PHYSICS FOR YOU | JUNE 157Vol. XXIIINo. 6June 2015Corporate Ofce : Plot 99, Sector 44 Institutional area, Gurgaon -122 003(HR).Tel : 0124-4951200e-mail : info@mtg.inwebsite : www.mtg.inRegd. Ofce406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.Managing Editor:Mahabir SinghEditor:Anil Ahlawat (BE, MBA)ContentsPhysics Musing (Problem Set-23)8AIPMT10Solved Paper 2015Exam Prep23Physics Musing (Solution Set-22)29Target PMTs31Practice Questions 2015Kerala PMT41Solved Paper 2015Brain Map46Ace Your Way CBSE XII51Series 1Core Concept62WB JEE65Solved Paper 2015Thought Provoking Problems76Concept Based FAQs80You Ask We Answer84Crossword85rial editSend D.D/M.O in favour of MTG Learning Media (P) Ltd.Payments should be made directly to :MTG Learning Media (P) Ltd, Plot No. 99, Sector 44, Gurgaon - 122003 (Haryana)We have not appointed any subscription agent.Subscribe online at www.mtg.inOwned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi.Readersareadvicedtomakeappropriatethoroughenquiriesbeforeactinguponany advertisementspublishedinthismagazine.Focus/Infocusfeaturesaremarketingincentives MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only.Editor : Anil AhlawatCopyright MTG Learning Media (P) Ltd.All rights reserved. Reproduction in any form is prohibited.Is the Advance of Science by Logic,Intuition or Experimentation?Onehadalwaysassumedthattheadvanceofscienceislogicaland based on assumptions that are veried by experimentation. This was true during the days of classical physics. However, modern physics in the twentiethcenturyhasgrownbyleapsandboundsoutingtheancient logicalmethods.Ifoneweretoanalysethemarchofphysics,Einstein, Bohr, Louis de Broglie, Dirac, Schrodinger, Max Born and many other did not wait for experimentation for the formulation of theories.AccordingtoEinsteinsphotoelectricity,photonisbothaparticleanda wave. Diffraction and interference of light as well as that of electrons were unied by Max Born. Max Born developed quantum mechanics based on theprobabilitywavessuggestedbyEinstein.Diracscontributionofhis idea of Dirac vacuum consisting of all particles and antiparticles is near theideaof SoonyaofIndianPhilosophy.Howeverourphilosophysays that it is Poorna or innity.Innity+innity=innity,accordingtomodernmathematics.But innityinnityisundened.Accordingtoourconcept,ifonetakes Poorna from Poorna, Poorna remains.Einsteins concept of c, the velocity of light and his enunciation that the massofabodyincreaseswithvelocityarebeyondclassicalphysicsand transcendsnormallogic.Einstein,Bohr,LouisdeBroglieandMaxBorn setrightthecontroversyofwhetherlightisaparticleorwave,bytheir correction that particles and light are simultaneously matter and wave.Intuitionhasprovedtobemorepowerfulthanlogic,assumptionsand experimentation.Anil AhlawatEditorIndividual Subscription Rates Combined Subscription Rates1 yr.2 yrs.3 yrs.1 yr.2 yrs.3 yrs.Mathematics Today330600775PCM90015001900Chemistry Today330600775PCB90015001900Physics For You330600775PCMB100018002300Biology Today330600775PHYSICS FOR YOU | JUNE 15 8SINGLE OPTION CORRECT TYPE1.AdiscofradiusRisrollingpurelyonafathorizontal surface, with constant angular velocity. Te angle between the velocity and acceleration vectors at point P is(a)zero (b)45 (c)135 (d)tan1(1/2)2.A solid ball of radius r rolls inside a hemispherical shell of radius R without slipping. Itisreleasedfromrestfrom pointAasshowninfgure. Te angular velocity of centre of the ball in position B about the centre of the shell is(a)25gR r ( ) (b) 107gR r ( ) (c) 25gR r ( ) (d) 52gR r ( ) 3.Anobjectismovingtowardsaconverginglensonits axis.Teimageisalsofoundtobemovingtowardsthe lens. Ten, the object distance u must satisfy (a)2f < u < 4f(b)f < u < 2f(c)u > 4f(d)u < f4.Auniformsolidbrasssphereofradiusa0andmassm issetspinningwithangularspeedw0aboutadiameter attemperatureT0.IfitstemperaturebeincreasedtoT withoutdisturbingthesphere,itsnewangularspeed, assuming that its new radius is a, will be(a)w = w0(b)w w = TT00(c)w w = aa020(d) w w = T TT0005.A metal ball A (density 3.2 g cm3) is dropped in water, while another metal ball B (density 6.0 g cm3) is dropped in a liquid of density 1.6 g cm3. If both the balls have the same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is(a)2.0(b)0.5(c)4.0(d)0.256.Twoglassplatesaretouchingatoneendandseparated by a thin wire at the other end. When a monochromatic parallelbeamofwavelength4200incidentnormally on the glass plates is refected, an interference pattern of 30 fringes is observed. If the wavelength of light used is taken7000insteadof4200,thenumberoffringes observed will be (a)50(b)40(c)30(d)187.Consider a YDSE that has diferent slits width, as a result, amplitudesofwavesfromslitsareAand2A,respectively. If I0 be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase diference between waves is f, is (a)I0cos2f(b) I0 23 2sinf(c) I095 4 [ cos ] + f (d) I095 8 [ cos ] + fSUBJECTIVE TYPE8.Twoidenticalsonometerwireshaveafundamental frequency of 500 Hz when kept under the same tension. What fractional increase in the tension of one wire would causeanoccurrenceof5beatspersecond,whenboth wires vibrate together?9.Findthetemperatureatwhichthefundamental frequencyofanorganpipeisindependentofsmall variationintemperatureintermsofthecoefcientof linear expansion (a) of the material of the tube.10.Auniformropeoflength12mandmass6kghangs verticallyfromarigidsupport.Ablockofmass2kgis attached to the free end of the rope. A transverse pulse of wavelength0.06 misproduced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?PhysicsMusingwasstartedinAugust2013issueofPhysicsFor YouwiththesuggestionofShriMahabirSingh. TheaimofPhysicsMusingis toaugmentthechancesofbrightstudentspreparingforJEE(MainandAdvanced)/AIIMS/OtherPMTswithadditionalstudymaterial.In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutionsoftheseproblemswillbepublishedinnextissueofPhysicsFor You.The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those whosendatleastfivecorrectsolutionswillbepublishedinthenextissue.WehopethatourreaderswillenrichtheirproblemsolvingskillsthroughPhysicsMusingandstandinbettersteadwhilefacingthecompetitive exams.By Akhil Tewari, Author Foundation of Physics for JEE Main & Advanced, Senior Professor Physics, RAO IIT ACADEMY, Mumbai.PHYSICSPHYSICS MUSINGMUSINGPHYSICS FOR YOU | JUNE 15 101.Tree blocks A, B and C, of masses 4 kg, 2 kg and 1kgrespectively,areincontactonafrictionless surface, as shown. If a force of 14 N is applied on the4kgblock,thenthecontactforcebetween A and B is BAC(a)8 N(b)18 N(c)2 N(d)6 N2.Ifradiusofthe 1327Al nucleusistakentobeRAl, then the radius of 53125Te nucleus is nearly(a) 35RAl(b) 13531 3/RAl(c) 53131 3/RAl(d) 53RAl3.Whichofthefollowingfguresrepresentthe variation of particle momentum and the associated de-Broglie wavelength?(a) p(b) p(c) p(d) pExam Q. No. MTG Book Q. No. P. No. Exam Q. No. MTG Book Q. No. P. No.1 AIPMT Guide 28 103 24 AIPMT Guide 39 6513 AIPMT Guide 33 650 27 AIPMT Guide 26 2946 AIPMT Guide 45 720 29 NCERT Fingertips 38 11912 AIPMT Guide 28 579 36 AIPMT Guide 81 21713 AIPMT Guide 56 449 37 AIPMT Guide 56 14118 AIPMT Guide 21 200 43 AIPMT Guide 120 2022 AIPMT Guide 1 314 44 AIPMT Guide 153 456Exam Q. No.MTG BookQ. No. P. No. Exam Q. No. MTG Book Q. No. P. No.2NCERT Fingertips17 291 20Physics For You May'1539 345AIPMT Guide118 615 32 NCERT Fingertips 54 947AIPMT Guide88 345 34 AIPMT Guide 73 2568 AIPMT Guide 21 609 39 NCERT Fingertips 38 2089AIPMT Guide34 535 41 AIPMT Guide 85 34515Physics For You Jan'1521 15 45 AIPMT Guide 18 446Here, the references of few are given :Exact QuestionsSimilar Questionsand more such questions PHYSICS FOR YOU | JUNE 15 114.Tetwoendsofametalrodaremaintainedat temperatures100Cand110C.Terateofheat fow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200C and 210C, the rate of heat fow will be(a)8.0 J/s(b)4.0 J/s(c)44.0 J/s(d)16.8 J/s5.Foraparallelbeamofmonochromaticlightof wavelengthl,difractionisproducedbyasingle slit whose width a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of the central maxima will be(a) Dal(b) 2Dal(c) 2Dal(d) Dal6.Whichlogicgateisrepresentedbythefollowing combination of logic gates?(a)AND(b)NOR(c)OR(d)NAND7.AparticleisexecutingSHMalongastraightline. Its velocities at distances x1 and x2 from the mean position are V1 and V2, respectively. Its time period is (a)212221222pV Vx x++(b)212221222pV Vx x(c)212221222px xV V++(d)222121222px xV V8.Twoidenticalthinplano-convexglasslenses (refractiveindex1.5)eachhavingradiusof curvatureof20cmareplacedwiththeirconvex surfacesincontactatthecentre.Teintervening spaceisflledwithoilofrefractiveindex1.7.Te focal length of the combination is(a)50 cm(b)50 cm (c)20 cm(d)25 cm9.An electron moving in a circular orbit of radiusr makesnrotationspersecond.Temagneticfeld produced at the centre has magnitude(a) m02ner(b) m02ner(c) mp02ner(d)Zero10.A particle of unit mass undergoes one-dimensional motion such that its velocity varies according tov(x) = bx2n,where b and n are constants and x is the position of the particle. Te acceleration of the particle as a function of x, is given by (a)2b2 x2n + 1(b)2nb2 e4n + 1(c)2nb2 x2n 1(d)2nb2 x4n 111.Teelectricfeldinacertainregionisacting radially outward and is given by E = Ar. A charge containedinasphereofradiusacentredatthe origin of the feld, will be given by(a)4pe0Aa3(b)e0Aa3(c)4pe0Aa2(d)Ae0a212.AradiationofenergyEfallsnormallyona perfectlyrefectingsurface.Temomentum transferred to the surface is (C = Velocity of light)(a) 22EC(b) EC2(c) EC(d) 2EC13.A, B and C are voltmeters of resistance R, 1.5R and 3R respectively as shown in the fgure. When some potential diference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. TenXYBCA(a)VA = VB VC(b)VA VB VC(c)VA = VB = VC(d)VA VB = VC14.ArodofweightWissupportedbytwoparallel knifeedgesAandBandisinequilibriumina horizontal position. Te knives are at a distance d from each other. Te centre of mass of the rod is at distance x from A. Te normal reaction on A is(a) Wd xx( ) (b) Wd xd( ) (c) Wxd(d) WdxPHYSICS FOR YOU | JUNE 15 1215.AwirecarryingcurrentIhastheshapeasshown inadjoiningfgure.Linearpartsofthewireare very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic feld at point O is ZRYOXIII (a) BIRi k = +mpp042^ ^(b) BIRi k = mpp042^ ^(c) BIRi k = +mpp042^ ^(d) BIRi k = mpp042^ ^16.Awindwithspeed40m/sblowsparalleltothe roofofahouse.Teareaoftheroofis250m2. Assumingthatthepressureinsidethehouseis atmospheric pressure, the force exerted by the wind ontheroofandthedirectionoftheforcewillbe (rair = 1.2 kg/m3)(a)2.4 105 N, upwards(b)2.4 105 N, downwards(c)4.8 105 N, downwards(d)4.8 105 N, upwards17.Inadoubleslitexperiment,thetwoslitsare 1mmapartandthescreenisplaced1maway. Amonochromaticlightofwavelength500nm isused.Whatwillbethewidthofeachslitfor obtainingtenmaximaofdoubleslitwithinthe central maxima of single slit pattern?(a)0.5 mm(b)0.02 mm(c)0.2 mm(d)0.1 mm18.A mass m moves in a circle on a smooth horizontal planewithvelocityv0ataradiusR0.Temassis attached to a string which passes through a smooth hole in the plane as shown.v0mR0Tetensioninthestringisincreasedgradually and fnally m moves in a circle of radius R02. Te fnal value of the kinetic energy is (a)202mv (b) 1202mv(c)mv02(d) 1402mv19.Keplersthirdlawstatesthatsquareofperiod ofrevolution(T)ofaplanetaroundthesun,is proportionaltothirdpowerofaveragedistance rbetweensunandplaneti.e.T2 =Kr3hereKis constant.IfthemassesofsunandplanetareMandm respectively then as per Newtons law of gravitation force of attraction between them is FGMmr=2, here G is gravitational constant.Te relation between G and K is described as (a)K = G(b)KG=1(c)GK = 4p2(d)GMK = 4p220.Tree identical spherical shells, eachofmassmandradiusrareplaced as shown in fgure. Consider an axis XX which is touching to two shells andpassingthroughdiameterof thirdshell.Momentofinertiaof the system consisting of these three spherical shells about XX axis is (a) 1652mr (b)4mr2(c) 1152mr (d)3mr221.AshipAismovingWestwardswithaspeedof 10kmh1andashipB100kmSouthofA,is movingNorthwardswithaspeedof10kmh1. Tetimeaferwhichthedistancebetweenthem becomes shortest, is (a)5 2 h(b)10 2 h(c)0 h(d)5 h22.TeratioofthespecifcheatsCCpv= intermsof degrees of freedom (n) is given by(a)12+n(b) 12+n(c)11+n(d) 13+nPHYSICS FOR YOU | JUNE 15 1323.If in a pn junction, a square input signal of 10 V is applied, as shown,then the output across RL will be (a) 5 V(b) 5 V(c) (d) 24.Acertainmetallicsurfaceisilluminatedwith monochromaticlightofwavelength,l.Te stoppingpotentialforphoto-electriccurrentfor this light is 3V0. If the same surface is illuminated with light of wavelength 2l, the stopping potential is V0. Te threshold wavelength for this surface for photo-electric efect is (a) p4(b) l6(c)6l(d)4l25.AparallelplateaircapacitorofcapacitanceCis connected to a cell of emf V and then disconnected fromit.AdielectricslabofdielectricconstantK, which can just fll the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?(a)Te change in energy stored is 12112CVK .(b)Te charge on the capacitor is not conserved.(c)Tepotentialdiferencebetweentheplates decreases K times.(d)Teenergystoredinthecapacitordecreases K times.26.Aparticleofmassmisdrivenbyamachinethat deliversaconstantpowerkwatts.Iftheparticle starts from rest the force on the particle at time t is (a)21 2mk t /(b) 121 2mk t /(c) mkt21 2 /(d)mk t1 2 /27.A Carnot engine, having an efciency of =110 as heatengine,isusedasarefrigerator.Ifthework doneonthesystemis10J,theamountofenergy absorbed from the reservoir at lower temperature is (a)90 J(b)1 J(c)100 J(d)99 J28.Onemoleofanidealdiatomicgasundergoesa transition from A to B along a path AB as shown in the fgure,AB5246P(in kPa)V(in m )3Techangeininternalenergyofthegasduring the transition is (a)20 J(b)12 kJ(c)20 kJ(d)20 kJ29.A block of mass 10 kg, moving in x direction with aconstantspeedof10ms1,issubjectedtoa retarding force F = 0.1x J/m during its travel from x = 20 m to 30 m. Its fnal KE will be(a)275 J(b)250 J(c)475 J(d)450 J30.Consider3rdorbitofHe+(Helium),usingnon-relativisticapproach,thespeedofelectroninthis orbit will be [given K = 9 109 constant, Z = 2 and h (Placks Constant) = 6.6 1034 J s](a)0.73 106 m/s(b)3.0 108 m/s(c)2.92 106 m/s(d)1.46 106 m/s31.AresistanceRdrawspowerPwhenconnected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes Z, the power drawn will be(a)PRZ(b)P(c) PRZ2(d)PRZ32.A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pullyattheedgeoftableandfromitsotherend anotherblockBofmassm2issuspended.Te coefcientofkineticfrictionbetweentheblock and the table is mk. When the block A is sliding on the table, the tension in the string is (a) m m gm mk 1 21 21 ( )( )++ m(b) m m gm mk 1 21 21 ( )( )+ m(c) ( )( )m m gm mk 2 11 2++m(d) ( )( )m m gm mk 2 11 2+mPHYSICS FOR YOU | JUNE 15 1433.Te refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). Te angle of minimum deviation is (a)90 A(b)180 + 2A(c)180 3A(d)180 2A34.On observing light from three diferent stars P, Q andR,itwasfoundthatintensityofvioletcolour is maximum in the spectrum of P, the intensity of greencolourismaximuminthespectrumofR and the intensity of red colour is maximum in the spectrum of Q. If TP, TQ and TR are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observations that(a)TP < TR < TQ(b)TP < TQ < TR(c)TP > TQ > TR(d)TP > TR > TQ35.Aconductingsquareframeofsideaandalong straightwirecarryingcurrentIarelocatedinthe sameplaneasshowninthefgure.Teframe moves to the right with a constant velocity V. Te emf induced in the frame will be proportional to (a) 122( ) x a +(b) 12 2 ( )( ) x a x a +(c) 12x(d) 122( ) x a 36.Two spherical bodies of mass M and 5M and radii Rand2Rarereleasedinfreespacewithinitial separationbetweentheircentresequalto12R.If theyattracteachotherduetogravitationalforce only, then the distance covered by the smaller body before collision is (a)7.5R(b)1.5R(c)2.5R(d)4.5R37.Two similar springs P and Q have spring constants KP and KQ, such that KP > KQ. Tey are stretched frstbythesameamount(casea),thenbythe same force (case b). Te work done by the springs WP and WQ are related as, in case (a) and case (b) respectively(a)WP > WQ ; WQ > WP(b)WP < WQ ; WQ < WP(c)WP = WQ ; WP > WQ(d)WP = WQ ; WP = WQ38.Twoparticlesofmassesm1,m2movewithinitial velocitiesu1andu2.Oncollision,oneofthe particles get excited to higher level, afer absorbing energy e. If fnal velocities of particles be v1 and v2 then we must have (a) 121212121 122 221 122 22m u m u m v m v + = + e(b) 121212121 2 1212 222 2122222mu mu mv mv + + = + e(c)mu mu mv mv121 222 121 222+ = + e(d) 121212121 122 221 122 22m u mu m v mv + = + e39.Teapproximatedepthofanoceanis2700m. Tecompressibilityofwateris45.41011Pa1 and density of water is 103 kg/m3. What fractional compression of water will be obtained at the bottom of the ocean? (a)1.2 102(b)1.4 102(c)0.8 102(d)1.0 10240.Figure below shows two paths that may be taken by a gas to go from a state A to a state C.In process AB, 400 J of heat is added to the system andinprocessBC,100Jofheatisaddedtothe system.Teheatabsorbedbythesysteminthe process AC will be (a)460 J(b)300 J(c)380 J(d)500 J 41.Te fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. Te length of organ pipe open at both the ends is(a)120 cm(b)140 cm(c)80 cm(d)100 cm42.Whentwodisplacementsrepresentedby y1 = a sin(wt) and y2 = b cos(wt) are superimposed the motion is(a)simple harmonic with amplitude a b2 2+(b)simple harmonic with amplitude ( ) a b +2(c)not a simple harmonic(d)simple harmonic with amplitude abPHYSICS FOR YOU | JUNE 15 1543.If energy (E), velocity (V) and time (T) are chosen asthefundamentalquantities,thedimensional formula of surface tension will be (a)[EV2T2](b)[E2V1T3](c)[EV2T1](d)[EV1T2]44.Across a metallic conductor of non-uniform cross sectionaconstantpotentialdiferenceisapplied. Tequantitywhichremainsconstantalongthe conductor is (a)drif velocity(b)electric feld(c)current density(d)current45.A potentiometer wire has length 4 m and resistance 8 W. Te resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is (a)44 W(b)48 W(c)32 W(d)40 WSOLUTIONS1.(d): Here, MA = 4 kg, MB = 2 kg, MC = 1 kg, F = 14 NNet mass, M = MA + MB + MC = 4 + 2 + 1 = 7 kgLet a be the acceleration of the system.Using Newtons second law of motion,F = Ma14 = 7a \ a = 2 m s2LetFbetheforceappliedonblockAbyblock B i.e. the contact force between A and B. Free body diagram for block AAgain using Newtons second law of motion,F F = 4a14 F = 4 214 8 = F\F = 6 N2.(d): Radius of the nucleus R = R0A1/3 \ RRAAAlTeAlTe= 1 3 /Here, AAl = 27, ATe = 125, RTe = ? RRAlTe= =27125351 3 / R RTe Al=533.(d): de-Broglie wavelength, l = hpor l 1pp , l =constantTis represents a rectangular hyperbola.4.(b): Rate of heat fow through a rod is given by dQdtKA dTdx= Let length of the rod be L.Case I : dTdxTx L L= ==DD110 100 10\ = dQdtKAL110(i)Also J s ,dQdt1 14 =(ii)Case II : dTdxTx L L= ==DD210 200 10\ = dQdtKAL210(iii)So, from equations (i), (ii) and (iii) dQdtdQdt2 1 14 = =J s5.(c) : Given situation isshown in the fgure.For central maxima, sinql=aAlso, q is very-very small so sin tan q q =yD\ yD a= l,yDa= lWidth of central maxima = = 22yDal.6.(a): Te Boolean expression of this arrangement is Y A B AB AB = + = = Tus, the combination represents AND gate.7.(d): InSHM,velocitiesofaparticleatdistances x1 and x2 from mean position are given by V a x12 2 212= w ( )(i) V a x22 2 222= w ( ) (ii)PHYSICS FOR YOU | JUNE 15 16From equations (i) and (ii), we get V V x x1222 22212 = w ( ) w =V Vx x12222212 \ =Tx xV V222121222p8.(a): Givencombinationisequivalenttothree lenses. In which two are plano-convex with refractive index1.5andoneisconcavelensofrefractive index 1.7.Using lens maker formula, 111 11 2f R R= ( ) mFor plano-convex lensR1 = , R2 = 20 cm, \ = = 1 11 5 11 1201 2f f( . ) = =0 520140.So,f1 = f2 = 40 cmFor concave lens,m = 1.7, R1 = 20 cm, R2 = 20 cm \ = 11 7 11201203f( . ) = = 0 72207100.So, f31007= cmEquivalent focal length (f eq) of the system is given by 1 1 1 11 3 2f f f feq= + + = ++1401100 7140 / = = = 12071002100150\feq = 50 cm9.(b): Current in the orbit,IeT=Ie e n ene = = = =( )( )2 222 p wwppp /Magnetic feld at centre of current carrying circular coil is given by BIrner= =m m0 02 210.(d): Accordingtoquestion,velocityofunitmass varies asv(x) = bx2n(i) dvdxnxn= 22 1b (ii)Acceleration of the particle is given by advdtdvdxdxdtdvdxv = = = Using equation (i) and (ii), we geta = (2nbx2n 1) (bx2n)= 2nb2 x4n 111.(a): According to question,electric feld varies as E = ArHere r is the radial distance.At r = a, E = Aa(i)Net fux emitted from a spherical surface of radius a is enet = qen0 = ( ) ( ) [ ] Aa aq420peUsing equation(i)\q = 4pe0Aa312.(d): Energy of radiation,E hhC= = ulAlso, its momentum ph ECpi= = =l p pECr i= = So, momentum transferred to the surface = = = p pECECECi r213.(c) : Te current fowing in the diferent branches of circuit is indicated in the fgure.VA = IRPHYSICS FOR YOU | JUNE 15 17VIR IRB = =2332VIR IRC = =33Tus, VA = VB = VC14.(b): Given situation is shown in fgure.N1 = Normal reaction on A N2 = Normal reaction on BW = Weight of the rod In vertical equilibrium,N1 + N2 = W(i)Torque balance about centre of mass of the rod,N1x = N2(d x)Putting value of N2 from equation (i)N1x = (W N1)(d x)N1x = Wd Wx N1d + N1xN1d = W(d x)\ =NW d xd1( ) 15.(a): Given situation is shown in the fgure.Parallel wires 1 and 3 are semi-infnite, so magnetic feld at O due to them B BIR k1 304= = mp^MagneticfeldatOduetosemi-circulararcin YZ-plane is given by BIRi204= m ^Net magnetic feld at point O is given by B B B B = + +1 2 3 = mpm mp0 0 04 4 4IRkIRiIRk^ ^ ^ = +mpp042IRi k ( )^ ^16.(a): By Bernoullis theorem, P v P v1 122 221212+ = + r rinside outside Assuming that the roof width is very smallPressure diference, P P v v1 2 221212 = r( )Here, r = 1.2 kg m3, v2 = 40 m s1, v1 = 0,A = 250 m2P P1 22 2121 2 40 0 = . ( )

