4E Phy Prelim 2021 Answer

1 BLS4EPhyPrelim2021P1Answer 1. D 2. D 3. B 4. D 5. C6. B 7. C 8. B 9. C 10. A11. C 12. C 13. D 14. C 15. D16. B 17. C 18. B 19. D 20. A21. B 22. A 23. C 24. D 25. C26. D 27. B 28. A 29. C 30. A31. B 32. B 33. B 34. B 35. B36. C 37. B 38. A 39. B 40. C 1 DTimefromX–Yistimeforhalfaperiod.Timefor1period=75s/25=3.0sTimeforhalfaperiod=3.0s/2=1.5s 2 D 3 BUsingF=ma,5000–T=500×2T=5000–1000=4000N 4 DOrder of movement of thecar: stationary, increasingvelocity, constant velocityanddecreasingvelocityinthesamedirection(soAiswrongasforthelastpart,thecarmovesintheoppositedirection). 5 CSinceF2 >F1 ,theobjectwillmovetowardstherightandfrictionactstowardtheleft.Fortheobjecttoaccelerate,theresultantforcemustbemorethanzero.F2 –F1 –F>0OnlyoptionCispossible. 6 B 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑏𝑙𝑜𝑐𝑘 = 5 𝑐𝑚× 4 𝑐𝑚× 10 𝑐𝑚 ( )– 2 𝑐𝑚× 3 𝑐𝑚× 10 𝑐𝑚( )

2 = 200 𝑐𝑚 3 − 60 𝑐𝑚 3 = 140 𝑐𝑚 3 𝑚𝑎𝑠𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦×𝑣𝑜𝑙𝑢𝑚𝑒 = 15 𝑔/𝑐𝑚 3 ×140 𝑐𝑚 3 = 2100 𝑔 7 CThecenterofgravityisatposition5,thecenterofthebeam.Inorderforthebeamtobebalanced, thepivot shouldbebetweenposition5andposition7(whereW2 is).OnlypossibleanswerisC. 8 BTakingmomentaboutZ,takingclockwisetobepositive.Resultantmoment=20×3–10×2–5×2=30Nmclockwise 9 CInfig.9.1,pressureoftrappedair=75cmHg+5cmHg=80cmHgInFig.9.2,Pressureoftrappedair=75cmHg–5cmHg=70cmHgComparethetwosituations,theproductofpressureandvolumeshouldbethesame.Sincethetubeusedarethesame,thecrosssectionalareawillbethesame.P1 V1 =P2 V2 P1 L1 =P2 L2 80×8=70×L2 L2 =9.1cm(2s.f.) 10 APressureatXandYarethesamesincetheyareatthesamelevel.PX =PY hX pX g=hY pY ghY =hX pX ÷pY =8cm 11 CUsefulworkdone=Workdoneonthebox–workdoneagainstfriction=Fd–Rd 12 CPowerisdefinedastheworkdoneperunittime.P=WD÷t=F×d÷t=F×v (assumingthewatermoveatconstantspeed,v=d÷t)=mg×v (forcerequiredistheweightofthewaterpumpedout,W=mg)=150×10×20 (for1second)=30000W=30kW 13 DOptionA:Sincemoregasisaddedintothevessel,thusthenumberofparticlesperunitvolumeincreases. Thefrequencyof collisionwiththewall of thevesselincreasesthuspressureincreases.OptionB:Theintermoleculardistancedecreasesduetothenumberofparticlesperunitvolumeincreases.

