Physics - Chapter 2 Motion in One-D (1-8) PDF

0000… CUET (ICAR), UPCATET, BCECE म सवािधक सेले न दे ने वाला सं ान “ ” For :- Maths, Bio Group [PCM, PCB, ACB, PCA]...

0000… CUET (ICAR), UPCATET, BCECE म सवािधक सेले न दे ने वाला सं ान “ ” For :- Maths, Bio Group [PCM, PCB, ACB, PCA] For :- CUET (ICAR) BCECE, UPCATET UP No. 1 Coaching Institute in PCM & PCB Mob. No :- 7905116713, 9454550886, 8574272684 Phone No :- 0512-3149209 SUBJECT physics CLASS 11 Th CHAPTER Motion in One Dimension NAME STUDENT NAME Best Offline & Online Class & Test Series are Available 1  physics Chapter – Motion in One Dimension (,d foeh; xfr) Topic Name – Distance and Displacement (nwjh rFkk foLFkkiu) 1. A body moves 6 m north m east and 10 m vertically upwards, 9. A cat moves from X to Y with a uniform speed VU and what is its resultant displacement from initial position : returns to X with a uniform speed Vd. The average speed ,d oLrq mÙkj fn”kk esa 6 ehVj iwoZ fn”kk esa 8 ehVj rFkk Å/okZ/kj Åij for this round trip is : dh vksj 10 ehVj xfr djrh gSA izkjafHkd fLFkfr ls oLrq dk ifj.kkeh ,d fcYyh ,d leku pky VU ls X ls Y rd tkrh gS vkSj ,dleku pky foLFkkiu gksxkA Vd ls Y ls X ij okil vkrh gSA blesa mldh vkSlr pky gksxhA (a) 10 √2m (b) 10m (a) (b) v v (c) 10/ √2m (d) 10 × 2m 2. A man goes 10m towards North, then 20m towards east (c) (d) displacement is : 10. One car moving on a straight road covers one third of the distance ,d O;fDr 10 ehVj mÙkj dh vksj rRi”pkr~ 20 ehVj iwoZ dh vksj pyrk with 20 km/hr and the rest with 60 km/hr. The average speed is : gS A mldk foLFkkiu gksxk A ,d lh/kh lM+d ij pyrh gqbZ ,d dkj nwjh dk ,d&frgkbZ Hkkx 20 (a) 22.5 m (b) 25m fdeh@?k.Vs dh pky ls rFkk “ks’k Hkkx 60 fdeh@?k.Vs dh pky ls iwjk (c) 25.5m (d) 30m djrh gSA bldh vkSlr pky gSA 3. A person moves 30m north and then 20m towards east and (a) 40 km/hr (b) 80 km/hr finally 30√2 m in south–west direction. The displacement (c) 46 km/hr (d) 36 km/hr of the person from the origin will be : 11. A car travels half the distance with constant velocity of 40 एक 30 मीटर उ र की ओर, िफर 20 मीटर पूव की ओर और अंत म 30√2 मीटर दि ण-पि म िदशा की ओर चलता है । का मूल ान से िव ापन kmph and the remaining half with a constant velocity of होगा: 60 kmph. The average velocity of the car in kmph is – (a) 10 m along north (b) 10 m long south ,d dkj xfr dh izFke vk/kh nwjh 40 fdeh0@?k.Vk ds fu;r osx ls rFkk “ks’k vk/kh nwjh 60 fdeh0@?k.Vk ds fu;r osx ls r; djrh gSA dkj dk (c) 10 m along west (d) Zero vkSlr osx fdeh@?k.Vk esa gksxk & 4. An aeroplane fillies 400 m north and 300 n south and then (a) 40 (b) 45 flies 1200m upwards then net displacement is : (c) 48 (d) 50 ,d ok;q;ku 400 m mÙkj dh vksj 300 m nf{k.k dh vksj rFkk 1200 m 12. A train has a speed of 60 km/hr for the first one hour and 40 Åij dh vksj xfr djrk gS rks dqy foLFkkiu gksxk A km/hr for the next half hour. Its average speed in km/h is : (a) 1200 m (b) 1300m izFke ,d ?kaVs rd fdlh Vªsu dh pky 60 fdeh-@?kaVk rFkk vxys vk/ks rd 40 (c) 1400 m (d) 1500m fdeh- ?kaVk jgrh gSA lEiw.kZ ;k=k esa Vªsu dh fdeh@?ka.Vk esa vkSlr pky gksxhA 5. An athlete completes one round of circular track of radius (a) 50 (b) 53. 33 R in 40 sec. What will be his displacement at the end of 2 (c) 48 (d) 70 min 20 sec : 13. Which of the following is a one dimensional motion : ,d f[kykM+h R f=T;k ds o`Ùkkdkj iFk esa ,d iw.kZ pDDj 40 lSd.M esa fuEu esa lss dkSu lh ,d foeh; xfr gSA yxkrk gSA 2 feuV 20 lSd.M Ik”pkr~ bldk foLFkkiu gksxk A (a) Landing of an aircraft (a) Zero (b) 2R i`Foh ij mrjrs gq, gokbZ tgkt dh xfr (c) 2𝛑R (d) 7𝛑R (b) Earth revolving around the sun lw;Z ds pkjksa vksj ?kwerh gqbZ i`Foh dh xfr (c) Motion of wheels of moving train pyrh gqbZ Vªsu ds ifg;ksa dh xfr Uniform Motion (d) Train running on a straight track lh/ks jsyekxZ ij nkSM+rh Vªsu dh xfr (,dleku xfr) 14. A 150 m long train is moving with a uniform velocity of 45 km/hr. The time taken by the train to cross a bridge of length 850 m is : 150 ehVj yach Vsªu 45 fdeh@?kaVk dh fu;r pky ls xfr eku gSA 850 6. The displacement – time graph for two particles A and B ehVj yacs iqy dks ikj djus esa bls le; yxsxkA are straight lines inclined at angles of 300 and 600 with the (a) 56 sec (b) 68 sec time axis. The ratio of velocities of VA : VB is : (c) 80 sec (d) 92 sec nks d.k A rFkk