= 121 2 1600 .= 960 N m2Force acting on the roof F = (P1 P2) A = 960 250= 2.4 105 N upwards17.(c) : For double slit experiment,d = 1 mm = 1 103 m, D = 1 m, l = 500 109 mFringe width bl= DdWidth of central maxima in a single slit=2lDaAs per question, width of central maxima of single slitpattern=widthof10maximaofdoubleslit pattern 210l l DaDd=ad= == =2102 10100 2 10 0 233. . m mm18.(a): Accordingtolawofconservationofangular momentummvr = mvrvR vR0 002= ; v =2v0(i)\ KKmvmvvv0022021212= = or KKvv0 0222 = = ( ) (Using (i))K = 4K0 = 2mv2019.(d): Gravitationalforceofattractionbetweensun andplanetprovidescentripetalforcefortheorbit of planet.\ =GMmrmvr22 vGMr2= (i)Time period of the planet is given by TrvTrv= =2 422 22p p, TrGMr22 24=p[ ] Using equation (i) TrGM22 34= p (ii)PHYSICS FOR YOU | JUNE 15 18According to question, T2 = Kr3(iii)Comparing equations (ii) and (iii), we get KGMGMK = \ =2442pp20.(b): Net moment of inertia of the system,I = I1 + I2 + I3Temomentofinertiaofashellaboutits diameter, I mr1223=Te moment of inertia of a shell about its tangent is given by I I I mr mr mr mr2 3 12 2 2 22353= = + = + =\ I mr mr = + 253232 2 = =123422mrmr21.(d): Given situation is shown in the fgure.Velocity of ship AvA = 10 km h1 towards westVelocity of ship BvB = 10 km h1 towards northOS = 100 kmOP = shortest distanceRelative velocity between A and B is v v vAB A B= + = 2 2 110 2 kmh cos ; 4512 100 = =OPOSOP OP = = =1002100 2250 2 kmTe time afer which distance between them equals to OP is given by tOPvtAB= = =50 210 25 h22.(a): For n degrees of freedom, CnRv =2Also, Cp Cv = R C C RnR Rp v= + = +2 CnRp = +21 = =+=+CCnRn Rnnpv2122( ) / \ = + 12n23.(b): Diodeisforwardbiasforpositivevoltage i.e. V > 0, so output across RL is given by24.(d): AccordingtoEinsteinsphotoelectric equation eVhc hcs = l l0where Vs = Stopping potential l = Incident wavelengthl0 = Treshold wavelengthor Vhces= 1 10l lFor the frst case 31 100Vhce= l l ...(i) For the second case Vhce00121= l l...(ii)Divide eqn. (i) by (ii), we get 31 112100=l ll l 3121 1 10 0l l l l = 323 1 10 0l l l l =