3 OptionC: Sincemoregasisadded, massof themassincreasesthustheweightincreases.Option D: Average kinetic energy remains constant as it is proportional to thetemperatureofthegas.Sincethetemperatureiskeptconstant,theaveragekineticenergywillnotchange. 14 CCeramictileisabetterheat conductorthanthewoolencarpet,thusheattransferisslowerthroughthewoollencarpetascomparedtotheceramictile.Thusthewoollencarpetwillfeelwarmerthantheceramictile. 15 DSinceradiationistheonlymethodofheattransferthatisonlytotravelinvacuum,heatfromthesuncanonlybetransfertoearththroughradiation. 16 BThethermometer showserrorinmeasuringtemperature(constantlymeasured1 o Clower) but if you are measuring the difference in temperature, the error will beeliminated.Differencebetweenmeltingiceandboilingwater=99–(−1)=100 o C 17 CΘ=(Rθ –R0 )÷(R100 –R0 )×10020=(Rθ –2)÷(2.5–2)×100Rθ =2.1Ω 18 BEnergyrequiredtomelt8.0goficecube= 8 𝑔1000 𝑔 ×336 𝑘𝐽 / 𝑘𝑔 = 2688 𝐽Timetakentomelttheicecube= 2688 𝐽1.3 𝐽 / 𝑠 = 2070 𝑠 19 DEnergygainedbythe2kgwater=energylostbythe4kgwaterM2 c ΔΘ2 =M4 c ΔΘ4 2×c×(Θ–10)=4×c×(70–Θ)Θ–10=140–2ΘΘ=50 0 C 20 A Sincethewavemovetotheleft,drawasecondwavedisplacetotheleft,thisrepresenthowthewavewillbelikeinthenextinstant.Theparticleswillmovefromthefirstwavetothesecondwavewithonlyupordownmotion.ParticlePwillmoveupandparticleQwillmovedown. 21 BFrequencyisdefinedasthenumberofcompletewaveperunittime.5completewavestakes6second.Frequency=5÷6=0.83Hz(2s.f.) 22 AWhenalightraytravelsfromwithinanopticallydensermediumtoamediumthatisopticallylessdenseatthecriticalangle,theangleofrefractionwillbe90°,anditwill

4 travelalongtheboundary. 23 CEverysecond,themirrorwillmoveadistanceof2m,theimagewillbe4mfurtherawayfromtheman.Thusthespeedwhichtheimageismovingis4m/s. 24 DDue to the high energy of ultraviolet radiation, it can beusetosterilizemedicalequipment. 25 CThefrequencyofsounddoesnotchangeasthewavetravelsfromonemediumtoanother.Usingv=fλ,320=f×0.5f=320÷0.5=640Hz 26 DLoudnessisproportionaltoamplitude,theamplitudeofthenewtraceshouldbelarger.Frequencyisproportional topitch, thefrequencyof thenewtraceshouldbehigher.Thismeansthatmorewavescanbeseenwiththesametime. 27 BStaticshockshappenwhenabuiltupofelectronsisdischarged.Movementofprotons,neutronsandcationsinvolvemovingthewholeatom/molecules. 28 ASincerodXandrodYrepels,XandYhavethesamecharge.OnlyoptionAandDarepossible.SincerodXandrodZattracts,XandZhaveoppositechargesorZisneutral.OnlyoptionAispossible. 29 CLampQhavethesamebrightnessasthepotentialdifferenceacrossQisnotaffectedbythejockeyconnectiononXY.WhenthejockeyslidesalongXtoY, ashortcircuitisintroducedinthecircuit.TheeffectiveresistancealongpathwithlampPwilldecrease.ThecurrentinthepathwithlampPwillincrease.ThepotentialdifferenceacrosslampPwillalsoincreaseduetothedecreaseineffectiveresistanceofXY. 30 AEffectiveresistanceofparallelresistors=(1÷3+1÷6) –1 =2ΩPotentialdifferenceacrosstheparallelresistors=12÷2 (Sharedequallysinceresistanceareequal)=6VCurrentthrough6Ωresistors=V÷R=6÷6=1AUsingQ=It=1×1=1C 31 BPotentialacross6Ωresistor,V=IR=4×6=24VPotentialdifferenceacross12Ωresistor=Potentialacross6Ωresistor

