Saakaar 2025 Physics Modern Physics Length Contraction, Time Dilation Q 1 Imagine a spaceship 100 meters long in its rest frame DPP: 1 Q 6 The average lifetime of a muon at rest is about 2.2μs. (when stationary) travels at a constant velocity of 0.6c A laboratory measurement of muons moving in a (where c is the speed of light) relative to Earth. The beam yields a lifetime of 10μs. If c is the velocity of length of the spaceship as measured by an observer on light, what is the speed of the muons in the beam ? Earth (in terms of m) is (A) 0.955 c (B) 0.975 c (C) 0.995 c (D) 0.935 c c Q 2 A rod of rest length L0 moves with speed 2 with respect to an observer such that velocity of the rod Q 7 A spaceship travels at a constant velocity of 0.8c makes 600 with its length. Find apparent length of the (where c is the speed of light) relative to Earth. A rod measured by the observer. clock on the spaceship displays a time interval of 1 (A) 15 L0 4 (B) √15L0 4 (C) √ 15 L0 4 (D) 5 L0 4 Q 3 Determine the X-component of the length of the rod of 10 m observed in a frame which is moving with hour. The corresponding time interval measured by an observer on Earth is ( In terms of hours) (A) 1.91 (B) 1.67 (C) 1.89 (D) 1.52 Q 8 What is the mean life of a meson travelling with velocity 0.6c in a direction making an angle 30 0.73c. If the proper mean life is 2. 3 × 10−8sec. degrees with the rod (A) 10m (A) 2. 2 × 10−7 sec (B) √3m (B) 3. 3 × 10−7 sec (C) 2√3m (D) 4√3m (C) 2. 2 × 10−8 sec Q 4 The Length contraction occur only ? (A) in perpendicular to the direction of motion (D) 3. 3 × 10−8 sec Q 9 Light takes 4 hours to cover the distance from Sun to (B) Along the direction of motion Neptune. If you travel in a spaceship at a speed 0.99c (C) Object's observed length is shorter than its proper (where c is the speed of light in vacuum), the time (in length. minutes) required to cover the same distance (D) Object's observed length is greater than its proper length. approximately. c Q 5 A rod of rest length L0 moves with speed 2 with respect to an observer such that velocity of rod makes 450 with its length. The apparant length of the rod measured by observer is measured with a clock on the spaceship will be √nL0 2√2 . Then the value of n is ........(Answer must be in interger). (A) 34 (B) 56 (C) 85 (D) 144 Q 10 A rod has length of 1m. It is moving in a spaceship with velocity 0.4c relative to the earth. The length of the rod measured by an observer in spaceship will be (A) 1m (B) 0.9m (C) 0.5m (D) 0.25m
Saakaar 2025 Physics Modern Physics Velocity Addition, Mass Energy Equivalence Q 1 The mass of a moving electron is 11 times its rest DPP: 2 Q 6 A rod of proper length l 0 oriented parallel to the x- mass. Calculate its kinetic energy and momentum axis moves with speed 2c/3 along the x-axis in the S- (A) 6MeV, 3 × 10 −21 N sec frame, where c is the speed of the light in free space. (B) 5. 1 MeV , 2. 98 × 10 −21 N sec The observer is also moving along the x-axis with (C) 7MeV, 2. 98 × 10 −21 N sec speed c/2 with respect to the S-frame. The length of (D) 5. 1 MeV , 3 × 10 −21 N sec the rod as measured by the observer is (A) 0. 