122400l ll l = = or25.(b): q = CVV = q/CDue to dielectric insertion, new capacitanceC2 = CKPHYSICS FOR YOU | JUNE 15 19Initial energy stored in capacitor,UqC122Final energy stored in capacitor,UqKC222Change in energy stored,AU = U2 U1 AUqC KCVK

,(

,(222111211New potential diference between plates VqCK VK26.(c) : Constant power acting on the particle of mass m is k watt.orP = k dWdtk dWkdt ;Integrating both sides, dW k dtW t0 0 W = kt(i)Using work energy theorem, W mv m 121202 2( ) kt mv 122[ ] Using equation (i) vktm2Acceleration of the particle,advdt akm tkmt 122 12Force on the particle, F mamkt 2 mkt21 2 /27.(a): For Carnot engine,Efficiency, ; 111011212TTTT TT121110910 (i)For refrigerator, . QQTT2121 orQ WQTT1121-QQ1110 109-[Using equation (i)]110 10910 109191 1- 1Q Q;Q1 = 90 JSo, 90 J heat is absorbed at lower temperature.28.(d): We know, AU = nCv AT

,( nRT T CRB A v5252( ) [ ] for diatomic gas,

,(52nR P VnRPVnRPV nRTB B A A[ ] 52( ) P V PVB B A A 522 10 6 5 10 43 3( ) 528 10 203( ) kJ29.(c) : Here, m = 10 kg, vi = 10 m s1Initial kinetic energy of the block isK mvi i 121210 10 5002 1 2( ) ( ) kg m s JWork done by retarding forceW F dx xdxxrxx

(( 120 1 0 122030 22030. .

(( 0 1900 400225 . JAccording to work-energy theorem,W = Kf KiKf = W + Ki = 25 J + 500 J = 475 J30.(d): Energy of electron in He+ 3rd orbitEZn32213 6 . eV 13 649. eV 13 6491 6 10 9 7 1019 19. . . J J As per Bohrs model, Kinetic energy of electron in the3rd orbit = E3. 9 7 101219 2. m ve v 2 9 7 109 1 101 46 1019316 1... msPHYSICS FOR YOU | JUNE 15 2031.(c) : Case I : P = VrmsIrms VVRrmsrms PVRV PR rmsrms22 ...(i)Case II : Power drawn in LR circuit P V I VVZRZrms rms rmsrmscos VRZrms22 PRRZ2

[Using eqn (i)] P P RZ22 32.(a): Given situation is shown in the fgure.Here, N = m1gf = kN = km1g(i)Let a be the acceleration of blocks.Equation of motion for A and BT f = m1a(ii)m2g T = m2a(iii)Adding equation (ii) and (iii), we getm2g f = (m1 + m2)a amg fm m-21 2 Put this value of a in equation (iii) T mg mmg fm m -2 221 2( )

----m mg m m gm mm m gm mk k 1 2 1 21 21 21 21 ( )(Using (i))33.(d): Aso-

,(,sinsinAA22cotsinsincotAAAA2222-

,(,

((o cossinsinsinAAAA2222,, -

,(,osin sin ;r o r o2 2 2 2 2 2 2 2

,( -

,( -A A A A.o = r 2A = 180 2A34.(d): According to Weins displacement law/mT = constant(i)For star P , intensity of violet colour is maximumFor star Q, intensity of red colour is maximum.For star R, intensity of green colour is maximum.Also, /r > /g > /vUsing equation (i), Tr < Tg < TvTQ < TR < TP35.(b): Here, PQ = RS = PR =QS = aEmf induced in the framer = B1(PQ)V B2(RS)V r02 2Ix aaV( ) /-r02 2Ix aaV( ) / -

((r022222Ix a x aaV( ) ( ) -

((r02222 2I ax a x aaV( )( ). -r12 2 ( )( ) x a x a36.(a)37.(a): Here, KP > KQCase (a) : Elongation (x) in each spring is same.W K x W K xP P Q Q 12122 2,.WP > WQPHYSICS FOR YOU | JUNE 15 21Case (b) : Force of elongation is same.So and , xFKxFKP Q1 2= =W KxFKP PP= =1212122W K xFKQ QQ= =1212222 \ WP < WQ38.(a): Total initial energy of two particles = +12121 122 22m u m uTotal fnal energy oftwo particles = + +12122 221 12m v m v eUsing energy conservation principle, 121212121 122 221 122 22m u m u m v m v + = + + e\ + = +121212121 122 221 122 22m u m u m v m v e39.(a): Depth of ocean d = 2700 mDensity of water, r = 103 kg m3Compressibility of water, K = 45.4 1011 Pa1 DVV = ?Excess pressure at the bottom, DP = rgd= 103 10 2700 = 27 106 PaWe know/,( )BPVV=DD D DDVVPBK P KB = = =. 1= 45.4 1011 27 106 = 1.2 10240.(a): As initial and fnalpoints are same so DUABC = DUAC AB is isochoric process. DWAB = 0DQAB = DUAB = 400 JBC is isobaric process.DQBC = DUBC + DWBC100 = DUBC + 6 104 (4 103 2 103)100 = DUBC + 12 10DUBC = 100 120 = 20 JAs, DUABC = DUACDUAB + DUBC = DQAC DWAC400 20 2 10 2 10122 10 4 104 33 4 = + DQAC()380 = DQAC (40 + 40), DQAC = 380 + 80 = 460 J41.(a): For closed organ pipe, fundamental frequency is given by ucvl=4Foropenorganpipe,fundamentalfrequencyis given by uovl= 22nd overtone of open organ pipeu = 3uo ; =u32vlAccording to question, uc = u vlvl 432=l = 6lHere, l = 20 cm, l = ?\l = 6 20 = 120 cm Your favourite MTG Books/Magazines available in KERALA atVisit MTG IN YOUR CITY on www.mtg.in to locate nearest book seller OR write to info@mtg.in OR call 0124-4951200 for further assistance.lDHYAN PRAKASHAN BOOK, OPP. VAZHUTHACAUD,TRIVANDRUMPH : 9497430066lIDEAL BOOK CENTRE, PULIMOODU, TRIVANDRUM Ph: 9645163611lH & C STORE, MAVOORROAD, CALICUT PH : 0495-2720620lH & C STORES-TDM HALL-ERNAKULAM, Ph: 0484-2352133/2351233lVIDYA BOOK HOUSE-KASARAGOD, Mobile : 9447727212l H & C STORE-KOLLAM, PH : 0474-2765421lH AND C STORE KOTTAYAM, PH : 0481-2304351l TBS PUBLISHERSAND DISTRIBUTORS,KOZHIKODE, PH : 0495-2721025,2720086,2721414l GIFTALIA BOOK BHAVAN-THRISSURE, PH : 0487-2336918PHYSICS FOR YOU | JUNE 15 2242.(a): Here, y1 = asinwty b t b t22= = +cos sin w wpHence,resultantmotionisSHMwithamplitude a b2 2+ .43.(a): Let S = kEaVbTcwhere k is a dimensionless constant.Writing the dimensions on both sides, we get[M1L0T2] = [ML2T2]a[LT1]b [T]c= [M L T ]2 2 a a + b a b + c Applying principle of homogeneity of dimensions, we get,a = 1... (i) 2a + b = 0... (ii) 2a b + c = 2... (iii)Adding (ii) and (iii), we getc = 2From (ii), b = 2a = 2\S = kEV2 T2or[S] = [EV2T2]44.(d): Te area of cross section of conductor is non uniform so current density will be diferent but the fow of electrons will be uniform so current will be constant.45.(c) : Required potential gradient = 1 mV cm1 =1101VmLength of potentiometer wire, l = 4 mSo potential diference across potentiometer wire = =1104 0 4 . V(i)In the circuit, potential diference across 8 W= I 8=+288R (ii)Using equation (i) and (ii), we get 0 4288 .=+R 4101688 40 =++ =RR ,\ R = 32 WnnHaryana police have nabbed four people, including two dentists and an MBBS student,fromRohtakforallegedlypassingonanswerkeystostudentsusing vests with SIM card units and bluetooth-enabled earpieces during the All India Pre Medical Test (AIPMT).Apartfromprobinghow90answerkeystothehighlycompetitiveall-India testwereleaked,policearealsoinvestigatingatleastninecandidateswho allegedlypaidthegangaroundRs15-20lakhforthe helpintheirbidto become doctors.Police said the gang may have spread its reach to other states too, particularly Bihar and Rajasthan. They added that the accused claimed they had purchased the engineered vests from a shop in New Delhi.Ofthefourarrested,policehaveidentiedtwoasBDSdoctorsSanchitand Bhupender, one as second-year MBBS student Ravi and the fourth as Rajesh. Thealleged kingpinofthisracket,RoopSinghDangi,isontherun,police said.AnotherMBBSdoctorhisidentityhasbeenwithheldisalsounderthe scanner for acting as a mediator between the accused and the students.ShrikantJadhav,InspectorGeneralofPolice(Rohtakrange),saidthearrests followedatip-off.Wealertedtheexaminationauthoritiesandordereda thoroughfriskingofeverystudent.Also,wereceivedconcreteinformation about the four accused who were staying in a hotel in Panipat. We tracked their mobile phones to the Jhajjar bypass in Rohtak and caught them, Jadhav told The Indian Express.Elaborating on the special vests, one of the investigating ofcers said, They hadsimcardunitslinkedtoearpiecesviabluetooth.Speciallycongured phones were also supplied to some students. Soon after the exam began, the accused started sending answer keys to the nine students from whom they had allegedlytakenRs15-20lakheach.Forstudentswithphones,answerkeys weresentthrough WhatsApp,andforthoseusingtheearpieces,theywere passed on through phone calls.Police said the engineered vests and keys to the 90 questions were recovered from the four accused who were produced in a Rohtak court they were sent to police custody for four days for further interrogation.So far, we have received a list of nine students who appeared in the AIPMT exams by allegedly paying money to the accused. Raids are being conducted to nab all those involved, Jadhav said.We have also sent teams to nab those who sold these vests to the accused. Each vest was purchased by the accused at a cost of approximately Rs 9,000 each, he added.Police also suspect that at least one of the accused Ravi, the MBBS student may have resorted to similar means to pass his own medical entrance test.He undertook PMT coaching from an institute in Kota, Rajasthan in 2005-06 and got through the exam in 2007. But after eight years, he is still in the second year. It appears from interrogation that he might have got through the exam using unfair means, the investigating ofcer said.As for Dangi, the alleged kingpin, police said he is a Meham resident who operates from Alwar in Rajasthan. They believe that he allegedly procured and supplied the answer keys to the four accused.During their interrogation, police said, the accused also claimed that the shop from where they purchased the vests had sold 700 such units in Bihar. There could be a possibility that the magnitude of this racket is much larger in Bihar and Rajasthan, Jadhav said.Courtesy : The Indian ExpressAIPMT keys leaked, students get answers via bluetooth vestsPHYSICS FOR YOU | JUNE 1523UNITS AND MEASUREMENT1.Which of the following units denotes the dimensions [ML2Q2], where Q denotes the electric charge?(a) weber (Wb) (b) Wb/m2(c) henry (H) (d) H/m22.Inanexperimentofsimplependulum,theerrors inthemeasurementoflengthofthependulum(L) and time period (T) are 3% and 2% respectively. TeTTmaximum percentage error in the value of LT2 is(a) 5% (b) 7% (c) 8% (d) 1%3.TedimensionsofabintheequationPa tbx=2,where P is pressure,PP x is distance andxx t is timett are(a) [M2LT3] (b) [ML0T2](c) [ML3T1] (d) [MLT3]4.Inthefollowingequation,x, xx tandtt FrepresentFFdisplacement, time and force respectively, F a btc d xA t = + ++ + +1sin( ) w fTe dimensional formula for Ad is dd(a) [T1] (b) [L1] (c) [M1] (d) [TL1]5.Whatisthenumberofsignifcantfguresin0.310 103?(a) 2 (b) 3 (c) 4 (d) 56.Testudyoftheearthssurfaceisnormally performed with(a) rectangular cartesian co-ordinates(b) gaussian system(c) cartesian co-ordinates, but spherical(d) none of these7.Whichoneofthefollowingisdimensionally incorrect?(a) Capacitance C = [M CC1L2T4A2](b) Magnetic feld induction B = [ML0T2A1](c) Coefcient of self-induction L = [ML2T2A1](d) Specifc resistance r = [M L3T3A2]8.AquantityXisgivenbyXXe0LVtDD,wheree0isthe permittivity of free space, L is length, DV is potentialVVdiference and Dt is time interval. Te dimensionalttformula for X is the same as that of XX(a) resistance (b) charge(c) voltage (d) current9.Temomentofinertiaofabodyrotatingabouta given axis is 12.0 kg m2 in the SI system. What is the value of the moment of inertia in a system of units in which the unit of length is 5 cm and the unit of mass is 10 g?(a) 2.4 103(b) 6.0 103(c) 5.4 105(d) 4.8 10510.Awirehasamass(0.30.003)g,radius (0.50.005)mmandlength(60.06)cm.Te maximum percentage error in the measurement of its density is(a) 1% (b) 2% (c) 3% (d) 4%11.MatchListIwithListIIandselectthecorrect answer :List I List IIA. spring constantspring constant spring constant 1. [M1L2T2]B. pascal2. [M0L0T1]C. hertz 3. [M1L0T2]D. joule 4. [M1L1T2]A B C D(a) 3 4 2 1(b) 4 3 1 2(c) 4 3 2 1(d) 3 4 1 2CHAPTERWISE MCQS FOR PRACTICEUseful for All National and State Level Medical/Engg. Entrance ExamsPHYSICS FOR YOU | JUNE 15 2412.Whichofthefollowingstatementsisincorrect regarding signifcant fgures?(a)All the non-zero digits are signifcant.(b)Allthezerosbetweentwonon-zerodigitsare signifcant.(c)Greaterthenumberofsignifcantfguresina measurement, smaller is the percentage error.(d)Te power of 10 is counted while counting the number of signifcant fgures.13.Telengthandbreadthofarectangularsheetare 16.2 cmand 10.1 cm, respectively. Te area of the sheet in appropriate signifcant fgures and error is (a)164 3 cm2(b)163.62 2.6 cm2(c)163.6 2.6 cm2(d)163.62 3 cm214.Te density of a material in CGS system of units is 4gcm3.Inasystemofunitsinwhichunitof length is 10 cm and unit of mass is 100 g, the value of density of material will be (a)0.04(b)0.4(c)40(d)40015.DistanceZtravelledbyaparticleisdefnedby Z = a + bt + gt2. Dimensions of g are(a)[LT1](b)[L1T](c)[LT2](d)[LT2]KINEMATICS16.Acricketercanthrowaballtoamaximum horizontal distance of 100 m. With the same speed how much high above the ground can the cricketer throw the same ball?(a)50 m(b)100 m(c)150 m(d)200 m17.A particle moving along the x axis has position given by x = (24t 2.0t3) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?(a)24 m s2(b)zero (c)12 m s2(d)48 m s218.Whentheangleofprojectionis75,aballfalls 10mshortofthetarget.Whentheangleof projectionis45,itfalls10maheadofthetarget. Bothareprojectedfromthesamepointwiththe samespeedinthesamedirection,thedistanceof the target from the point of projection is(a)15 m(b)30 m(c)45 m(d)10 m19.Sixvectors, athroughf havethemagnitudesand directionsindicatedinthefgure.Whichofthe following statements is true?(a) b c f += (b) d c f +=(c) d e f +=(d) b e f += bcfe da20.Aparticleisprojectedverticallyupwardsfroma pointAontheground.Ittakestimet1toreacha point B, but it still continues to move up. If it takes furthertimet2toreachthegroundfrompointB. Ten height of point B from the ground is(a) 121 22gt t ( ) + (b)gt1t2(c) 181 22gt t ( ) + (d) 121 2gt t21.Arectangularvesselwhenfullofwater,takes 10mintobeemptiedthroughanorifceinits bottom. How much time will it take to be emptied when half flled with water?(a)9 min(b)7 min(c)5 min(d)3 min22.Timetakenbythe projectiletoreach fromAtoBist,then thedistanceABis equal to(a)2ut(b)3 ut(c) 32ut(d) ut323.A ball A is thrown up vertically with a speed u and at the same instant another ball B is released from a height h. At time t, the speed of A relative to B is(a)u(b)2u(c)u gt(d)u gt224.A car covers the frst one-third of a distance x at a speed of 10 km h1, the second one-third at a speed of20kmh1andthelastone-thirdataspeedof 60kmh1.Findtheaveragespeedofthecarover the entire distance x.(a)10 km h1(b)12 km h1(c)18 km h1(d)20 km h125.Teco-ordinatesofamovingparticlearex=at2, y = bt2 where a and b are constants. Te velocity of the particle at any moment is(a)22 2t a b + (b)2t a b +(c)22 2t a b (d)22 2a b +uB6030APHYSICS FOR YOU | JUNE 152526.Tespeedofaprojectilewhenitisatitsgreatest height is2 5 /times its speed at half the maximum height. What is its angle of projection? (a)30(b)60(c)45(d)027.A passenger is walking on an escalator at a speed of 6 km/h relative to escalator. Te escalator is moving at 3 km/h relative to ground and has a total length of 120 m. Te time taken by him to reach the end of the escalator is(a)16 s(b)48 s(c)32 s(d)80 s28.Abodyisprojectedsuchthatitskineticenergyat the top is (3/4)th of its initial kinetic energy. What is the angle of projection with the horizontal?(a)30(b)60(c)45(d)12029.Aparticleismovingon circular path as shown in the fgure.Tendisplacement from P1 to P2 is(a)22r cos 0(b)22r tan 0(c)2r sin0(d)22r sin 030.A particle moves in x-y plane. Te position vector of particle at any time t is r t i t j - {( ) ( ) } 2 22 m. Te rateofchangeof0attimet=2s(where0isthe angle which its velocity vector makes with positive x-axis) is(a) 2171 rad s(b) 1141 rad s(c) 471 rad s(d) 651 rad sSOLUTIONS1.(c):[ML2Q2] = [ML2T2A2][Wb] = [ML2T2A1]Wbm=[M A ]22