A ds foLFkkiu le; xzkQ ljy js[kk;sa gSa tks le; v{k 15 A particle moves along a semicircle of radius 10m in 5 ds lkFk Øe”k( 300 o 600 ds dks.k cukrh gSA buds osxksa dk vuqikr VA seconds. The average velocity of the particle is : ,d d.k 10 ehVj f=T;k ds v}Zo`Ùk esa 5 lSd.M esa ?kwerk gSA d.k dk : VB gksxkA vkSlr osx gksxkA (a) 1:2 (b) 1 : √3 (a) 2𝛑 ms-1 (b) 4𝛑 ms-1 (c) √3 : 1 (d) 1 : 3 (c) 2 ms-1 (d) 4 ms-1 7. A bullet emerges from a barrel of length 1.2 m with a speed of 16. The ratio of the numerical values of the average velocity 640 ms-1 Assuming constant acceleration, the approximate and average speed of a body is always : time that it spends in the barrel after the gun is fired is : fdlh oLrq ds fy;s vkSlr osx rFkk vkSlr pky ds la[;kRed ekuksa dk cUnwd dh 1.2 m yEckbZ dh ufydk ls 640 ms-1 dh pky ls md vuqikr lnSoA xksyh fudyrh gSA Roj.k fu;r gksus ij Qk;j gksus ds i”pkr~xksyh }kjk (a) Unity / bdkbZ gksrk gS ufydk esa yxk yxHkx le; gSA (b) Unity or less / bdkbZ vFkok bdkbZ ls de gksrk gS (a) 4 ms (b) 40 ms (c) Unity or more / bdkbZ vFkok bdkbZ ls vf/kd gksrk gS (c) 400 𝛍s (d) 1 s (d) Less than unity / bdkbZ ls de gksrk gS 8. A boy walks to his school at a distance of 6 km with 17. A person travels along a straight road for the first half time constant speed of 2.5 km/hour and walks back with a with a velocity v1 and the next half time with a velocity v2 constant speed of 4 km/hr. His average speed for round The mean velocity V of the man is : trip expressed in km/ hour, is - ,d O;fä ,d lh/kh lM+d ij xfr ds izFke vk/ks le; esa v1 osx ls ,d yM+dk 6 fdeh0 nwj fLFkr vius Ldwy 2-5 fdeh@?k.Vs dh fu;r rFkk vxys vk/ks le; esa v2 osx ls xfe djrk gSA O;fä dk vkSlr osx pky ls tkrk gS rFkk 4 fdeh@?k.Vs dh fu;r pky ls ykSVrk gSA iwjh V gksxkA ;k=k ds fy;s yM+ds dh vkSlr pky fdeh@?k.Vs esa gksxh & (a) = + (b) V = (a) 24 / 13 (b) 40/ 13 (c) 3 (d) 1 / 2 (c) V =√ v v (d) V= 2 18. If a car covers 2/5th of the total distance with v1 speed and 27. The acceleration 𝛂 of a particle starting from rest varies 3/5th distance with v2 then average speed is : with time according to relation a = 𝛂t + 𝛃. The velocity of ,d dkj r; dh xbzZ dqy nwjh dk 2@5 Hkkx v1 pky ls rFkk “ks’k 3@5 the particle after a time t will be : Hkkx v2pky ls r; djrh gS A bldh vkSlr pky gS A fojke ls xfreku ,d d.k dk Roj.k 𝛂 lEcU/k a = 𝛂t + 𝛃 ds vuqlkj (a) √v v (b) le; ds lkFk ifjofrZr gksrk gSA le; t ds i”pkr~ d.k dk osx gksxkA (a) (b) (c) (d) +β + βt (c) 𝛂t2 + βt (d) ( ) 19. The numerical ratio of displacement to the distance covered is always : 28. A car starts from rest and accelerates uniformly to a speed r; fd;s x;s foLFkkiu ,ao nwjh ds vuqikr dk vakfdd eku ges”kk gksrk gSA of 180 kmh-1 in 10 seconds. The distance covered by the (a) Less than one / ,d ls de car in this time interval is (b) Equal to one / ,d ,d dkj fojke ls xfr izkjEHk djrh gs rFkk mldh pky 10 lsd.M esa (c) Equal to or less than one 180 kmh-1 rd ,dleku :i ls Rofjr gksrh gSA dkj }kjk bl le; ,d ds cjkcj vFkok ,d ls de vUrjky esa r; dh xbZz nwjh gSA (d) Equal to or greater than one (a) 500 m (b) 250 m ,d ds cjkcj vFkok ,d ls vf/kd (c) 100 m (d) 200 m 20. A particle covers half of its total distance with speed v1 29. The initial velocity of a particle is u (at t = 0) and the acceleration and the rest half distance with speed v2. Its average speed (f) is given by at. Which of the following relation is valid : during the complete journey is : fdlh d.k dk izkjfEHkd osx u (t = 0 ij) ,ao Roj.k ( f) dk eku gSA ,d d.k viuh dqy ;k=k dh vk/kh nwjh pky v1ls vkSj “ks’k vk/kh nwjh pky fuEu esa ls dkSu lk dFku lR; gSA v2 ls r; djrk gSA lEiw.kZ ;k=k ds nkSju bl d.k dh vkSlr pky gksxhA (a) v = u + at2 (b) v = u + a (a) (b) (c) v = u + at (d) v = u (c) (d) 30. A body starting from moves with constant acceleration. The ratio of distance covered by the body during the 5th 21. A car moves a distance of 200m. It covers first half of the sec to that covered in 5 sec is : distance at speed 60 kmh-1 and the second half at speed v. ,d oLrq fojkekoLFkk los fu;r Roj.k ls xfr djuk izkjaHk djrh gSA If the average speed is 40kmh-1, the value of v is : oLrq }kjk 5 osa lSd.M esa r; dh xbZ nwjh rFkk 5 lSd.M esa r; dh xbZ ,d dkj 200m dh nwjh pyrh gSA ;g izFke vk/kh nwjh dks pky 60 dqy nwjh dk vuqikr gksxkA kmh-1 ls rFkk “ks’k vk/kh nwjh dks pky v ls r; djrh gSA ;fn vkSlr (a) 9/25 (b) 3/5 pky 40kmh-1 gks rc v dk eku gksxkA (c) 25/9 (d) 1/25 (a) 30 kmh-1 (b) 13 kmh-1 31. A particle moves along x–axis as x = 4 (t – 2) + a (t – 2)2 (c) 60 kmh -1 (d) 40 kmh-1 which of the following is true : ,d d.k dh fLFkfr x&v{k ds vuqfn”k x = 4 (t – 2) + a (t – 2)2 ls nh tkrh gS rks fuEu esa ls dkSu lk dFku lR; gSA (a) The initial velocity of particle is 4 Non-uniform Motion d.k dk izkjafHkd osx 4 gS (b) The acceleration of particle is 2a (vleku xfr) d.k dk Roj.k 2a gS (c) The particle is at origin at t = 0 d.k t = 0 ij ewy fcUnq ij gS 22. A particle has an initial velocity of 3ı̂ + 4ȷ̂ and an (d) None of these / buesa ls dksbZ ugh acceleration of 0.4ı̂ + 0.3ȷ̂. Its speed after 10 s is ; 32. A man is 45 m behind the bus when the bus start accelerating ,d d.k dk izkjfEHkd osx 3ı̂ + 4ȷ̂ vkSj Roj.k 0.4ı̂ + 0.3ȷ̂ gS A 10 from rest with acceleration 2.5 m/s2. With what minimum lsd.M ds i”pkr bldh pky gS A velocity should the man start running to catch the bus : (a) 10 units (b) 7√2 units ,d O;fä cl ls 45 m ihNs gSA cl fojke ls pyuk izkjaHk dj 2.5 (c) 7 units (d) 8.5 units m/s2 ds Roj.k ls Rofjr gksrh gSA O;fä dks U;qure fdrus osx ls nkSM+uk 23. A particle starts its motion from rest under the action of a pkfg, rkfd og cl dks idM+ ldsA constant force. If the distance covered in first 10 seconds (a) 12 m/s (b) 14 m/s is S1 and that covered in the first 20 seconds is S2. then : (c) 15 m/s (d) 16 m/s ,d d.k vpj cy ds izHkko esa fojkekoLFkk ls xfr izkjaHk djrk gSA ;fn igys 33. The initial velocity of a body moving along a straight line is 7 10 lsd.M esa pyh nwjh S1 rFkk igys 20 lsd.M esa pyh nwjh S2 gks rksA m/s. If has a uniform acceleration of 4 m/s2. The distance (a) S2 = 2S1 (b) S2 = 3S1 covered by the body in the 5th second of its motion is : ,d ljy js[kk esa xfr”khy fi.M dk izkjfEHkd osx 7 eh@lS gSA bldk (c) S2 = 4S1 (d) S2 = S1 Roj.k ,d leku rFkk 4 eh@lS2 gSA xfr ds 5 osa lSd.M esa fi.M }kjk 24. A particle moves along a straight line OX, At a time t (in pyh gqbZ nwjh gksxhA seconds ) the distance x (in metres) of the particle from O (a) 25 m (b) 35 m is give by x = 40 + 12t – t3 : (c) 50 m (d) 85 m How long would the particle travel before coming to rest : 34. The velocity of a body depends on time according to the ,d d.k OX js[kk ds vuqfn”k xfr djrk gSA fcUnq O ls le; t 2 equation v = 20 + 0.1 t. The body is undergoing : ¼lSd.M esa½ ij d.k dh nwjh x ¼ehVj esa½ fuEu gS x = 40 + 12t – t3 ;fn fi.M ds osx dh le; ij fuHkZjrk lehdj.k v = 20 + 0.1 t2 }kjk fojke fLFkfr esa vkus rd d.k }kjk r; nwjh gksxhA O;Dr gS rks fi.M xfr”khy gSA (a) 24 m (b) 40 m (a) Uniform acceleration (c) 56 m (d) 16 m (b) Uniform retardation 25. A Particle moves in a straight line with a constant (c) Non – uniform acceleration acceleration. It changes its velocity from 10 ms-1 to 20 (d) Zero acceleration ms−1 while passing through a distance 135 m in t second. 35. A particle moving with a uniform acceleration travels 24 The value of t is : m and 64 m in the first two consecutive intervals of 4 sec ,d d.k ljy js[kk esa fu;r Roj.k ls xfr dj jgk gSA t lsd.M esa each. Its initial velocity is – 135 m nwjh r; djus ds ckn bldk osx 10 ms-1 ls 20 ms-1 gks एक समान रण के साथ गितमान एक कण 4 से कंड के पहले दो लगातार अंतराल tkrk gSA rc t dk eku gSA म 24 मीटर और 64 मीटर की या ा करता है । इसका ारं िभक वेग है - (a) 12 (b) 9 (a) 1 m/sec (b) 10 m/sec (c) 10 (d) 1.8 (c) 5 m/sec (d) 2 m/sec 26. The distance travelled by a particle starting from rest and 36. A body starting from rest moves with uniform acceleration. moving with an acceleration 4/3 ms-2, in the third second is : The distance covered by the body in time t is proportional to : fojke ls xfr izkjEHk djrs gq, d.k dk Roj.k 4@3 ms-2 gS d.k }kjk ,d fi.M fojkekoLFkk ls fu;r Roj.k ls xfr izkjEHk djrk gS rc le; rhljs lsd.M esa r; nwjh gksxhA esa d.k }kjk pyh x;h nwjh lekuqikrh gksxh A (a) 10/3 m (b) 19/3 m (a) √t (b) t2/3 2 (c) 6m (d) 4 m (c) t (d) t3 3 37. The velocity of a particle is v = v0 +b gt + ft2. If its position is 49. What is the relation between displacement, time and x = 0 at t = 0, then displacement after unit time (t = 1) is : acceleration in case of a body having uniform acceleration ;fn fdlh d.k dk osx v = v0 +b gt + ft2 