55l 0 (B) 0. 48l 0 Q 2 Find the velocity of a 0.1MeV electron according to (C) 0. 87l 0 (D) 0. 97l 0 classical and relativistic mechanics. (A) 3 × 10 8 m/s (B) 1. 64 × 10 9 m/s (C) 3 × 10 6 m/s (D) 1. 64 × 10 8 m/s Q 7 At what velocity will the mass of a body is 2.25 times its rest mass? (A) 3. 2 × 10 9 m/s Q 3 The momentum of an electron (mass m) which has the (B) 2. 68 × 10 8 m/s same kinetic energy as its rest mass energy is (c is (C) 4. 8 × 10 10 m/s velocity of light) (D) 5 × 10 6 m/s (A) √3m 0 c 2 (C) m 0 c (B) √2m 0 c (D) m0c/√2 Q 4 In a certain inertial frame two light pulses are emitted at point 5km apart and separated in time by 5μs an observer moving at a speed v along the line joining these points notes that the pulses are simultaneous. Therefore, v is (A) 0.7c (D) 0.9c 3 its rest mass. The linear momentum of the particle is (A) 2m 0 c (B) 2m0 (D) √3 m0 c √3 Q 9 Find the speed that a proton must be given if its mass is to be twice its rest mass 1. 67 × 10 −27 & What energy must be given to proton to achieve this speed? (A) v = √3 2 (B) v = c 1 2 c Q 10 Two events separated by a (spatial) distance 6 × 10 9 Q 5 The mass m of a moving particle is 2m0 , where m 0 is √ (C) m 0 c of 2.000 MeV/c. The total energy in MeV is (C) E = 1. 27 × 10 −10 J (D) E = 1. 27 × 10 −12 J (B) 0.8c (C) 0.3c Q 8 An electron (m = 0. 511 MeV /c 2 ) have momenta m, are simultaneous in one inertial frame. The time interval between these two events in a frame moving with a constant speed of 0.8c (where the speed of light c = 3 × 10 8 m/s) is (A) 40 s 3 (B) 80 s 3 (C) 40s (D) 80s
citsivitaler sti fo oitar eht ,nehT .ygrene ssam tser sti eciwt si elcitrap gnivom a fo ygrene latoT 8Q c c 31 7 31 11 )D( )B( c c 31 9 31 21 = xv 8 01 8 xv m 52.1 )C( m8.0 )D( .htrae eht no revresbo na ot sraeppa ti sa htgnel sti etaluclaC .c6.0 yticolev a htiw htgnel sti gnola gnivom si dor gnol m 0.1 A 2Q ces )D( 2− 01 = 't m5 = ' z )C( 8 8 .ces/m × 7.2 fo deeps a htiw tekcor a ni deirrac si ti nehw htgnel sti dniF .m 0.2 fo htgnel a sah dor A 3Q )C( yticolev s/m 01 × 2 = m 1 )A( m 2.1 )B( )A( si noitcerid-X ni B ot tcepser htiw A fo yticolev ehT .s/m 801 × 2− sah B elcitrap rehtona dna yticolev a sah A elcitrap eno yrotarobal a nI 7Q s/m 01 × 999 .2 = v 6 s/m 01 × 3 = v 6 s/m 01 × 999 .2 = v s/m 01 × 4 = v m01 = 'y )B( m00926− = 'x ces )A( 4− )D( 01 = 't m01 = 'y eht fo ssam tser eht lauqe ssam sti evah m5 = ' z ?notorp m00072− = 'x ot redro ni evom nortcele na tsum tsaf woH 6Q ces c 28.0 )C( c 48.0 )D( )C( 01 = 't m01 = 'y ces 4− )B( 01 = 't m5 = ' z m01 = 'y m00962− = 'x %0.8 )C( % 6.5 )D( m 58.0 )C( m 27.0 )D( 2− m5 = ' z c 668.0 )A( c 667.0 )B( m00072− = 'x ?revresbo eht ot evitaler dor fo deeps eht si tahW .tser ta nehw htgnel sti fo flah eb ot dnuof si dor a fo htgnel ehT 5Q % 2.7 )A( % 2.8 )B( .htgnel nwo sti ot °06 ta noitcerid a ni thgil fo yticolev eht semit 8.0 yticolev a htiw gnivom dor a fo noitcartnoc egatnecrep eht etaluclaC 4Q m 279.0 )A( m 278.0 )B( )A( .noitamrofsnart naelilag gnisu sixa 'XX nommoc eht gnola F emarf eht ot tcepser htiw ces / m 8 01 × 7 .2 yticolev htiw gnivom si hcihw 'F emarf ecnerefer a ni tneve eht fo setanidrooc eht etaluclaC .F emarf ecnerefer a ni dnoces 4− 01 × 1 = t dna m 5 = z ,m 01 =y ,m 001 = x ta srucco tneve nA 1Q snoitseuQ tnatropmI emoS scisyhP nredoM 3 :PPD scisyhP 5202 raakaaS IIT JAM