(( T1[henry] = [ML2 T2 A2]Hm[MT A ]22 2

(( Obviously henry (H) has dimensions MLQ.22

(((P2OrrP12.(b) :Time period of simple pendulum isTLg 2Squaring both sides, we get. TLg2 24 or gLT 422...(i)Te maximum percentage error in g is A A A ggLLTT -

,( 100 100 2 100= 3% + 2 2% = 7%From (i), we getLTg2 24Te maximum percentage error in LT2 isAALTLTgg22100 100 7

,( %3.(b) :Pa tbx2[a] = [T2], as t2 is subtracted from a.From,Pa tbxtbx2 2[ ] b

((( tPx2 21 2[TML T L][ ][ ] = [M1L0T4].

(( ab[T[M L T021 4]] [ML0T2]4.(b) :F a btc d xA t - -- - -1sin( ) c oAssin(ct+o)isdimensionless,thereforeAhas dimensions of force..[A] = [F] = [MLT2]As each term on RHS represents force.- 1c d xF 1cF . [ ][ ][ ][ ] cF1 121 1 2MLTM L TPHYSICS FOR YOU | JUNE 15 26Ascisaddedtodx,thereforedimensionsofcare same that of dx..[dx] = [c]orM L TLM L T [ ][ ][ ][ ][ ][ ] dcx 1 1 21 2 2Te dimensional formula for Ad is[Ad] = [MLT2][M1L2T2] = [L1]5.(b)6.(c):Spherical cartesian co-ordinates are used with latitudes and longitudes.CartealsomeansmapinFrench.Tisisderived from Descrate the great mathematician.7.(c):[ ][ ][ ][ ][ ][ ] CqW 2 22 21 2 4 2ATML TM L T A[ ][ ][ ][ ][ ][ ][ ][ ] BFI l MLTA LML T A022 1[ ][ ][ ][ ][ ][LdidtWqti

,(

((

((r ML TATTAML T2 22 22 2A][ ][ ][ ][ ][ ][ ][ ][ ] p R ALML T A LLML T A2 3 2 23 3 2Choice (c) is dimensionally wrong.8.(d) : As, CqVAA or rr00ALqVCAL

,(AA or r0 ( )( )AAq LA V ...(i) X LVt r0AA( ) Given . Xq LA VLVt( )( )AAAA(Using (i))But [A] = [L]2 . XqtAAcurrent9.(d) :n nMMLLTTa b c2 1121212

,(

,(

,(...(i)Dimensional formula of moment of inertia = [ML2T0].a = 1, b = 2, c = 0Here, n1 = 12.0, M1 = 1 kg, M2 = 10 gL1 = 1 m, L2 = 5 cm, T1 = 1 s, T2 = 1 sn21 2 012 01101511

,(

,(

,(.kggmcmss(Using (i))

,(

,( 12100010100511 2ggcmcm = 12 100 400 = 4.8 10510.(d) :Here, A A A mmrrLL 0 0030 30 0050 50 066..,..,. As pmr L ( )2. A A A A pp

,( - -

,( 1002100mmrrLL --

,(0 0030 32 0 0050 50 066100..... = 1 + 2 + 1 = 4%11.(a): Spring constant[MLT[L]M L T Fx21 0 2][ ]pascal = unit of pressure FA[MLT[LM L T221 1 2]][ ]hertz = unit of frequency = 1T= [M0L0T1]joule = unit of work = force distance =[MLT2] [L]= [M1L2T2].12.(d) :Tepowerof10isirrelevanttothe determination of signifcant fgures.13.(a) :Let length and breadth of a rectangular sheet are measured by using a metre scale as 16.2 cm and 10.1cmrespectively.Eachmeasurementhasthree signifcant fgures..Length l can be written asl = 16.2 0.1 cm = 16.2 cm 0.6%Similarly, the breadth b can be written as b = 10.1 0.1 cm = 10.1 cm 1%Area of the sheet, A= l b = 163.62 cm2 1.6% =163.62 2.6 cm2Terefore,asperrule,areawillhaveonlythree signifcantfguresanderrorwillhaveonlyone signifcant fgure. Rounding of, we get A = 164 3 cm214.(c):As n1u1 = n2u2410010323gcmgcmn( ) n2 = 4015.(c):ByhomogeneityofdimensionsofLHSand RHS,Distance (LHS) = [] [T]2 . [ ] [ ] LT2PHYSICS FOR YOU | JUNE 152716.(a) :Here, Rmax = 100 mRugmax =2 (where u is the velocity of the projection of ball)oru2 = 100g...(i)Using v2 u2 = 2as(0)2 100g = 2 (g) (H)[Using (i)]orH = 50 m17.(a) :Given : x = 24t 2.0t3 mVelocity, ms vdxdtddtt t = = ( . ) 24 2 03 1= 24 6t2 m s1Acceleration, ms adxdtddtt = = 222 224 6 ( ) = 12t m s2For v = 0, we get24 6t2 = 0 or t = 2 sHence, at t = 2 s, the acceleration will bea = 12(2) m s2 = 24 m s2Its magnitude is 24 m s2.18.(b) :Let d be distance of the target from the point of projection.\ = ugd22 7510sin( )orugd2210 = ...(i)andugd22 4510sin( ) = +orugd210 = + ...(ii)Divide (i) by (ii), we get dd+=101012or d = 30 m19.(c):Asperthelawsof vector addition,d e f += dfeTis is as shown in adjacent fgure.20.(d) :Time taken for the particle to reach the highest point is t t1 22+.Asv = u gtAt highest point, v = 0Terefore, initial velocity of the particle isu gt t=+ 1 22...(i)Terefore, height of point B from the ground ish ut gt gt tt gt = =+ 1 121 21 1212212(Using (i))or h gt t tgt = +121 2122 212 or h gt t =121 221.(b) :If A0 is the area of orifce at the bottom below the free surface and A that of vessel, time t taken to empty the tank, tAAHg=02\ttHHHH1212112= =/\ tt2121027 = = minmin22.(d) :Refer the fgure below. Horizontal component of velocity at A.BCAu6030u uuAC u tutH H= = \ = = cos602 2AB ACut ut= = = sec 30223 323.(a) :At time t,BuB= 0Ahu uA=Velocity of A, vA = u gt (upwards)Velocity of B, vB = gt (downwards)= gt (upwards)Relative velocity of A with respectto B isvAB = vA vB = (u gt) (gt) = u24.(c):For frst one-third of distanceDistance covered= x3 kmspeed = 10 km h1.Te time taken for the journey,tx x1310 30= =/h hFor the next one-third of distance :Distance covered= x3 km.PHYSICS FOR YOU | JUNE 15 28Speed = 20 km h1Te time taken for travel istx x2320 60 /h hFor the last one-third of distance :Distance covered x3 km.Speed is 60 km h1Te time taken for travel istx x3360 180 /h h.Average Speed = total distancetotal time- -xx x x30 60 18018010xx = 18 km h125.(a) :vdxdtat vdydtbtx y 2 2 ;. - - v v v at btx y2 2 2 2 2 24 4 - 22 2t a b26.(b) :Maximum height, HuggHu 2 2 2 22 2sin sin 0 0or ...(i)Velocity at highest point, vH = u cos0Let vx, vy be the horizontal and vertical velocity of projectile atheight H2. Tenvx = ucos0andv u gHu gHy2 2 2 2 222 sin sin 0 0 v uu uy2 2 22 2 2 22 2 sinsin sin00 0(Using (i)). -( )Net velocity at height Hv vx y22 21 2 /As per question, 25252 21 22 2 2v v v v v vx y H x y H-( ) -( ) /oror25 22 222 2 2uuu cos sin cos 0 0 0 -