5 =24VUsingV=IR,24=I×12I=2ACurrentthroughammeter=4+2=6A 32 BWhentheresistanceof Rincreases, theeffectiveresistorconnectedtovoltmeterV2 increases.VoltmeterV2 willthenregisterahighreadingandthereadingonvoltmeterV1 willdecreasedsincethesumofV1 andV2 mustremainsconstant. 33 BOptionA:Electricalenergy=Pt=(50÷1000)×2=0.1kWhOptionB:Electricalenergy=Pt=(3000÷1000)×(20÷60)=1kWhOptionC:Electricalenergy=I 2 Rt=(0.1) 2 ×100×10=10kWhOptionD:Electricalenergy=Pt=(50÷1000)×(30÷60)×2=0.05kWh 34 BWhenlivewiretouchesthecase,duetothelowresistanceoftheearthwire,currentflowwillincrease.Thiswillcausethefusetomeltandbreakthecircuit. 35 BDirectionofmagneticfieldlinesstartfromnorthpolepointingtosouthpole. 36 CDirectionofmagneticfieldlinesstartfromnorthpolepointingtosouthpole.IfcompassAandBarecorrect,theendnearertocompassAisasouthpole.TheendnearertocompassDwillbeanorthpole,whichmeanscompassDispointedcorrectly. 37 BUsingFleming’sleft handrule,Indexfingerrepresentmagneticfieldpointingintothepaper.Themiddlefingerwhichrepresentsthecurrent(flowofpositivecharge)pointstowardtheright.Thethumbwhichrepresentsforcewillbepointingup.Thepositiveionwillexperienceanupwardsforce. 38 AUnlikepolesattract,thusmagnetYwillattractmagnetX.SinceZissoftiron,magnetYwillcauseZtobecomeaninducedmagnetthusattractingZ. 39 BUsingtheright-handgriprule,themagneticfieldofawireflowingintothepaperisclockwise. Thus,thewireswillmovetowardseachother.

6 40 CNS ÷NP =VS ÷VP NS =12÷240×1200=60Thenumberofturnsofcoilinsecondarycoilis60. P2SectionA ForExaminerUse 1 (a) scale:1cm:200N[1]Resultantforce=1732N+/-40N[1]Direction:Upwards[1](Nonorth)Correctdrawingusingparallelogramortiptotailmethodwithcorrectarrows[2] 2 (a) LabelWandindicatedistanceof1.5m. [1] (b) TakingmomentsaboutX,Totalclockwisemoments=Totalanticlockwisemoments(T2 x2.3)+(20x0.2)=(10x1.3)+(40x1.6) [1]T2 =31.7N [1] (c) TotalUpwardsForce=TotalDownwardsForceT1 +31.7=70 [1]T1 =38.3N [1] 3 (a) Force=10÷2.0x2.8=14N [1] (b) Pressure=force÷area=14/3.0×10 –5 [1]=467000N/m 2 (3sf) orPa [1]

7 (c) Anypossiblemethods● Pistonwithsmallercrosssectionalarea● Increasetheforcetocompressthespringperunitlength (d) Whenthegaugeisattached,thereisaslightincreaseintheoverallvolumeofthetyre. [1] Thesamenumberofmoleculeshitthelargerwallsofthetyreswiththesameforceresultinginthefallinpressure. [1] 4 (a)Thearrangementoftheatomschangesfromonethatiscloselypackedtogetherinaregular arrangement toonewhichiscloselypackedbut without anyorder. Theatomsarenowabletomoveabout,slidingovereachotherfromvibratingaboutafixedposition[1]. (b)Inconduction,heatistransferthroughthesubstancethroughvibrationofmoleculesandfreeelectronsformetal. [1] Inconvection,heatistransferduetodensitychangesoftheparticles. [1] 5 (a) (i) energy=Pxt=200x(4x60) [1]=48000J [1] (ii) Heatlossbyheater=heatgainedbysubstanceX200x(6x60)=0.4xlf lf =72000/0.4=1800