3 c 4 c ≈ ≈ xu )D( xu )B( 2 c ≈ c ≈ xu )C( xu )A( emarf bal ni yticolev eht fo tnenopmoc -X ehT .noitcerid-X evitisop gnola 'S emarf tser a ni c8.0 yticolev htiw metsys etanidrooc a ni s/m 01 × ) k2 + j 4 + i 3( = 'u ˆ ˆ ˆ 7 yticolev sah elcitrap A 01Q )regetni na eb tsum rewsnA( ?n fo eulav eht eb lliw tahW . n eb lliw noitisop lanigiro sti ot c snruter ti nehw )emarf yrotarobal eht ni( elcitrap eht fo deeps eht ,emarf yrotarobal eht ni 3 c fo deeps a htiw sevom won erauqs eritne eht fI .deeps ni egnahc yna tuohtiw noitcerid segnahc ti ,renroc hcae tA .noitisop lanigiro sti ot snruter ti taht os erauqs eht fo sedis ruof eht sesrevart dna erauqs a fo renroc eno morf 2 c deeps a htiw gnivom strats tser ta elcitrap a ,emarf yrotarobal eht nI 9Q . si a fo eulav ehT . 2 a√ eb lliw muccav ni thgil fo deeps eht ot deeps IIT JAM
c ereH .S revresbo yranoitats a ot tcopser htiw noitcerid emas eht ni ,ylevitcepser ,c6.0 dna 4/c 3√ – )D( 2/c )B( 4/c 7√ – )C( 2√/c – )A( c8.0 sdeeps citsivitaler htiw gnivom era hcae m05 htgnel reporp fo B dna A spihsecaps owT 6Q ecafrus s'htrae eht hcaer c 899.0 naht retaerg deeps a ta gnilevart snouM )D( sμ 8 .43 si ecnerefer fo emarf s'htrae ni snoum fo emitefil ehT )C( mk 69.0 si ecnerefer fo emarf s'noum ni erehpsomta s'htrae fo ssenkciht tnerappa ehT )B( :si S morf nees sa elcitrap eht fo deeps ehT .erugfi eht deeps ta gnivom S emarf laitreni na redisnoC 3Q eht ,htrae no revresbo na roF .c 899.0 fo deeps gnola S emarf laitreni rehtona morf yawa 2/c fo ecafrus eht evoba erehpsomta eht fo thgieh fo deeps eht si c erehw ,sixa'x-x nommoc eht gniwollof eht fo hcihW .mk 4.01 si htrae eht gnivom si elcitrap a ,S morf devresbo sA .thgil ?eurt era stnemetats ni nwohs sa ,noitcerid 'y eht ni 2/c deeps htiw ecafrus s'htrae hcaer reven nac snoum ehT )A( a ta ecafrus s'htrae eht sdrawot yllacitrev gnillevart era hcihw snoum redisnoC .sμ2 fo emit efil a evah yehT .erehpsomta reppu eht ni decudorp selcitrap yratnemele era snouM 5Q si dor eht fo htgnel eht ni noitcartnoc egatnecrep ehT .htgnel nwo sti ot c4.0 )C( c5.0 )D( o06 ta noitcerid a ni c8.0 fo deeps a htiw gnivom si dor A 2Q c2.0 )A( c3.0 )B( 5√ )D( c 2√ si ,emarf s’htrae eht ni pihsecaps eht fo yticolev etamixorppa ehT c mk 21 )C( 2 c 3√ )B( c 01 × 80 .1 )A( fo ecnatsid dexfi nac taht metsys troppus efil a sah werc ecaps A 1Q sdrawot gnivom si ecaps retuo morf notorp A 4Q deeps muminim tahW .sruoh 0001 rof ylno tsal s’htrae ni derusaem sa c99.0 yticolev htiw htrae werc eht fo levart efas rof deriuqer eb dluow eht ot lellarap gnilevart ,pihsecaps A .emarf a yb detarapes snoitats ecaps owt neewteb . c79.0 eb ot yticolev s’notorp serusaem ,notorp QYP scisyhP nredoM 4 :PPD scisyhP 5202 raakaaS IIT JAM