(( orsin20 = 3cos20 or sin cos 0 0 3or tan tan 0 3 60or0 = 6027.(b) :Velocityofthepassengerwithrespectto groundvPG = vPE + vEG = 6 + 3 = 9 km h1Time takenmm ss s txvPG 1209518240548128.(a) : According to the given problem 1234122 2m u mu ( cos ) 0 orcos2340 or cos cos 0 3230or0 = 3029.(d) : PPOrr21xAccording to cosine formula, cos 0 - r r xr2 2 222or 22 2 2 2r r r x cos 0 - or x2 = 2r2 2r2cos0 = 2r2 [1 cos0]

((2 222 2r sin0Displacement from P1 to P2 is x =22r sin 030.(a) :Given,r t i t j m - {( ) ( ) }^ ^2 22Comparingitwithstandardequationofposition vector, r x i y j -^ ^, we get x = 2t and y = 2t2 vdxdtx 2and vdydtty 4.tan0 vvttyx422Diferentiating with respect to time we get, (sec )22 00 ddtor( tan ) 1 22- 00 ddtor( ) 1 4 22- tddt0or ddtt0 -21 42at s, =rad s tddt

,(-221 4 2217210( )SOLUTION SET-221.(b): Te woman experiences three forces; mg, her weightactingverticallydownwards;N1,reaction duetoherweight;N2,horizontalreactionwhich provides the centripetal acceleration.From Newtons second law, = = F Nmvrx 22SFy = N1 mg = 0v= (2pr)u(where u is frequency)= (2p 9) (6/60)= 1.8p m s1Terefore, N2250 1 89178 = =( )( . ) pNN1 = mg = 490 NTemagnitudeofherweightisthemagnitudeof the resultant force exerted on her by the chair. N N N = +1222

= + 490 1782 2= 521 N2.(c) : LetTbethetimeoffightandubetheinitial velocity of the stone then AB2Ru 222Rug=sinq...(i)andTug=2 sinq...(ii)Eliminating u, we gettanq q = == \ = gTR2 2410 104 250 313303.(b): GBORRW AAs a a /sin sin( )2 p q= = \ =sin sin sin ( sin ) q q 2 214.(b): At NTP, temperature = 273 K and pressure = 105 N m2

vRTMRTMP= =g, = = == vP vPgg2 25330 1 3101 4( ) ..and g = 1 + (2/f) f = 2/(g 1) f = 5 5.(d): Along the vertical direction, Net impulse due to normal = change in momentumNdt m v mv= + ( ) cos /2 q ...(i)(where N is the normal by the foor on the ball)Along the horizontal direction,Friction force, f = mNLet horizontal velocity of the ball afer collision = vNet impulse due to friction = change in momentum in horizontal direction = m q Ndt mv mv sinm[m(v/2) + mv cosq] = mv mv sinq (Using eqn. (i))\v = v sin q m(v/2) + v cos q}6.(c) : Tsinq =mg2T = Tcosq=mg2sincosqq\T =mg2tan q7.(c) : Tension T3 required to move third block = mmg. Tension T2 to move 2nd block = mmg + 2T3

= 3mmg. Force F required to move the frst block= 2T2 + mmg = 7mmg. PHYSICS FOR YOU | JUNE 15298.(d): ChangeinpositionofCMinvertical direction h l l

,( -cos302 CMl lcos30= l 3/260ABCFl60Work done by gravity -

,(mg h mgll 34 -

,(mgl3 449.(b): 1 kg10 N s10 cos30 N s30301 kgJTJTHere 10cos30 JT = JT

[Boththemasseswillmovewithsamevelocity along the string] JT5 30523 cos N s10.(b): Equation of path,

yx220dydtx dxdt10, vxvy x10At x = 10 vy = vxTotal v v vx y -2 2, 6 22 vxvx236218 , vx 3 2vy 3 2Nowsincethemagnitudeofvelocityis6ms1 constant, tangential acceleration will be zero. It has only normal acceleration.Now yx220

dydxx x 220 10. dydxx

((101

d ydx22110Radius of curvature at the given point is Rdydxd ydx- ,

((-11 111010 223 2223 23 2///( )( )

10 2 2 2 20 2Normal acceleration avRRT 23620 29 210unitsSOLUTION OF MAY 2015 CROSSWORDWINNERS (May 2015) Guneet KaurHarsh GuptaRizwan Khan Solution Senders (April 2015)Aditya SrivastavaChandra Shekhar PanigrahiRajneesh KumarSolution Senders of Physics MusingSET-221.Swayangdipta Bera 2.Debajyoti Dash (West Bengal)3.Sayantan Bhanja (West Bengal)SET-211.Komal Khatri (Pune) 2.Girish Ranjan (Bihar)PHYSICS FOR YOU | JUNE 15 30PHYSICS FOR YOU | JUNE 15311.A wave is represented by y(x, t) = asin(kx wt + f). Te phase of the wave is(a)f(b)kx wt(c)wt + f(d)kx wt + f2.Name of units of some physical quantities are given in List I and their dimensional formulae are given inListII.MatchListIwithListIIandselectthe correct answer.List I List IIP. Pa s 1. [M0L2T2K1]Q. N m K12. [MLT3K1]R. J kg1K13. [ML1T1]S. W m1 K14. [ML2T2K1]PQRS(a)4312(b) 3241(c)3142(d) 34123.Teapparentweightofapersoninalifmoving downwards is half his apparent weight in the same lifmovingupwardswiththesameacceleration. Te acceleration of the lif is (a)g(b) g4(c) g2(d) g34.Te ratio of the angular velocities of the earth about its own axis and the hour hand of a watch is(a)1 : 2(b)2 : 1(c)1 : 12(d)12 : 15.Consider the following statements :Nuclear force is 1.charge independent2.long range3.centralWhich of the above statements is/are correct?(a)1 only(b)1 and 2(c)2 and 3(d)1 and 36.Te magnetic susceptibility of a material of a rod is 299. Te permeability of the material of the rod is(m0 = 4p 107 H m1)(a)3771 105 H m1 (b)3771 106 H m1(c)3771 107 H m1 (d)3771 108 H m17.A particle motion on a space curve is governed by x = 2sint, y = 3cost and z =5sint. Te speed of the particle at any instant is(a)3 2 sint(b)3 2 cos t(c)3 2 sin t (d)independent of time8.ATVtransmittingantennais128mtall.Ifthe receivingantennaisatthegroundlevel,the maximumdistancebetweenthemforsatisfactory communication in LOS mode is (Radius of the earth = 6.4 106 m)(a)64 10 km(b) 12810km(c)128 10 km(d) 6410km9.A battery of emf 3 V and internal resistance 0.2 W is being charged with a current of 5 A. What is the potentialdiferencebetweentheterminalsofthe battery?(a)2 V(b)3 V(c)3.5 V(d)4 V10.When NaCl is added to water, the surface tension of water(a)increases(b)decreases(c)remains constant(d)nothing can be said11.A person walks on a straight road from his house to a market 2.5 km away with a speed of 5 km h1. Finding the market closed, he instantly turns back and reaches his house with a speed of 7.5 km h1. Te average speed of the person isPHYSICS FOR YOU | JUNE 15 32(a) 1431ms(b) 531ms(c) 561ms(d) 131ms12.Tehalflifeofaradioactiveelementis10h.Te fractionofinitialactivityoftheelementthatwill remain afer 40 h is(a) 12(b) 116(c) 18(d) 1413.Light travels from air to water, from water to glass andthenagainfromglasstoair.Ifxrepresents refractiveindexofwaterwithrespecttoair,y represents refractive index of glass with respect to water and z represents refractive index of air with respect to glass, then which one of the following is correct?(a)xy = z(b)yz = x(c)zx = y(d)xyz = 114.Which of the following parameters is the same for molecules of all gases at a given temperature?(a)Mass(b)Speed(c)Momentum(d)Kinetic energy15.In a meter bridge experiment, the length AB of the wire is 1 m. Te resistors X and Y have values 5 W and 2 W respectively. When a shunt resistance S is connected to X, the balancing point is found to be 0.625 m from A. Ten, the resistance of the shunt is (a)5 W(b)10 W(c)7.5 W(d)12.5 W16.Teequationoftrajectoryofaprojectileisy x x = 10592m. Te range of the projectile is(a)36 m(b)24 m(c)18 m(d)9 m17.Arayoflighttravelsfromanopticallydenser medium towards a rarer medium. Te critical angle forthetwomediaisC.Temaximumpossible angle of deviation of the ray is(a) p2 C(b)p 2C(c)2C(d) p2 +C18.Tesuccessiveresonancefrequenciesinanopen organ pipe are 1944 Hz and 2600 Hz. If the speed of sound inair is 328 m s1, then the length of the pipe is (a)0.40 m(b)0.04 m(c)0.50 m(d)0.25 m19.A machine gun fres 240 bullets per minute with a velocityof600ms1.Ifthemassofeachbulletis 10 g, the power of the gun is(a)7.2 kW(b)72 kW(c)3.6 kW(d)36 kW20.Tesensitivityofagalvanometerthatmeasures currentisdecreasedby140timesbyusingshunt resistance of 10 W. Ten, the value of the resistance of the galvanometer is(a)400 W(b)410 W (c)30 W(d)390 W21.Whichoneofthefollowingiscorrectabout wave-particle duality?(a)Wave-particle duality holds for matter particles but not for light.(b) Wave-particle duality holds for light but not for matter particles. (c)Wave-particledualityholdsforelectronsbut not for protons.(d) Wave-particle duality holds for light as well as for matter particles.22.Workdonetoincreasethetemperatureof onemoleofanidealgasby30C,ifitis expandingundertheconditionV T2/3is (R = 8.31 J mol1K1)(a)116.2 J(b)136.2 J (c)166.2 J(d)186.2 J23.Electric charges A and B attract each other. Electric charges B and C also attract each other. But electric chargesCandDrepeleachother.IfAandDare held close together then which one of the following is correct?(a)Tey cannot afect each other.(b) Tey attract each other.(c)Tey repel each other.(d) Cannot be predicted due to insufcient data.24.Inatransistorifavariesbetween 2021100101and ,then the value of b lies between(a)110(b)0.950.99 (c)20100(d)200300PHYSICS FOR YOU | JUNE 153325.Teminimumforcerequiredtomoveabodyup an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefcient of friction between the body and the inclined plane is 12 3, the angle of the inclined plane is(a)60(b)45(c)30(d)1526.Te threshold frequency of the metal of the cathode in a photoelectric cell is 1 1015 Hz. When a certain beam of light is incident on the cathode, it is found thatastoppingpotential4.144Visrequiredto reducethecurrenttozero.Tefrequencyofthe incident radiation is(h = 6.63 1034 J s)(a)2.5 1015 Hz(b)2 1015 Hz(c)4.144 1015 Hz(d)3 1016 Hz27.A50mFcapacitorisconnectedtoanacsource V = 220 sin50t where V is in volt and t is in second. Te rms current is(a)0.55 A(b) 0 552.A (c) 20 55 .A(d)2 A28.Of the following, NAND gate is(a) (b) (c) (d) 29.On a temperature scale Y, water freezes at 160Y and boils at 50Y. On this Y scale, a temperature of 340 K is(a)106.3Y(b)96.3Y (c)86.3Y(d)76.3Y30.Asatelliteisrevolvingveryclosetoaplanetof density r. Te period of revolution of the satellite is(a) 3prG (b) 32prG(c) 3prG (d) 3prG31.If the ratio of maximum and minimum intensities of an interference pattern is 36 : 1, then the ratio of amplitudes of the two interfering waves will be(a)3 : 7(b)7 : 4(c)4 : 7(d)7 : 532.A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is(a) p4A m2(b) 14 A m2 (c)p A m2(d)4p A m233.Teelectricfeldforanelectromagneticwavein freespaceis E kz t i = 30 6 108 1cos( ) .^Vm Te magnitude of wave vector is(a)2 rad m1(b)3 rad m1(c)4 rad m1(d)6 rad m134.Twoparallelplanesheets1and2carryuniform charge densities s1 and s2(s1 > s2) as shown in the fgure. Te magnitude of the resultant electric feld in the region marked I is12(a) se102(b) se202(c) s se1 202+(d) s se1 20235.InMillikansoildropexperiment,achargedoil dropofmass3.21014kgisheldstationary between two parallel plates 6 mm apart, by applying apotentialdiferenceof1200Vbetweenthem.How many electrons does the oil drop carry?(g = 10 m s2)(a)7(b)8(c)9(d)1036.Two long wires each parallel to the z-axis and each carrying current I, are at (0, 0) and (a, b). Te force per unit length of each wire is(a) mp022 22Ia b ( ) +(b) mp022 22I a ba b( )( )++(c) mp022 2 3 22Ia b ( )/+(d) mp022 2 1 22Ia b ( )/+PHYSICS FOR YOU | JUNE 15 3437.Two blocks of masses 1 kg and2kgareconnectedbyametal wiregoingoverasmoothpulley asshownintheadjacentfgure. Te breaking stress of the metal is403106 2p Nm . Whatshouldbe the minimum radius of wire used if it should not break? (g = 10 m s2)(a)0.5 mm(b)1 mm(c)1.5 mm(d)2 mm38.A rectangular coil is rotating in a uniform magnetic feldB.Teemfinducedinthecoilismaximum when the plane of the coil(a)is parallel to B (b) makes an angle 30 with B (c)makes an angle 45 with B (d) is perpendicular to B39.A steady dc current is fowing through a cylindrical conductor. Which of the following statements is/are correct?1.Te electric feld at the axis of the conductor is zero.2.Te magnetic feld at the axis of the conductor is zero.Selectthecorrectanswerusingthecodegiven below :(a)1 only(b)2 only(c)both 1 and 2(d)neither 1 nor 240.Te ratio between kinetic and potential energies of abodyexecutingsimpleharmonicmotion,when itisatadistanceof 1Nofitsamplitudefromthe mean position is(a)N2 + 1(b) 12N(c)N2(d)N2 141.PQRisarightangledtriangularplateofuniform thicknessasshowninthefgure.IfI1,I2andI3 are moments of inertia about PQ, QR and PR axes respectively, then(a)I3 < I2 < I1(b)I1 = I2 = I3(c)I2 > I1 > I3(d)I3 > I1 > I242.A tank of height 5 m is full of water. Tere is a hole ofcross-sectionalarea1cm2initsbottom.Te volume of water that will come out from this hole per second is (g = 10 m s2)(a)103 m3 s1(b)104 m3s1(c)10 m3s1(d)102 m3s143.Tetemperatureofaperfectblackbodyis727C anditsareais0.1m2.IfStefansconstantis 5.67 108 W m2 K4, then heat radiated by it in 0.3 minutes is(a)1701 J(b)17010 J(c)102060 J(d)1020 J44.Te plates in a parallel plate capacitor are separated by a distance d with air as the medium between the plates.Inordertoincreasethecapacityby66%,a dielectric slab of dielectric constant 5 is introduced betweentheplates.Whatisthethicknessofthe dielectric slab?(a) d4(b) d2(c) 58d(d)d45.AsimplependulumisexecutingSHMwitha period of 6 s between two extreme positions B and CaboutapointO.IfthelengthofthearcBCis 10cm,howlongwillthependulumtaketomove from position C to a position D towards O exactly midway between C and O?(a)0.5 s(b)1 s(c)1.5 s(d)3 sSOLUTIONS1.(d) :Te argument of the sine function is called the phase.Tus the phase of the wave is kx wt + f.2.(d) :P.Dimensional formula of Pa s = [ML1T2][T] = [ML1T1]Q.Dimensional formula of N m K1 = [MLT2][L][K1] = [ML2T2K1]R.Dimensional formula of J kg1 K1 = [ML2 T2][M1][K1] = [M0L2T2K1]S.Dimensional formula of W m1 K1 = [ML2 T3][L1][K1] = [MLT3K1]Tus P 3, Q 4, R 1, S 2PHYSICS FOR YOU | JUNE 15353.(d) :Let a be acceleration of the lif.Whenthelifismovingdownwardswith accelerationa,thentheapparentweightofthe person inside it,Wapp = m(g a)where m is the mass of the person.When it is moving upwards with same acceleration a, then his apparent weight isWapp = m(g + a)As W Wapp app= 12 (given)\ m g a m g a ( ) ( ) = +122(g a) = g + aor2g 2a = g + a 3a = g or ag=34.(a) :Te earth completes one rotation about its own axis in 24 h. Its angular velocity isw1 = 224p radhTe hour hand completes one rotation in 12 h. Its angular velocity isw2 = 212p radhTeir corresponding ratio is wwpp12224212122412= = =radhradh5.(a) :Nuclearforceischargeindependent,short range and non-central.6.(c) :Te permeability (m), permeability of vacuum (m0)andmagneticsusceptibility(c)arerelatedas m = m0(1 + c)Here, m0 = 4p 107 H m1, c = 299\m = 4p 107 H m1(1 + 299)= 4 227 107 300 H m1= 3771 107 H m17.(d) :Here, x = 2sint, y = 3cost and z =5sint\vx = dxdtddt= (2sint) = 2costvy = dydtddt= (3cost) = 3sintand vz = dzdtddtt t = = ( sin ) cos 5 5Te speed of the particle at any instant isv = v v vx y z2 2 2+ += ( cos ) ( sin ) ( cos ) 2 3 52 2 2t t t + +=4 9 52 2 2cos sin cos t t t + +=4 5 4 52 2 2 2cos sin sin cos t t t t + + += 5 5 4 42 2 2 2sin cos sin cos t t t t + + += 5 5 42 2 2 2(sin cos ) (sin cos ) t t t t + + +=5 4 + ( sin2 q + cos2 q = 1)= 3Hence it is independent of time.8.(b) :Here, Radius of the earth, R = 6.4 106mHeight of transmitting antenna, hT = 128 mTe maximum distance (dM) between the transmitting and receiving antennas for satisfactory communication in LOS mode isdM =2 2 Rh RhT R+where hR is the height of the receiving antenna.As the receiving antenna is at the ground levelso, hR = 0.\dM =2 2 6 4 10 1286RhT= ( . m)( m)=12 8 128 106 2. m = 128 128 101062 m= 128 1010128103= m km9.(d) :Te potential diference between the terminals of the battery during charging isV = e + IrHere, e = 3 V, r = 0.2 W, I = 5 A\V = (3 V) + (5 A)(0.2 W) = 3 V + 1 V = 4 V10.(a) :AsNaCliscompletelysolubleinwater,so when NaCl is added to water, the surface tension of water increases.11.(b) :Time taken by person to reach the market is t1 = 2 55121. kmkmhh=and time taken by him to return back to his house ist2 = 2 57 5131. km. kmhh=\Total time taken = t1 + t2 = 121356h h + = hTotal distance travelled = 2.5 km + 2.5 km = 5 kmTe average speed of the person isPHYSICS FOR YOU | JUNE 15 36vav = Total distance travelledTotal time taken = 556kmh= 6 km h1 = 6 518m s1 = 53m s112.(b) :IfA0beinitialactivityoftheelement,then fraction of initial activity afer time t is AAt T0121 2= //where T1/2 is the half life.Here, t = 40 h, T1/2 = 10 h\ AA040 10 41212116= = =h h /13.(d) :As amw wmg gma = 1where the subscripts a, w and g represent air, water and glass respectively.Here, amw = x, wmg = y and gma = z\ xyz = 114.(d) 15.(b) : When the resistance S is connected in parallel with X, the balance point is obtained at 0.625 m (= 62.5 cm) from A. \ XSX SY+==62 5100 62 562 537 5.( . )..cmcmcmcm ( )( )55253WWWSS += ( )( )52 553WW WSS +=(15 W)S = (10 W)(5 W + S)(15 W)S = (10 W)(5 W) + (10 W)S(15 W)S (10 W)S =(10 W)(5 W)(5 W)S = (10 W)(5 W) S = =( )( )( )10 5510W WWW16.(c) :Comparing the equation y x x = 10592 with the equation of trajectory of a projectile y xgxu= tancosqq22 22whereuisthevelocityofprojectionandqisthe angle of projection with the horizontal. We gettanq = 10...(i)and gu 2592 2cos q= ...(ii)Dividing eqn. (i) by eqn. (ii), we get tancosqqgu 210592 2= sincoscosqqqgu 29052 2= 2 9052ugsin cos q q=