gSA bldh t = 0 ij fLFkfr ,dleku Roj.k ls xfr”khy fdlh oLrq ds foLFkkiu] le; rFkk Roj.k esa x = 0 gS rc bdkbZ le; (t = 1) i”pkr~ bldk foLFkkiu gksxkA lgh lEcU/k gksxkA (a) v0 + 2g + 3f (b) v0 + g/2 + f /3 (a) S = ut + f t2 (b) S = ( u + f ) t (c) v0 + g + f (d) v0 + g /2 + f (c) 2 S=v -2fs (d) None of these 38. A car moving with a velocity of 10 m/s can be stopped by the 50. A body is moving according to the equation x = at + bt2 – application of a constant force F in a distance of 20 m. If the ct3 where x = displacement and a, b and c are constants. velocity of the car is 30 m/s, it can be stopped by this force in The acceleration of the body is : ,d fu;r cy F ds vuqiz;ksx ls 10 eh@lS ds osx ls pyrh gqbZ dkj fdlh d.k dk foLFkkiu (x) le; (t) ls fuEu izdkj lacf/kr gS x = at dks 20 eh dh nwjh esa jksdk tk ldrk gS A ;fn dkj dk osx 30 eh@lS gks rks bl cy ds }kjk bls fdruh nwjh esa jksdk tk ldrk gS A + bt2 – ct3 ;gk¡ a, b rFkk c fu;rkad gSA d.k dk Roj.k gksxkA (a) 20/3 m (b) 20 m (a) a + 2bt (b) 2b + 6ct (c) 60 m (d) 180 m (c) 2b – 6ct (d) 3b – 6ct2 39. The displacement of a particle is given by y = a + bt + ct2 51. A particle travels 10 m in first 5 sec and 10 m in next 3 − dt4. The initial velocity and acceleration are respectively sec. Assuming constant acceleration what is the distance travelled in next 2 sec : fdlh d.k ds foLFkkiu dk lehdj.k y = a + bt + ct2 − dt4 gSA ;fn dksbZ d.k izFke 5 lSd.M esa 10 ehVj rFkk vxys 3 lSd.M esa Hkh 10 izkjfEHkd osx rFkk Roj.k Øe”k( gksaxsA ehVj dh nwjh r; djrk gSA ;fn Roj.k dks fu;r ekuk tk;s rks vxys 2 (a) b – 4d (b) −b, 2c lSd.M esa r; dh x;h nwjh gksxhA (c) b, 2c (d) 2c, - 4d (a) 8.3 m (b) 9.3 m 40. Speed of two identical cars are u and 4u at a specific (c) 10.3 (d) None of these instant. The ratio of the respective distances in which the 52. The motion of a particle is described by the equation u = at. two cars are stopped from that instant is : The distance travelled by the particle in the first 4 seconds : fdlh fof”k’V {k.k ij nks leku izdkj dh dkjksa ds osx u rFkk 4u gSaA nksuks fdlh d.k dk xfr dk lehdj.k u = at }kjk fn;k tkrk gS izFke 4 dkjks ds }kjk fojke esa vkus ls iwoZ pyh xbZ nwfj;ksa dk vuqikr gksxkA lSd.M esa d.k }kjk r; dh xbZ nwjh gksxh A (a) 1:1 (b) 1:4 (a) 4a (b) 12a (c) 1:8 (d) 1: 16 (c) 6a (d) 8a 41. A body moves from rest with a constant acceleration of 5 53. A body is moving with uniform acceleration describes 40 m in the m/s2. Its instantaneous speed (in m/s) at the end of 10 sec is : first 5 sec and 65 m in next 5 sec. Its initial velocity will be : 2 ,d oLrq fu;r Roj.k 5 [email protected] ls fojke izkjEHk djrh gSA 10 osa ,d oLrq ,d lkeku Roj.k ls izFke 5 lSd.Mksa esa 40 eh rFkk vxys 5 lSd.M ds vUr esa rkr{kf.kd pky ¼[email protected] esa½ gksxhA lSd.M esa 65 eh pyrh gSA oLrq dk izkjafHkd osx gksxkA (a) 50 (b) 5 (a) 4 m/s (b) 2.5 m/s (c) 2 (d) 0.5 (c) 5.5 m/s (d) 11 m/s 42. A body starts from rest.What is the ratio of the distance 54. The velocity of a bullet is reduced from 200m/s to 100 m/s travelled by the body during the 4th and 3rd second : while travelling through a wooden block of thickness ,d oLrq fojkekoLFkk ls pyuk izkjEHk djrh gS] blds }kjk pkSFks rFkk 10cm. The retardation, assuming it to be uniform, will be : rhljs lSd.M esa r; dh x;h nwfj;ksa dk vuqikr gksxkA ,d 10 lseh eksVkbZ ds ydM+h ds xqVds dks ikj djus ij xksyh dk osx (a) 7/5 (b) 5/7 200 [email protected] ls ?kVdj 100 [email protected] jg tkrk gSA enau ;fn (c) 7/3 (d) 3/7 ,dleku gks rks bldk eku gksxkA 43. The acceleration ‘a’ in m/s2 of a particle is given by a = 3t2 +2t + (a) 10 × 104 m/s2 (b) 12 × 104 m/s2 2 where t is the time. If the particle starts out with a velocity u = (c) 13.5 × 104 m/s2 (d) 15 × 104 m/s2 2m/s at t = 0, then the velocity at the end of 2 second is : 55. Starting from rest acceleration of a particle is a = 2 (t – 1). fdlh d.k dk Roj.k a = 3t2 +2t + 2 [email protected] }kjk fn;k tkrk gS The velocity of the particle at t = 5s is : ;gk¡ t le; gSA ;fn d.k t = 0 ij eh u = 2 lS ls pyuk izkjEHk djrk fojkekoLFkk ls xfr izkjEHk djus okys ,d d.k dk Roj.k a = 2 (t – 1) gS rc 2 lSd.M i”pkr~ d.k dk osx gksxkA gSA t = 5s ij d.k dk osx gksxkA (a) 12 m/s (b) 18 m/s (a) 15 m /sec (b) 25 m/sec (c) 27 m/s (d) 36 m/s (c) 5 m/sec (d) None of these 44. A particle moves