1E 2E etisbeW WP | ppA SOi | ppA diordnA )sregetni ni eb tsum rewsnA( si r dna ,q ,p fo mus ehT .s 8−01 × r si elcitrap eht fo emarf tser eht ni elcitrap eht ssorc ot dor eht yb nekat emit dna ,s 8−01 × q si ,dor eht fo emarf tser eht ni dna emarf-S eht .)secalp lamiced owt ot ,nehT .semarf owt eht ni elcitrap eht fo seigrene evitcepser eht eb 2E dna 1E teL .1S ot tcepser htiw i c8 .0 ˆ yticolev htiw gnivom si 2S emarF .1S emarf laitreni na ni )thgil fo deeps eht si c( j c8 .0 ˆ htiw gnivom si elcitrap A 01Q )regetni ni eb tsum rewsnA( si γ dna ,β ,α fo mus ehT .] ) γ 2 c v − 1( − 1[ 2 v β α L = tΔ ni dor eht ssorc ot elcitrap eht rof nekat emit eht dna p eb elcitrap eht fo emarf tser eht ni )c fo smret ni( dor eht fo yticolev evitaler ehT .m6.3 si emarf-S eht ni derusaem sa dor eht fo htgnel ehT .ecaps eerf ni thgil fo yticolev eht si c dna ,S emarf laitreni na ni derusaem era sdeeps eht htoB .woleb erugfi eht ni nwohs sa c8.0 deeps a htiw noitcerid-x evitagen eht gnola gnivom si elcitrap a dna ,noitcerid-x evitisop eht ni htgnel sti gnola c4.0 fo deeps a htiw gnivom si dor A 7Q 1 si c thgil fo deeps eht dna V ,L fo smret ni tΔ fo noisserpxe ehT .L si setanidrooc owt eht neewteb ecnereffid eht taht sdnfi revresbo ehT .0 > tΔ ecnereffid emit htiw ,semit tnereffid owt ta sdne owt sti fo setanidrooc noitisop eht serusaem revresbo dnuorg A .emarf etanidrooc a fo sixa-x eht gnola 0 > V deeps mrofinu htiw sevom L htgnel reporp fo dor latnoziroh A 9Q )regetni ni eb tsum rewsnA( si Q + P fo eulav ehT .sμ Q si B pihsecaps eht ni revresbo na yb nees sa B pihsecaps eht ekatrevo yletelpmoc ot A pihsecaps eht yb nekat emit eht dna sμ P si S revresbo eht yb nees sa B fo tnorf eht htiw enil ni si A fo kcab eht si taht( B pihsecaps eht ekatrevo yletelpmoc 58 )C( 441 )D( ot A pihsecaps eht yb nekat emit ehT 43 )A( 65 )B( yletamixorppa eb lliw pihsecaps eht no kcolc a htiw derusaem ecnatsid emas eht revoc ot deriuqer )setunim ni( emit eht ,)muucav ni thgil fo deeps eht si c erehw( c99.0 deeps a ta pihsecaps a ni levart uoy fI .enutpeN ot nuS morf ecnatsid eht revoc ot sruoh 4 sekat thgiL 8Q .woleb nwohs sa B pihsecaps eht dniheb tsuj si A pihsecaps eht ,0=t tA .thgil fo deeps eht si IIT JAM
owt eht fo rebmun emas sah elpmas a , 0 syad 83 )A( syad 54 )B( syad 95 )C( syad 96 )D( )4 .1 ≈ 4 nl ,1 .1 ≈ 3 nl ekaT( yletamixorppa elpmas evitcaoidar a fo ytivitca ehT 9Q tA .ylevitcepser λ dna λ9 = t tnatsnoc yaced evah B dna A secnatsbus evitcaoidar owT 5Q ces/snoitargetnisid 4 ces/snoitargetnisid si elpmas eht fo )syad ni( efil-flah ehT .syad 03 retfa eulav laitini eht fo %57 ot desaerced si ?sraey ces/noitargetnisid ces/snoitargetnisid .sraey sti ,%210.0 si K 04 9 01 01 4 4 4 × 6.3 )D( 01 01 01 × 3 )C( × 5.2 )B( × 160.1 )A( × 13.1 si efil flah fo ecnadnuba ehT .muissatop naM .gk 001 ni woH .sraey 03 fo efil-flah a sah 731-sC .smarg si thgiew s’ydob eht fo %53.0 taht gnimussa 09 retfa niamer lliw 731-sC fo smarg hcum K 01 sah yllaitini )731-sC( 731-muiseC fo elpmas A 8Q 04 fo ytivitca eht etaluclaC 4Q 342 )A( 053 )C( 002 )D( g05.0 )C( g52.0 )D( 003 )B( g57.0 )A( g00.1 )B( si sraey 02 retfa tnuoma gniniamer ehT .g1 si ssam laitini sti dna sraey .lom/g 891 eb ot uA cimota ekaT . uA 891 uA 891 fo thgiew fo efil-flah ehT 3Q eseht fo enoN )D( mg 9 )C( mg 81 )D( 891 fo gm 00.1 fo ytivitca eht etaluclaC .syad 