftheearth.
Solution: Consideranobjectofmassmonthesurfaceoftheearth.LetthemassofearthbeM.IfRistheradiusoftheEarth,thenaccordingtotheuniversallawofgravitation,thegravitationalforce(F)actingbetweentheEarthandtheobjectisgivenbytherelationF=GMmR2,whereGistheuniversalgravitationconstantgivenby,G=6.67×10-11Nm2kg-2.
Q.7. Whatdoyoumeanbyaccelerationduetogravity?
Solution: Anobjectmovingundergravityexperiencesaforceandhenceacceleration.Theaccelerationoftheobjectduetotheforceofgravityiscalledaccelerationduetogravity.Thevalueofaccelerationduetogravitychangesfromplacetoplaceonthesurfaceoftheearthandalsoaswegoupfromthesurface.Onthesurface,itsvalueisnearlyconstantequalto9.8m/s2.
Q.8. Whyistheweightofanobjectonthemoon16thitsweightontheearth?
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Solution: Weightofanobjectistheproductofmassandtheaccelerationduetogravity.Theaccelerationduetogravityg=GMR2.
LetMeandMmbethemassesoftheearthandthemoon,respectively.And,ReandRmbetheirradii.TheweightofanobjectofmassmontheearthsurfaceWg=GMemRe2
TheweightofanobjectofmassmonthemoonsurfaceWm=GMmmRm2
HencetherequiredratioWgWm=GMemRe2÷GMmmRm2=MeMm×Rm2Re2
Weknowthat
CelestialBody Mass(kg) Radius(m)Earth 5.98×1024 6.37×106Moon 7.36×1022 1.74×106
Hence,WgWm=5.98×10247.36×1022×1.74×1066.37×1062=6Theweightonthesurfaceofthemoonwillbe1/6timesthatontheearth.
Q.9. Whatdoyoumeanbybuoyancy?
Solution: Whenanobjectisimmersedinafluid,eitherpartiallyorfully,itexperiencesanupwardforcebythefluid.Thisforceiscalledbuoyancyforceorupthrust.Thisisthenetreactionforcebythefluidontheobject.
Q.10. Youhaveabagofcottonandanironbar,eachindicatingamassof100kgwhenmeasuredonaweighingmachine.Inreality,oneisheavierthanother.Canyousaywhichoneisheavierandwhy?
Solution: Measuredweight=Actualweight–BuoyantforceHence,Actualweight=Measuredweight+Buoyantforce
Asvolumeofcottonismore,thebuoyantforcewillbemore.Hence,theactualweightofthecottonwillbemorethanthatofanironbar.
Q.11. Whydoesanobjectfloatorsinkwhenplacedonthesurfaceofwater?
Solution: Ifthedensityofanobjectisgreaterthanthedensityofwater,itwillsinkandifthedensityofanobjectislessthanthedensityofwater,thenitwillfloat.
Note:Densityisdefinedasmassperunitvolumeofanobject.
NCERTScienceGrade9 Chapter10Gravitation
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NCERTScienceGrade9 Chapter10Gravitation
PracticemoreonGravitation Page8 www.embibe.com