...(iii)Te range of the projectile isRugug== =22222 2sinsin cos( sin sin cos )qq qq q q = 18 m(using (iii))17.(b) 18.(d) :Let L be the length of the open pipe.Te resonance frequencies of vibration in the open pipe are unnvL=2; n = 1, 2, 3, ........where v is the speed of sound in air. And the diference between successive frequencies is

D D unvLnvL= =2 2 ( Dn = 1)or Lvn=2DuHere, v = 328 m s1Dun = 2600 Hz 1944 Hz = 656 Hz\ L = =3282 65614msHz)m=0.25 m1(19.(a) :Here,Mass of the bullet, m = 10 g = 10 103 kgVelocity of the bullet, v = 600 m s1Te number of bullets fred per second is n = =240604PHYSICS FOR YOU | JUNE 1537Te power of the gun is P nmv =122=12 4 10 103 (600)2 W = 72 102 W = 7.2 kW20.(d) :Let the resistance of the galvanometer be G.As shunt resistance, SI GI Igg= ( )\ =GI I SIgg( )Here, I I Sg= =14010 , W\

GIII=401040( ) W= (39)(10 W) = 390 W21.(d) 22.(c) :Work done, dW = PdVAccording to an ideal gas equation PV = nRTor PnRTV=\ dW nRTdVV= ...(i)As V T2/3 (given)\V = KT2/3where K is a constant of proportionality.Diferentiating both sides, we get dV K T dT =231 3 /\ == =dVVK T dTKTT dTdTT23 23231 32 31//Putting this value in eqn. (i), we get dW nRTdTTnRdT = =2323\ = = = W nRdT nR dT nR T TTTTT23232312122 1( )Here, n = 1, R = 8.13 J mol1 K1 T2 T1 = 30C = 30 K\ = = W231 8 31 30 166 21 1( )( . )( ) . J mol K K J23.(c)24.(c) :Te a and b are related by the relation baa= 1For a =2021 b == =202112021202112120For a =100101,

b == =10010111001011001011101100Tus the value of b lies between 20-100.25.(c) :Here,m =12 3Letmbemassofthebodyandqbeangleofthe inclined plane.Te minimum force required to move the body up the inclined plane isF1 = mgsinq + mmgcosq = mg(sinq + mcosq)and the minimum force required to prevent it from sliding down the plane isF2 = mgsinq mmgcosq = mg(sinq mcosq)As F1 = 3F2 (given)\mg(sinq + mcosq) = 3mg(sinq mcosq)sinq + mcosq = 3sinq 3mcosq2sinq = 4mcosqtanq = 2m =212 313 = q = = tan1133026.(b) :Here,Treshold frequency, u0 = 1 1015 HzStopping potential, V0 = 4.144 VPlancks constant, h = 6.63 1034 J sAccording to Einsteins photoelectric equationKmax = hu f0where u is the frequency of incident radiation.orhu = Kmax + f0But f0 = hu0 and Kmax = eV0\hu = hu0 + eV0or u u = +00eVh

= +1 101 6 10 4 1446 63 10151934HzC VJ s( . )( . )( . ) = 1 1015 Hz + 1 1015 Hz= 2 1015 HzPHYSICS FOR YOU | JUNE 15 3827.(b) :On comparing V = 220 sin50t withV = V0 sinwt, we getV0 = 220 V and w = 50 rad s1Te capacitive reactance isXCC = = 1 150 50 101 6w( )( ) rad s C=10254WTe rms current isIVXVXVVC Crmsrmsrms= = =00 22( ) =220210254VW=220 2510 20 5524A= A.28.(d) :Te symbol for NAND gate is 29.(c) : If TY be temperature on the Y scale corresponding to 340 K on kelvin scale, then TY =( )( )16050 160340 273373 273YY YK KK K TY + =16011067100YYKKTY + 160Y = 0.67(110Y)TY + 160Y = 73.7YTY= 73.7Y 160Y = 86.3Y30.(c) :As the satellite is revolving very close to the planet, its period of revolution is TRGM= 23p ...(i)whereMisthemassoftheplanet,Risitsradius and G is the universal gravitational constant.As rp=MR433 or M R =433p rPutting this value of M in eqn.(i), we getTRG R=24333pp r= 234pprG