along a straight line such that its 56. A particle moves along X-axis in such a way that its displacement at any time t is given by S = t3 – 6t2 + 3t + 4 coordinate X varies with time t according to the equation The velocity when the acceleration is zero is : X = (2 – 5t + 6t2) M. The initial velocity of the particle is : ,d d.k ljy js[kk esa b; izdkj xfr”khy gS fd blds foLFkkiu dk ,d d.k v{k dh vksj bl izdkj xfreku gS fd mldk funsZ”kkad le; ds lehdj.k S = t3 – 6t2 + 3t + 4 ehVj gSA bl d.k dk osx D;k gksxk lkFk ehVj ds vuqlkj ifjofrZr gksrk gSA d.k dk izkjfEHkd osx gksxkA tcfd Roj.k “kwU; gSA (a) − 5m/s (b) 6 m/s (a) 3 ms-1 (b) −12 ms-1 (c) 3m/s (d) 3 m/s (c) 42 ms-1 (d) −9 ms-1 57. An object accelerates from rest to a velocity 27.5 m/s in 10 45. If a body starts from rest and travels 120 cm in the 6th sec then find distance covered by object in next 10 sec : second, then what is the acceleration : ,d oLrq fojkekoLFkk ls Rofjr gksdj 10 lSd.M esa 27-5 eh@lS dk osx ,d dkj fojke ls izkjEHk gksdj NBosa lSd.M esa 120 lseh dh nwjh r; izkIr djrh gS rks vxys 10 lSd.M esa oLrq }kjk r; dh xbZ nwjh gSA djrh gSA dkj dk Roj.k gksxkA (a) 550 m (b) 137.5 m (a) 0.20 m/s2 (b) 0.027 m/s2 (c) 412.5 m (d) 275 m (c) 0.218 m/s2 (d) 0.03 m/s2 58. If the velocity of a particle is given by v = (180 – 16 x) 46. If a car at rest accelerates uniformly to a speed of 144 1/2 m/s, then its acceleration will be : km/h in 20 s. Then it covers a distance of : ;fn fdlh d.k dk osx v = (180 – 16 x)1/2 eh@lS gks rks bldk ,d dkj fojke ls Rofjr gksdj 20 lSd.M esa 144 fdeh@?kaUVk dk osx Roj.k gksxkA izkIr dj ysrh gSA blds }kjk r; nwjh gksxhA (a) zero (b) 8 m/s2 (a) 20 m (b) 400 m (c) −8 m/s2 (d) 4 m/s2 (c) 1440 m (d) 2880 m 59. The displacement of a particle is proportional to the cube 47. The position x of a particle varies with time t as x = at2 – bt3. of time elapsed. How does the acceleration of the particle The acceleration of the particle will be zero at time t equal to : depends on time obtained : fdlh d.k dh fLFkfr x le; t ds lkFk fuEu izdkj ls nh tkrh gS x = fdlh d.k dk foLFkkiu le; dh r`rh; ?kkr ds lekuqikrh gS] rks Roj.k at2 – bt3 fdl le; ij d.k dk Roj.k “kwU; gksxkA le; ds lkFk fdl izdkj ifjofrZr gksxkA (a) a/b (b) 2a / 3b (a) a ∝ t2 (b) a ∝ 2t (c) a/ 3b (d) Zero (c) a ∝ t3 (d) a∝t 48. If a train travelling at 72 km ph is to be brought to rest in a 60. A particle moves a distance x in time t according to equation distance of 200 metres, then its retardation should be : x = (t + 5)−1. The accelartin of particle is proportional to – 72 fdeh@?kUVk ls xfr”khy Vªsu 200 ehVj pydj fojkekoLFkk esa vk एक कण समीकरण x = (t + 5)−1 के अनुस ार समय t म x दू री तय करता है । कण tkrh gSA bl ih dk;Z djus okyk voeanu gksxk A का रण आनुपाितक है - (a) 20 ms-2 (b) 10 ms-2 (a) (velocity)2/3 (b) (velocity)3/2 (c) 2 ms-2 (d) 1 ms-2 (c) (velocity)2 (d) (velocity)−2 4 61. An object moving with a speed of 6.25 m/s, is decelerated at a 70. A 210 m long train is moving due North at a speed of rate given by = − 2.5√v where v is the instantaneous 25m/s. A small bird is flying due South a little above the speed. The time taken by the object to come to rest would be : train with speed 5 m/s. The time taken by the bird to cross 6-25 m/s dh pky ls xfr”khy ,d oLrq ds eUnu dh nj blls nh the train is – 210 मीटर लंबी एक टे न 25 मीटर/से कड की गित से उ र की ओर बढ़ रही है । एक tkrh gS] = − 2.5√v tgk¡ v rkR{kf.kd pky gS oLrq dks fojke छोटा प ी 5 मीटर/से कड की गित से टे न से थोड़ा ऊपर दि ण की ओर उड़ रहा voLFkk esa vkus esa yxk le; gSA है । प ी को टे न पार करने म लगा समय है - (a) 1s (b) 2s (a) 6s (b) 7s (c) 4s (d) 8s (c) 9s (d) 10s 62. A body moves with initial velocity 10ms-1 If it covers a distance of 20m in 2s, Then acceleration of the body is : 10ms-1 ds izkjafEHkd osx ls ,d fi.M xfr djrk gSA ;fn ;g 2 lsdUM Motion Under Gravity esa 20 m dh nwjh r; djrk gS rc fi.M dk Roj.k gksxkA (a) Zero (b) 10 ms-2 (xq:Ro ds v/khu xfr) (c) 5 ms-2 (d) 2ms-2 63. A particle has initial velocity (2𝚤̂ + 3𝚥̂) and acceleration (0.3𝚤̂ + 0.2𝚥̂). The magnitude of velocity after 10 seconds 71. A body is thrown vertically upwards. If air resistance is to will be – be taken into account, then the time during which the body एक कण का ारं िभक वेग (2𝚤̂ + 3𝚥̂) और रण (0.3𝚤̂ + 0.2𝚥̂). होता है । 