7.2 si 01 si ecnatsbus evitcaoidar a fo efil flah ehT 7Q mg 3 )A( mg 6 )B( 1− 1− ?dnuof evah 1− uoy dluow lairetam hcum woh ,reilrae setunim 02 tnemerusaem eht enod dah uoy fI .setunim 06 retfa mg 1 ylno dnuof uoy dna lairetam s s s 7− 6− 6− 01 × 5 )C( 01 × 4 .3 01 × 9 .2 )B( )A( .lom/g 891 eb ot uA 891 fo thgiew cimota ekaT .tnatsnoc yaced eht etaluclaC .syad 7.2 si evitcaoidar fo mg 72 fo elpmas a dah uoY 6Q 5 2 )sregetni 3 1 uA 891 fo efil-flah ehT 2Q )D( )C( 4 1 )B( )A( 2 1 niamer lliw ytivitca lanigiro fo noitcarf tahW emoceb ot ielcun fo rebmun ?sruoh 04 retfa ni eb tsum rewsnA(?n fo eulav eht eb lliw tahW . λn eb lliw 1 ) 2 e 1 ( eht fo oitar eht rof nekat emit ehT .ielcun .sruoh 02 si edilcun evitcaoidar a fo efil-flah ehT 1Q ytivitcaoidaR scisyhP nredoM 5 :PPD scisyhP 5202 raakaaS IIT JAM
etisbeW WP | ppA SOi | ppA diordnA 1 )H ( 1 mota negordyH fo ,) H 1 ( mota noretued fo ssam derusaem ehT 9Q sessam dna N 71 F 71 ( M ,4054800 .71 = ) N 71 dna O 71 )D( )C( , u872269 .04 = a C fo sessam 02 14 832 )D( 04 29 U eF 65 )B( eH )D( )A( 0 1 832 832 (m 02 24 )D( )C( )B( )A( fo ssam detaluclaC fo tcefed ssam eht etaluclaC 4Q Ve M 1 .0 )B( Ve M 5− )C( 01 )A( fo ecnadnuba egral ehT 2Q eF 65 VeM 612 )A( VeM 9.0 )B( VeM 4.9 )C( VeM 408 )D( )A( )C( Ve M 84 .11 )D( 01 :si suelcun a ni noelcun rep ygrene gnidnib raelcun mumixam eht )A( sah suelcun 832 sti esuaceb si )C( 4 Ve M 48 .21 )B( epotosi muiclac eht fo suelcun 1 ehT 5Q Vek60.374 Ve M60.374 Ve M603.74 Ve M6037.4 587050 ⋅ 832 = ) U U Ve M 3− ygrene gnidnib eht fo edutingam fo redro ehT 3Q snortuen dna snotorp fo rebmun neve na )D( ygrene yrtemmys llams a )C( ygrene gniriap egral a )B( fo ygrene gnidnib ehT .ylevitcepser 02 1 = p , u566800 .1 = n 91 , u226859 .14 = a C cimota )B( )A( (M (M - Ve M 8 , N srabosi 71 VeM 0061 )C( VeM 0081 )D( 29 U 602 aC .u672700 .1 dna u528169 .04 = a C 14 2 F 71 71 O 71 N 71 4590200 .71 = ) F 4131999 . 61 = )O 71 : era sessam cimotA ?elbats tsom eht eb ot tcepxe uoy dluow , F 71 eht fo hcihw ,sessam detalubat fo sisab eht nO 8Q VeM 021 )A( VeM 0051 )B( )VeM 46.139 = .u.m.a 1 gnikat( ot lauqe yletamixorppa erofereht si 8050.832 dna 7800.1,8700.1 yb nevig era dna nortuen ,mota negordyh a fo sessam ehT 7Q bP aC rof mumixam si noelcun rep ygrene gnidniB 6Q Ve M 48 .41 Ve M 84 .01 si 24 eht morf nortuen a evomer ot dedeen ygrene nehT : si ssecorp eht ni ygrenE gnidniB eht ni niag latot ehT .VeM 5.8 si stnemgarf fo taht elihw VeM6.7 si ielcun detnemgarfnu fo noelcun rep ygrene gnidnib eht ,021 rebmun ssam fo hcae stnemgarf owt otni skaerb 042 rebmun ssam htiw suelcun A 1Q ygrenE gnidniB scisyhP nredoM 6 :PPD scisyhP 5202 raakaaS IIT JAM
snoelcun/ Ve M 8 .9 snoelcun/ Ve M 6 .8 )D( )C( snoelcun/ Ve M 6 .7 snoelcun/ Ve M 7 .6 .suelcun U 29 832 )B( )A( fo noelcun rep ygrene gnidnib eht si tahW .ylevitcepser u8050.832 dna u7800.1, u8700.1 yb nevig era mota U 29 832 dna nortuen ,mota negordyh a fo sessam ehT 01Q Ve M 6532 .2 Ve M 6532 .3 )D( )C( Ve M 6534 .4 Ve M 5634 .2 )B( )A( .suelcun noretued eht fo ygrene gnidnib eht si tahW .u7800.1 dna u8700.1 , u1410.2 IIT JAM
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