= =3 4432( ) pprpr G G31.(d) :LetA1andA2beamplitudesofthetwo interfering waves. Ten IIA AA Amaxmin( )( )=+1 221 22ButIImaxmin( ) =361given\ ( )( )A AA A1 221 22361+= A AA A1 21 261+=A1 + A2 = 6(A1 A2)A1 + A2 = 6A1 6A25A1 = 7A2or AA1275=32.(a) :When a wire of length l is bent into a circular coil of N turns of radius r, then l = N2prorrlN=2pHere, l = 6.28 m,N = 2\r == =6 282 2 3 14120 5.( . ).mm mTe area of the coil is A = pr2 = p(0.5 m)2 = 0.25p m2Te magnetic moment of the coil, M = NIA = (2)(0.5 A)(0.25p m2) = 0.25p A m2= p42Am33.(a) :Comparing the given equation with E E kz t i = 0cos( )^wwe get, w = 6 108 rad s1Te velocity of electromagnetic wave in free space isc = 3 108 m s1Te magnitude of wave vector is kc= ==w6 103 1028 18 11rad smsrad m34.(c) :In the region I, the electric felds E E1 2anddue to sheets 1 and 2 act in the same direction.PHYSICS FOR YOU | JUNE 1539\Temagnitudeoftheresultantelectricfeldin the region I isEI = E1 + E2 = + =+ seses se10201 202 2 235.(d) :Here,Mass of the drop, m = 3.2 1014 kgDistance between the plates, d = 6 mm = 6 103 mPotential diference between the plates,V = 1200 V Te electric feld between the plates is EVd= =12006 103VmLet the charge on the drop be q.As the drop is held stationary, \ Upward force on drop = Weight of the dropdue to electric feldqE = mg

qVdmg =

qmgdV=

= ( . )( )( ) 3 2 10 10 6 10120014 2 3kg ms mV = 1.6 1018 CTe number of electrons the drop carries is

nqe= ==1 6 101 6 10101819..CC36.(d) :Te force per unit length of each wire is

fI Id= mp0 1 22whereI1andI2arethecurrentsinthetwowires respectively and d is the distance between them.Here, I1 = I2 = I,d a b = +2 2\ =+fIa bmp022 2237.(b) :Here, m1 = 1 kg, m2 = 2 kg, g = 10 m s2Breaking stress =403106 2p NmTe tension in the wire isTm m gm m=+=+2 2 1 2 101 21 21 22( )( )( )( )(kg kg mskg kg) =403NTe stress in the wire is =TensionArea of cross-sectionTo avoid breaking, this stress should not exceed the breaking stress.Iftheminimumradiusneededtoavoidbreaking is r, thenBreaking stress =Tr p2 40310403 6 22pp =NmNr r26 26 2403403101 10 = = NNmmpp r = 103m = 1 mm38.(a) :Te emf induced in the coil at any instant t is e = NBAwsinwtTisismaximumwhensinwt=1orwt= p2i.e. when the plane of the coil is parallel to the feld.39.(c) 40.(d) :Tekineticenergyofthebodyexecuting simpleharmonicmotionatadistancexfromthe mean position is K m A x = 122 2 2w ( )and the potential energy is U m x =122 2wwhere A is the amplitude, w is the angular frequency and m is the mass of the body.AtxAN= , K m AAN= 122 22w and U mAN=1222w Teir corresponding ratio is KUm AANmAN=12122 2222ww=AANAN22222PHYSICS FOR YOU | JUNE 15 40