10 rises is : से कंड के बाद वेग का प रमाण होगा - ,d fi.M dks Å/okZ/kj Åij dh vksj Qsadk tkrk gSA ;fn ok;q dk (a) 9√2 units (b) 5√2 units izfrjks/k ux.; u ekuk tk;s rks oLrq ds Åij tkus dk le; & (c) 5 units (d) 9 units (a) Equal to the time of fall 64. The motion of a particle along a straight line is described oLrq ds uhps fxjus ds le; ds cjkcj gksrk gSA by equation X = 8 + 12 t – t3. Where x is in metre and t in (b) Less than the time to fall second. The retardation of the particle when its velocity oLrq ds uhps fxjus ds le; ls de gksrk gSA becomes zero, is : (c) Greater than the time to fall ,d ljy js[kk ds vuqfn”k fdlh d.k dh xfr dks lehdj.k x = 8 + oLrq ds uhps fxjus ds le; ls vf/kd gksrk gSA 12 t – t3 }kjk ifjHkkf’kr ¼izdV½ fd;k tkrk gSA tgk¡ x ehVj esa rFkk t (d) Twice the time of fall lsdaM esa gSA osx “kwU; gksaus ij d.k dk eanu gSA oLrq ds uhps fxjus ds le; dk nksxquk gksrk gSA (a) 24 ms-2 (b) Zero 72. A ball p is dropped vertically and another ball Q is thrown (c) 6 ms-2 (d) 12 ms-2 horizontally with the same velocities from the same height 65. Potential energy of a particle is related to x coordinate by and at the same time. If air resistance is neglected then : equation x2 – 2x. Particle will be in stable equilibrium at : ,d xsan p dks Å/okZ/kj uhps dh vksj rFkk ,d vU; Q xsan dks {kSfrt ,d oLrq dh foHko ÅtkZ] lehdj.k x2 – 2x ds x funsZ”kkad ls lacf/kr fn”kk esa leku osx o leku Å¡pkbZ ls ,d gh le; ij iz{kSfir fd;k gSA oLrq LFkk;h lkE;oLFkk esa gksxh ;fn A tkrk gSA ;fn gok dk izfrjks/k ux.; eku ys rks A (a) x = 0.5 (b) x =1 (a) Ball P reaches the ground first (c) x=2 (d) x=4 xsan P i`Foh ij igys igq¡psxh (b) Ball Q reaches the ground first xsan Q i`Foh ij igys igq¡psxh Relative Motion (c) Both reach the ground at the same time nksuks xsansa ,d gh le; ij i`Foh ij igq¡psxh (vkisf{kd xfr) (d) The respective masses of the two balls will decide the time xsanksa dk igys igq¡puk xsanks ds nzO;eku ij fuHkZj djrk gS 73. An object is projected upwards with a velocity of 100 m/s. It will strike the ground after (approximately) : 66. Two trains each 50 m long are travelling in opposite direction ,d oLrq dks 100 eh@lS ds osx ls Å/okZ/kj Åij dh vksj iz{ksfir fd;k with velocity 10 m/s and 15 m/s. The time of crossing is : x;k gSA bls i`Fohij Vdjkus esa yxk le; gksxkA ¼yxHkx½A 50 ehVj yEckbZ dh nks jsyxkfM+;k¡ ijLij foijhr fn”kk esa 10 eh@ (a) 10 sec (b) 20 sec lSd.M rFkk 15 eh@ lSd.M ds osx ls xfr”khy gSA os fdrus le; esa (c) 15 sec (d) 5 sec ,d nwljs dks ikj dj ysaxh A 74. A stone dropped from the top of the tower touches the (a) 2s (b) 4s ground in 4 sec. The height of the tower is about : (c) 2 √3s (d) 4 √3s ehukj dh pksVh ls ,d iRFkj ij fxjk;k tkrk gS tks tehu rd igq¡pus 67. A 120 m long train is moving in a direction with speed 20 esa 4 lSd.M dk le; ysrk gSA ehukj Å¡pkbZ yxHkx gksxh m/s A train B moving with 30 m/s in the opposite (a) 80 m (b) 40 m direction and 130 m long crosses the first train in a time : (c) 20 m (d) 160 m 120 ehVj yEch jsyxkM+h A 20 [email protected] ds osx ls fdlh fn”kk esa 75. A body is released from the top of a tower of height h. It xfr”khy gSA ,d vU; jsyxkM+h B tks fd 130 eh yEch gS foijhr fn”kk takes t sec to reach the ground. Where will be the after esa 30 [email protected] ds osx ls xfr”khy gSA ;g igyh Vªsu A dks fdrus time t /2 sec : le; esa ikj dj ysxh A ,d fi.M Å¡pkbZ dh ehukj ls LorU=rkiwoZd fxjk;k tkrk gSA ;g fi.M (a) 6s (b) 36 s i`Foh rd igq¡pus esa t lSd.M dk le; ysrk gSA lSd.M ds i”pkr~ (c) 38 s (d) None of these fi.M dh fLFkfr dgk¡ gksxhA 68. A body is sent across a river with a velocity of 8 km / hr. (a) At h/2 from the ground /tehu ls h/2 dh nwjh ij If the resultant velocity of boat is 10 km/ hr, then velocity (b) At h/4 from the ground /tehu ls h/4 dh nwjh ij if the river is : (c) Depends upon mass and volume of the body ,d uko 8 fdeh@?k.Vs ds osx ls unh ikj djrh gSA ;fn uko dk ;g fi.M ds nzO;eku o vk;ru ij fuHkZj djsxk ifj.kkeh osx 10 fdeh@?k.Vk gks rc unh dk osx gksxkA (d) At 3h/4 from the ground / tehu lss 3h/4 dh nwjh ij (a) 10 km/ hr (b) 8 km/ hr 76. A particle is projected up with an initial velocity of 80 ft / (c) 6 km/ hr (d) 4 km/ hr sec. The ball will be at a height of 96ft from the ground after: 69. A train of 150 m length is going towards north direction at ,d d.k 