== ANNANN22222211( )41.(a) : Te centre of mass of a triangular lamina lies at the centroid. But the distance of a side of a triangle fromitscentroidisinverselyproportionaltothe length of that side, sorPQ >rQR >rPR\I1 >I2 >I3orI3 R(d) R25.A 20 cm long capillary tube is dipped vertically in water and the liquid rises upto 10 cm. If the entire system is kept in a freely falling platform, the length ofwater column in the tube will be(a)5 cm(b)10 cm(c)15 cm (d)20 cm26.A train is moving with a uniform speed of 33 m/s andanobserverisapproachingthetrainwiththe same speed. If the train blows a whistle of frequency 1000Hzandthevelocityofsoundis333m/s, thentheapparentfrequencyofthesoundthatthe observer hears is(a)1220 Hz(b)1099 Hz(c)1110 Hz(d)1200 Hz27.Aphotonofwavelength300nminteractswitha stationaryhydrogenatomingroundstate.During theinteraction,wholeenergyofthephotonis transferred to the electron of the atom. State which possibilityiscorrect.(Consider,Planckconstant =41015eVs,velocityoflight=3108m/s, ionization energy of hydrogen = 13.6 eV)(a)Electron will be knocked out of the atom(b)Electron will go to any excited state of the atom(c)Electron will go only to frst excited state of the atom(d)Electron will keep orbiting in the ground state of atom28.Block B lying on a table weighs W. Te coefcient of static friction between the block and the table is m. Assume that the cord between B and the knot is horizontal. Te maximum weight of the block A for which the system will be stationary is(a) Wtanqm(b)mW tanq(c)m q W 12+ tan (d)mW sinq29.Teinputstothedigitalcircuitareshownbelow. Te output Y is(a)A B C + + (b)( ) A BC +(c)A B C + + (d)A B C + +30.TwoparticlesAandBhavingdiferentmasses areprojectedfromatowerwithsamespeed.A isprojectedverticallyupwardandBvertically downward. On reaching the ground(a)velocity of A is greater than that of B.(b)velocity of B is greater than that of A.(c)both A and B attain the same velocity.(d)the particle with the larger mass attains higher velocity.Category II (Q31 to Q35)Eachquestionhasonecorrectoptionandcarries 2marks,foreachwronganswer1/2markwillbe deducted.31.Two cells A and B of e.m.f. 2 V and 1.5 V respectively, are connected as shown in fgure through an external resistance 10 W. Te internal resistance of each cell is 5 W. Te potential diference EA and EB across the terminals of the cells A and B respectively arePHYSICS FOR YOU | JUNE 15 68(a)EA = 2.0 V, EB = 1.5 V (b)EA = 2.125 V, EB = 1.375 V(c)EA = 1.875 V, EB = 1.625 V(d)EA = 1.875 V, EB = 1.375 V32.Achargeqisplacedatonecornerofacube.Te electric fux through any of the three faces adjacent tothechargeiszero.Tefuxthroughanyoneof the other three faces is(a)q/3e0(b)q/6e0(c)q/12e0 (d)q/24e033.Inthecircuitshownbelow,theswitchiskeptin positionaforalongtimeandisthenthrownto position b. Te amplitude of the resulting oscillating current is given by(a) E L C /(b)E/R(c)infnity(d) E C L /34.Te pressure p, volume V and temperature T for a certain gas are related bypAT BTV=2,where A and B are constants. Te work done by the gas when thetemperaturechangesfromT1toT2whilethe pressure remains constant, is given by(a) A T T BT T ( ) ( )2 1 2212 + (b) AT TV VB T TV V( ) ( )2 12 122122 1(c) A T T BT T ( ) ( )2 1 2212 (d) AT TV V( )2 222 135.Acylinderofheighthisflledwithwaterandis kept on a block of height h/2. Te level of water in the cylinder is kept constant. Four holes numbered 1,2,3and4areatthesideofthecylinderandat heights 0, h/4, h/2 and 3h/4 respectively. When all four holes are opened together, the hole from which water will reach farthest distance on the plane PQ is the hole no.(a)1(b)2(c)3(d)4Category III (Q36 to Q40)Eachquestionhasoneormorecorrectoption(s), choosing which will fetch maximum 2 marks on pro ratabasis.However,choiceofanywrongoption(s) will fetch zero mark for the question.36.Consider two particles of diferent masses. In which ofthefollowingsituationstheheavierofthetwo particles will have smaller de Broglie wavelength?(a)Both have a free fall through the same height.(b)Both move with the same kinetic energy.(c)Both move with the same linear momentum.(d)Both move with the same speed.37.Acirculardiscrollsonahorizontalfoorwithout slippingandthecentreofthediscmoveswitha uniformvelocityv.Whichofthefollowingvalues thevelocityatapointontherimofthedisccan have?(a)v(b)v(c)2v(d)zero38.Aconductingloopintheformofacircleis placedinauniformmagneticfeldwithitsplane perpendicular to the direction of the feld. An e.m.f. will be induced in the loop if(a)it is translated parallel to itself.(b)it is rotated about one of its diameters.(c)it is rotated about its own axis which is parallel to the feld.(d)the loop is deformed from the original shape.39.Findtherightcondition(s)forFraunhofer difraction due to a single slit.(a)Sourceisatinfnitedistanceandtheincident beam has converged at the slit.(b)Source is near to the slit and the incident beam is parallel.(c)Sourceisatinfnityandtheincidentbeamis parallel.(d)Source is near to the slit and the incident beam has converged at the slit.PHYSICS FOR YOU | JUNE 15 6940.Two charges +q and q are placed at a distance a inauniformelectricfeld.Tedipolemomentof the combination is2qa i j (cos sin ),^ ^q q +where q is the angle between the direction of the feld and the line joining the two charges. Which of the following statement(s) is/are correct?(a)Tetorqueexertedbythefeldonthedipole vanishes.(b)the net force on the dipole vanishes.(c)Tetorqueisindependentofthechoiceof coordinates.(d)Te net force is independent of a. SOLUTIONS1.(b) :For frst lens, f1 = 2 m, u1 = 4 m, v1 = ?Using lens formula, 1 1 11 1 1v u f =1 14121v=; 1 1214141v= =v1 = 4 mFor second lens, f2 = 1 mObject distance, u2 = |v1| 3 = 4 3 = 1 mImage distance, v2 = ?Again using lens formula, 1 11112v =or11 1 22v= + = ;v2 = 0.5 mSo, distance of fnal image from the source point= 4 + 3 + v2 = 4 + 3 + 0.5 = 7.5 m2.(b) :Given situation is shown in the fgure.L = length of pendulum m = mass of the bobq=anglebetween pendulum and verticalv = speed of pendulumLet T be the tension in the string at given situation.Acceleration of the bob= vL2Using Newtons second law of motion, FmvLnet =2 T mgmvL = cos q2 T mgmvL= + cos q23.(c):Let the unstretched length of the wire be L.Area of cross-section of the wire = A (say)Youngs modulus,Y =StressStrainCase I :Length of wire = L1Tension in the wire = T1Extension in the wire = (L1 L)\ YT AL L LT LA L L==1111// ( ) ( )...(i)Case II :Length of wire = L2Tension in wire = T2Extension in wire = (L2 L)\ YT AL L LT LA L L==2222// ( ) ( )...(ii)From eqns (i) and (ii), T LA L LT LA L L1122( ) ( ) =T1(L2 L) = T2(L1 L)T1L2 T1L = T2L1 LT2L(T2 T1) = T2L1 T1L2\LT L T LT T=2 1 1 22 14.(b) :Here, density of metal = r Mass of hollow sphere= (Surface area) (thickness) r= 4pR2 t rSphere will foat in water if ( ) 4432 3p r p r Rt g R gw tRw3rr\ tR3r[rw = 1 g cm3]5.(c):AsRlAl lA llV= ==r r r2Here, r and V are constant.\R l2\ RRll121222=Given,l2 = 2l1Hence, RRll121212414= =R2 : R1 = 4 : 1PHYSICS FOR YOU | JUNE 15 706.(d) :In the circuit, diode D1 is reversebiasedandofers infniteresistance,diodeD2 isforwardbiasedandofers 50 W resistance. Equivalent circuit is redrawn.Total resistance of the circuit, R = 50 + 50 + 150 = 250 WV = 10 V, I = ?IVR= = =102500 04 .ASo, current through the resistance 150 W is 0.04 A.7.(b) :Assuming magnetic feld (B), conductor length (l)anditsvelocity(v)aremutuallyperpendicular so,emfinducedbetweenthetwoendsofthe conductore = BlvHere,B = 0.1 T, l = 0.1 m, v = 15 m s1, e = ?e = 0.1 0.1 15 = 0.15 V8.(b) :Given situation is shown in the fgure.Angle of refraction, r = (90 i)Using Snells law at the interfacem1 sin i = m2 sin r1 sin i = m sin(90 i) = m cos itan i = m9.(a) :Totalangularmomentumoftwoparticles about the point O L L Lnet A B= += + r p r pA A B BHere, r j r iA B= = 1 5 2 8 . , . m m p mv i iA A= = = ( . )( . ) ( . ) 6 5 2 2 14 3N s p mv j jB B= = = ( . )( . ) ( . ) 3 1 3 6 11 16N s\ L j i i jnet= + ( . ) ( . ) ( . ) ( . ) 1 5 14 3 2 8 11 16= + 21 45 31 248 9 8 . ( ) . ( ) . k k k Lnet=9 82 1.kg m s10.(b) :Here, v iA=( ) 101 m s v i jB= + ( cos sin ) 20 60 20 601 m s=+= +( ) ( ) ( 2012203210 10 31i j i j ) m sRelative velocity of B with respect to A is given by v v vBA B A= = + =( ) 10 10 3 10 10 31i j i j ms11.(c):Whenlightisrefracted/refectedfroma surface,thenfrequencyoflightdoesnotchange because it depends on the source of light.12.(*):Rateofheatradiationemittedbyabodyat temperature t1 C (= (t1 + 273) K).u1 = e s A (t1 + 273)4Rateofheatradiationabsorbedbyabodydueto surrounding temperature t2C (= (t2 + 273) K)u = e s A (t2 + 273)4Net rate of heat absorbed by the body= u u1 = e s A [(t2 + 273)4 (t1 + 273)4]*None of the given options is correct.Option(a)wouldbecorrectift1andt2werein kelvin.13.(b) :Range of work function of metals = 2 5 eV hc = 4 1015 eVs 3 108 m s1= 1200 eV-nmAs,l = hcE lmin= =1200240eV nm5 eV nm- lmax= =1200600eV nm2 eV nm-Hence light of wavelength 650 nm cannot be used for photoelectric efect.14.(d) :Here, m = 1.6, n = 7, l = 600 nmt = ?According to question, shif in central point due to insertion of plastic sheet = 7bor ( ) m l = 17tDdDd\t == 717 600 101 6 19lm .PHYSICS FOR YOU | JUNE 15 71 = =7 6 100 67 1076. mt = 7 mm15.(c):Let length of open organ pipe = lolength of closed organ pipe = lcAlso, lo = 2lcFundamental frequency of the open pipe,uo = 100 HzAlso,uoovl= =2100 vlvlo c= = 2002200 ; vlc= 400Fundamental frequency of closed organ pipe uccvl= = =44004100 HzSothefrequencyofthirdharmonicoftheclosed organ pipe = 3uc = 300 Hz.16.(c):Here,C1 = 5 mF C2 = 10 mFV = 300 VU = ?Equivalent capacitance of the circuit, CC CC C=+=+= = 1 21 265 105 1010310310 mF F U CV = = =121210310 300 0 152 6 2( ) .J17.(c):Given situation is shown in the fgure.Gravitational force, FGm mr=1 22Acceleration of particle of mass m1, aFmGmr1122= =\a1 m218.(b) :Here m1 = m2 = m, m3 = 3ms1 = s2 = s3 = s, T1 = 40 C, T2 = 50 C, T3 = 60 CLetTbethefnaltemperatureatthethermal equilibrium.Using principle of calorimetry,Heat gained by m1 and m2 = heat lost by m3m1s1(T T1) + m2s2(T T2) = m3s3(T3 T)ms(T 40) + ms(T 50) = 3ms(60 T)T 40 + T 50 = 3(60 T)2T 90 = 180 3T5T = 270; \T = 54C19.(a) :As potential energy of satellite VGMmr= Kinetic energy of satellite,KGMmr=2Total energy of satellite, E = K + V EGMmrGMmrGMmr= = 2 2orK VVKV+ = \ = 2 2;20.(d) :Electric feld due to charged conducting sheet, E = se0 Let T be the tension in the string.Ball is in equilibrium soT cos q = mgT sin q = qE\ =TTqEmgsincosqq tanqse= =qEmgqmg021.(a) :Given circuitcan be redrawn as shown in fgure.Here, frst two cells are in parallel so net emf,eeeqrr= =++=1222212122 Vrreq = = + =1 12121WCurrent in the circuit,I =++= =2 21 2431 33 .A22.(c):As,vRTMrms =3For oxygen molecules at temperature T, v vRTMrms = =3 ...(i)Now, temperature is doubled and oxygen molecules PHYSICS FOR YOU | JUNE 15 72dissociateintooxygenatom(molarmassbecomes M/2) then rms speed will be = = = vR TMRTMvrms3 22232( )( ) /[using eqn. (i)]23.(c):Radius of circular orbit of a charged particle in a magnetic feld is given by, rmvqBpqBmKqB= = =2...(i)Here, K = Kinetic energy of charged particleChargeqisacceleratedthroughsomepotential diference V. So kinetic energy of charge, K = qVFrom eqn. (i), rmqVqB=2For given q, V and B, r m rrmmABAB= or mmrrRRABAB= = 212224.(b) :Suppose along arc AB there is a large number of particles. If arc AB 0the position of centre of mass will be at distance R from the origin.If arc length AB increases, centre ofmassofthesystemstarts moving down (< R).25.(d) :Capillaryriseinanacceleratedlifusedasa plateform, hSr aeff=2 cos qrHere, aef = 0(for free falling plateform)\h = It means length of water column in the tube will be 20 cm.26.(a) :AccordingtoDopplersefectofsound, apparent frequency heard by observer, u u =+v vv vs00Here, u0 = 1000 Hz, v0 = vs = 33 m s1v = 333 m s1, u = ? u =+ = =333 33333 3310003663001000 1220 Hz27.(d) :Here, wavelength of photon, l = 300 nmh = 4 1015 eVs, c = 3 108 m s1Ionization energy of hydrogen = 13.6 eVEnergy of photon,Ehc=l = ( )( )( )4 10 3 10300 1015 8 19eVsm sm= 4 eVFor hydrogen atom,Enn = 13 62.eVFirst excitation energy = E2 E1 = 13 6213 612 2. .= 3.4 + 13.6 = 10.2 eVSinceenergyofphotonislessthan10.2eV,sono excitation is possible, i.e., electron will keep orbiting in the ground state of atom.28.(b) :Weight of block B = WWeight of block A = W(say)T = tension in the stringNormal on block B, N = WFriction force on block B, f = mN = mWSystem will be in equilibrium, ifT cos q = f = mW...(i)T sin q = W...(ii)Divide eqn (ii) by eqn (i), we get sincosqq m= WW\W = mW tan q29.(c): Output of NAND gate, Y AB1 =Output of NOT gate, Y C2 =Output of OR gate,Y = Y1 + Y2= + AB C = + + A B C ( ) AB A B = +PHYSICS FOR YOU | JUNE 15 7330.(c)31.(c):Suppose current I is 5 10 5 1.5 V2 VIAB fowing through the circuit. UsingKirchhofsvoltage law in the circuit2 1.5 5I 10I 5I = 00.5 = 20I;I 0 520140.ATerminal potential diference across the cell A,EA = 2 I r 21405 1 875 . VTerminal potential diference across the cell B,EB = 1.5 + I r - 1 51405 1 625 . .V32.(d)33.(d) :During connection of switch with point a for long time, capacitor gets fully charged.Charge on capacitor, q = CEEnergy stored in capacitor,UqC22When switch is thrown to position b, there is an LC oscillating circuit. Suppose amplitude of current in LC circuit is I0.Using energy conservation principle,Maximum electrical energy = Maximum magnetic energyqCLI202212CECLI I ECL2 2020 . ;34.(c):For a certain gas,pressure, pAT BTV2or, VAT BTp2... (i)Work done by the gas at constant pressure,W = pAV = p(V2 V1)(((pAT BTpAT BTp2 221 12[Using eqn (i)]= A(T2 T1) B(T22 T12)35.(b) :Suppose there is a hole in the cylinder at depth y below water level.Velocity of efuxv gy 2Time taken by water to reach on the plane PQ will be thh ygh yg- ,(22 3 2( ).Horizontal distance x travelled by the liquid is x vt gyh yg 23 2 x y h y 2 3 2 ( )For x to be maximum, dxdy 0 12 2 3 22 3 4 0y h yh y( )( ) or,3h 4y = 0 . yh 34Hence, x will be maximum at yhhh ,(34 4where hole number 2 is present.36.(a,b,d) : de-Broglie wavelength of a particle is given by,/ hphmvhmKK2; kinetic energy(a)If both the particles are allowed to fall through same height, their speed will be same. v gh 2. / 1mSoheavierblockwillhavesmallerde-Broglie wavelength. i.e., option (a) is correct.(b)For same kinetic energy (K), / 1mAgainheavierblockwillhavesmallerde-Broglie wavelength, i.e., option (b) is correct.PHYSICS FOR YOU | JUNE 15 74(c)For same momentum, both particles have same de Broglie wavelength.So option (c) is incorrect.(d)For same speed, l 1mAgain, heavier particle will have smaller de-Broglie wavelength, i.e., option (d) is correct.37.(a, c, d) : In case of pure rolling,v = wRSo, velocity at the point of contact,vg = v wR = 0Velocity at the top,vt = v + wR = 2 v Velocityatapointon therimofdisc,vr=vis possible if v and wR are at 120.Velocity at a point on the rim of disc, vr = v is not possible because 1 cosq 1.38.(b, d) : An emf will be induced in the loop if there is change in magnetic fux through it.Flux changes through the loop only in options (b) and (d).39.(b, c) : In Fraunhofer difraction, the incident rayson the slit are parallel. Tis can be achieved by keeping :(a)point source at the focus of a converging lens(b)point source at infnite distance40.(b, c, d) : Net force on the dipole = +qE qE = 0So, option (b) is correct.Torque on the dipole,t = pEsinq 0 Here,p,Eandqareindependentofchoiceofco-ordinates.So, options (c) and (d) are correct.MTG o f f e r s C l a s s r o o m S t u d y Ma t e r i a l f o r J EE( Mai n&Advanc ed) ,AI PMTand FOUNDATIONMATERIALforClass7,8,9, 10,11&12with YOURBRANDNAME& COVER DESIGN.This study material will save you lots of money spentonteachers,typing,proof-readingand printing.Also,youwillsaveenormoustime. Normally, a good study material takes 2 years to develop. But you can have the material printed with your logo delivered at your doorstep.Prot from associating with MTG Brand the most popular name in educational publishing for JEE (Main & Advanced)/AIPMT/PMT ....Order sample chapters on Phone/Fax/e-mail.Phone : 0124-495120009312680856, 09717933372e-mail: sales@mtg.in | www.mtg.inATTENTIONCOACHINGINSTITUTES:a great offer from MTGCONTENTPAPERPRINTINGEXCELLENTQUALITYCLASSROOM STUDY MATERIALYour logo herePHYSICS FOR YOU | JUNE 15 75PHYSICS FOR YOU | JUNE 15 761.A boy standing in an elevator observes a screw falls from the ceiling. Te ceiling is 3 m above the foor. (a)Iftheelevatorismovingupwardwithaspeed of2.2ms1,howlongdoesittakeforthescrew to hit the foor? (b) How long is the screw in air if theelevatorstartsfromrestwhenthescrewfalls, and moves upwards with a constant acceleration of a = 4 m s2?2.Anenemyfghterjetisfyingataconstantheight of250mwithavelocityof500ms1.Tefghter jetpassesoverananti-aircrafgunthatcanfre atanytimeandinanydirectionwithaspeedof 100ms1.Determinethetimeintervalduring which the fghter jet is in danger of being hit by the gun bullets.3.A particle is to be projected from a point P so that it maystriketheinclineperpendicularly.Determine the required velocity of projection if it is horizontal initially.4.Asteelballwhichcanslideonasmoothrodof length L is attracted by an electromagnet. Te force ofattractionimpartsanaccelerationa=K/(Lx)2, whereKisconstantandxisdistancetravelled. If the ball is released from rest at x = 0, determine the velocity v with which it strikes the pole face.5.Te current velocity of a river grows in proportion tothedistancefromitsbankandreachesthe maximumvaluev0inthemiddle.Nearthebanks the velocity is zero. A boat is moving along the river in such a manner that it is always perpendicular to current and the speed of the boat in still water is u. Find the distance through which the boat crossing the river will be carried away by the current if the width of river is c. Also trace the trajectory of boat.6.Agas-flledsphericalballoonisexpanding. Te radius of sphere at time t is r. Find the radius whentheratesofincreaseofthesurfaceareaand the radius are numerically equal.7.Aprojectileisfredwithvelocityufromagun adjustedforamaximumrange.Itpassesthrough twopointsPandQwhoseheightsabovethe horizontalareheach.Showthattheseparation between the two is ugu gh24 .8.Aparticleismovinginacircleofradiusrinsuch awaythatatanyinstant,thetotalacceleration makes an angle of 45 with radius. Initial speed of particleisu.Findthetimetakentocompletethe frst revolution.By : Prof. Rajinder Singh Randhawa**Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699PHYSICS FOR YOU | JUNE 1577SOLUTIONS1.(a) Consider the elevator foor to be origin. It moves with a constant velocity. Equation for fooryf = vft = 2.2t ...(i)Te screw falls withacceleration due togravityy h t gts = + 2 2122. ...(ii)At time t, ys = yf i e t h t gt . ., . . 2 2 2 2122= + or thg= ==2 2 39 80 78.. s(b)Equation for foor is y a tf f=122Equation for screw is y h gts = 122When the screw meets the foor, ys = yf\ =123122 2a t gtfor s tg af=+=2 30 66 .2.Te equation of trajectory of bullet isy xgxu= tan sec q q12222 = + xgxutan ( tan ) q q121222 ...(i)Foragivenvalueofx,maximumycanbedeter-mined fromdydxgxu(tan )( tan )qq = =122 022or tanq =ugx2...(ii)Putting (ii) in (i), we get, yuggxumax = 222212Te bullet can hit an area defned by yuggxu 222212On substituting given values, we get xx22000250 500 2 500 2 + orTefghterjetcantravel 1000 2 mwhileitcan behit.Sotheplaneisindangerforaperiodof 1000 25002 2 = s3.Letvbethevelocity whentheparticlestrikes theincline.Tevelocity components are vx = v0 and vy = gt\ = = = tan cot q qvvvgttvgxy0 0Since, x = v0t, we have xvg=02cot q ...(i)Te equation of trajectory of particle is y hgxv= 12202 ...(ii)Te incline is a straight line, its equation isy= xtanq...(iii)On solving (i), (ii) and (iii) we get vgh0222=+ cot q4.Acceleration of ball, aKL x= ( )( )2givenSince, advdtdvdxdxdtvdvdx= = =\ =vdvdxKL x ( )2 v dvKdxL x= ( )2On integrating, v dvK dxL xvL D= 0202( )/ vKL xvKD LL D 202 22122 1== /or\ = v KD L22 1 PHYSICS FOR YOU | JUNE 15 785.Current velocity, v = ky v = v0 whenyc=2,\ = = vkc kvc0022ory ut vvcut = \ =20\Rate of change of velocity = =vtv uc20\Accelerationoftheboatalongthewater current,av ucx =20\ = +x tv uct 01220 2 xv ucyuvucy y ut = = =0220 2( ) \ = yucvx20It represents the equation of a parabola.When ycxcvu= =2140,Hence, drif =220xcvu=6.Surface area of sphere is S = 4pr2,So,dsdtrdrdt= 8pAs dsdtdrdt= (given)Hence 8pr = 1or r =18 p7.Te trajectory of projectile is given by y xgxu= + tan ( tan ) q q22221Gun is adjusted for maximum range, i.e.,q = 45\ = y xgxu22 For y = h, we have,h xgux xugxugh = + =22 22 20 or If x1 and x2 are roots of the above equation, x xugx xugh1 221 2