80 ft / sec ds izkjfEHkd osx ls Åij dh vksj iz{ksfir fd;k tkrk a speed of 10 m/sec. A parrot flies at the speed of 5 m/sec gSA fdrus le; i”plr~ ;g d.k i`Foh ls 96ft dh Å¡pkbZ ij gksxkA towards south direction parallel to the railway track. The (a) 2.0 and 3.0 sec (b) Only at 3.0 sec time taken by the parrot to cross the train is ; (c) Only at 2.0 sec (d) After 1 and 2 sec ;fn 150 ehVj yEch jsyxkM+h mÙkj dh vksj 10 m/sec dh pky ls xfr dj 77. A body falls from rest, its velocity at the end of first jgh gSA ,d rksrk jsy ds iFk ds lekUrj ,ao nf{k.k dh vksj 5 m/sec ds second is (g = 32 ft/sec) – osx ls mM+ jgk gS] rks rksrs }kjk jsyxkM+h dks ikj djus esa fy;k x;k le; gS एक िपंड आराम से िगरता है , पहले से कंड के अंत म इसका वेग है (g = 32 ft/sec) (a) 12 sec (b) 8 sec (a) 16 ft/sec (b) 32 ft/sec (c) 15 sec (d) 10 sec (c) 64 ft/sec (d) 24 ft/sec 5 78. A stone thrown upward with a speed u from the top of the 88. A stoneis shot straight upward with a speed of 20 m/sec tower reaches the ground with a velocity 3u. The height from a tower 200 m high. The speed with which it strikes of the tower is : the ground is approximately – fdlh ehukj ls ,d iRFkj dks izkjfEHkd osx u ls Åij dh vksj iz{ksfir 200 मीटर ऊंचे टॉवर से एक प र को 20 मीटर/से कंड की गित से सीधे ऊपर की fd;k tkrk gS] ftldk i`Foh ij igq¡pus ij osx 3u gks tkrk gSA ehukj ओर फका जाता है । िजस गित से यह जमीन से टकराता है वह लगभग है - dh Å¡pkbZ gSA (a) 60 m/sec (b) 65 m/sec (a) 3 u2/g (b) 4 u2/g (c) 70 m/sec (d) 75 m/sec (c) 6 u2/g (d) 9 u2/g 89. Velocity of a body on reducing the point from which it 79. A ball is thrown vertically upwards with a velocity of was projected upwards, is – 25ms from the top of a tower of height 30m. How long िकसी िपंड का वेग उस िबंदु को कम करने पर होता है जहां से उसे ऊपर की ओर will it travel before it hits ground ेिपत िकया गया था - ,d 30 ehVj Å¡ph ehukj ds “kh’kZ ls xsan dks Å/okZ/kj Åij dh vksj 25 (a) υ=0 (b) υ = 2u eh@lS ds osx ls Qsadk tkrk gSA rc dszn dks lrg rd igq¡pus esa yxk (c) υ = 0.5 u (d) υ = u le; gksxkA (a) 6s (b) 5 s 90. Time taken by an object falling from rest to cover the (c) 4s (d) 12 s height of h1 and h 2 is respectively t1 and t2 then the ratio 80. An object start siding on a frictionless inclined plane and of t1 to t2 is : from same height another object start falling freely : fdlh oLrq dks h1 rFkk h2 Å¡pkbZ;ksa rd fxjus esa yxus okys le; Øe”k ,d oLrq ?k’kZ.k jfgr urlery ij fQlyuk izkjEaHk djrh gS rFkk ,d t1 o t2 gksa rks t1: t2 gksxkA vU; oLrq leku Å¡pkbZ ls Lora=rkiwoZd fxjk;h tkrh gSA (a) h1 : h2 (b) h : h (a) Both will reach with same speed (c) h1 : 2h2 (d) 2h1 : h2 nksuks leku pky ls i`Foh ry ij igq¡prh gS (b) Both with reach with same acceleration 91. A particle is thrown vertically upwards. It is velocity at nksuks leku Roj.k ls i`Foh ry ij igq¡prh gS half of the maximum height is 10 m/s, then maximum (c) Both will reach in same time height attained by it is (Take g = 10 m/s2) – nksuks leku le; ij igq¡prh gS एक कण को उ ाधर ऊपर की ओर फका जाता है । अिधकतम ऊंचाई के आधे (d) None of these / buesa ls dksbZ ugh पर इसका वेग 10 मीटर/से कड है, तो इसके ारा ा अिधकतम ऊंचाई है (g = 81. When a ball is thrown up vertically with velocity V0, It 10 मीटर/से कड2 ल) - reaches a maximum height of ‘h’. If one wishes to triple (a) 8m (b) 10 m the maximum height then the ball should be thrown with (c) 12 m (d) 16 m velocity : 92. Two balls A and B of same masses are thrown from the tc ,d xsan dks Å/okZ/kj Åij dh vksj V0 osx ls Qsadk tkrk gS] rc top of the building. A, thrown upward with velocity V blds }kjk izkIr vf/kdre Å¡pkbZ ‘h’ dks rhu xquk djus ds fy, xsan and B, thrown downward with velocity V, then : dks Qsadus dk osx gksuk pkfg,A leku nzO;eku dh nks xsans A rFkk B fdlh ehukj dh pksVh ls Qsadh (a) √3 V0 (b) 3V0 tkrh gSA ;fn A dks Å/okZ/kj Åij dh vksj V osx ls tcfd B dks (c) 9 V0 (d) 3/2 V0 Å/okZ/kj ukps dh vksj V osx ls Qsadk tk, rksA 82. Water drops fall at regular intervals from a tap which is (a) Velocity of A is more than B at the ground 5m above the ground. The third drop is leaving the tap at i`Foh ry ij A dk osx B ls T;knk gksxk the instant the first drop touches the ground. How far (b) Velocity of B is more than A at the ground above the ground is the second drop at that instant : i`Foh ry ij B dk osx A ls T;knk gksxk ,d uy i`Foh iy ls 5 eh dh Å¡pkbZ ij gSA blls fu;fer le; vUrjky ij ty dh cw¡ns fxjrh gSaA ftl {k.k rhljh cw¡n uy ls fxjus (c) Both A & B strike the ground with same velocity yxrh gS] igyh cw¡n /kjrh ij igq¡p tkrh gSA mlh {k.k ij nwljh cww¡n i`Foh ry ij A o B nksuksa ds osx leku gksxsa i`Foh ls fdruh Å¡pkbZ ij gksxhA (d) None of these / buesa ls dksbZ ugh (a) 2.50 m (b) 3.75 m 93. A cricket ball is thrown up with a speed of 19.6 ms-1. The (c) 4.00 m (d) 1.25 m maximum height it can reach is : 83. A ball is thrown vertically upwards from the top of a ,d fØdsV xsan dks 19-6 [email protected] ds osx ls Å/okZ/kj Åij dh tower at 4.9 ms−1. It strikes the pond near the base of the vksj Qsadk tkrk gSA blds }kjk izkIr vf/kdre Å¡pkbZ gksxhA tower after 3 seconds. The height of the tower is – (a) 9.8 m (b) 19.6 m एक गद को एक टावर के शीष से 4.9 ms−1 पर लंबवत ऊपर की ओर फका जाता (c) 29.4 m (d) 39.2 m है । यह 3 से कंड के बाद टावर के आधार के पास तालाब से टकराता है । टावर की 94. Two balls are dropped from height h and 2h respectively ऊंचाई है - from the earth surface. The ratio of time of these balls to (a) 73.5 m (b) 44.1 m reach the earth is : (c) 29.4 m (d) None of these 84. A bullet is fired with a speed of 1000 m/sec in order to hit a h rFkk 2h Å¡pkbZ;ksa ls nks xsansa NksM+h tkrh gSA bu xsanksa dks i`Foh rd target 100 m away. If g = 10 m/s2, the gun should be arrived igq¡pus esa yxus okys le; dk viqikr gksxkA 100 मीटर दू र िकसी ल को भेदने के िलए 1000 मीटर/से कंड की गित से एक (a) 1 : √2 (b) √2 : 1 गोली चलाई जाती है । यिद g = 10 m/s2, तो बंदूक आनी चािहए - (c) 2:1 (d) 1 : 4 (a) Directly towards the target 95. The acceleration due to gravity on the planer A is 9 times (b) 5 cm above the target the acceleration due to gravity on planer B. A man jumps (c) 10 cm above the target to a height of 2m on the surface of A. What is the height (d) 15 cm above the target of jump by the same person on the planet B - 85. A body starts to fall freely under gravity. The distances fdlh xzg A ij xq:Roh; Roj dk eku ,d vU; xzg B dh rqyuk esa 9 covered by it in first, second and third second are in ratio: xquk gSA ;fn dksbZ O;fDr xzg A dh lrg ij 2 ehVj rd mNyrk gS ,d oLrq xq:Roh; izHkko esa LorU=rkiqoZd fxjuk izkjEHk djrh gSA mlds }kjk izFke] f}rh; o r`rh; lSd.M esa r; dh x;h nwfj;ksa dk vuqikr gksxkA rks xzg B ij og fdruk ÅWpk mNy ldrk gS & (a) 1:3:5 (b) 1 : 2 : 3 (a) 18 m (b) 6 m (c) 1:4:9 (d) 1 : 5 : 6 (c) 2/3 m (d) 2/9 m 86. A body falling for 2 seconds covers a distance S equal to 96. A body falls from rest in the gravitational field of the that convered in next second. Taking g = 10 m/s2, S = earth. The distance travelled in the fifth second of its एक िपंड 2 से कंड के िलए िगरकर अगले से कंड म तय की गई दू री के बराबर S motion is (g = 10 m /s2) : दू री तय करता है । g = 10m/s2 लेते ए, S = ,d fi.M i`Foh ds xq:Roh; {ks= esa fojkekpLFkk ls fxjrk gS xfr ds (a) 30 m (b) 10 m ik¡pos lSd.M esa blds }kjk r; dh xbZ nwjh gksxh (g = 10 m /s2) : (c) 60 m (d) 20 m (a) 25 m (b) 45 m 87. A stone is thrown with an initial speed of 4.9 m/s from a (c) 90 m (d) 125 m bridge in vertically upward direction. It falls down in 97. If a body is thrown up with the velocity of 15 m/s then water after 2 sec. The height of the bridge is – एक प र को एक पुल से ऊ ाधर ऊपर की िदशा म 4.9 मीटर/सेकड की maximum height attained by the body is (g = 10 m/s2) : ारं िभक गित से फका जाता है । 2 से कंड के बाद यह पानी म िगर जाता है । पुल ,d xsan dks i`Foh dh lrg ls Åij dh fn”kk esa dh 15m/s pky Qsadk की ऊंचाई है - tkrk gSA rc xsan }kjk izkIr vf/kdre Å¡pkbZ gSA (g = 10 m/s2) : (a) 4.9 m (b) 9.8 m (a) 11.25 m (b) 16.2 m (c) 19.8 m (d) 24.7 m (c) 24.5 m (d) 7.62 m 6 98. A balloon is rising vertically up with a velocity of 29 Standard Question ms−1. A stone is dropped from it and it reaches the (egRoiw.kZ iz’u) ground in 10 seconds. The height of the balloon when the stone was dropped from it is (g = 9.8 m-2) ,d xqCckjk i`Foh dh lrg ls 29 eh@lS0 ds osx ls Åij mB jgk gSA 108. A particle moves along the sides AB, BC, CD of a square blesa ,d iRFkj fxjk;k tkrk gS tks 10 lSd.M esa i`Foh ry ij igqWprk gSA xqCckjs dh ml le; ÅWpkbZ D;k Fkh tc iRFkj fxjk;k x;k FkkA of side 25 m with a velocity of 15 ms -1. Its average (a) 100 m (b) 200 m velocity is. ,d d.k 25 eh0 Hkqtk okys oxZ dh Hkqtkvksa (c) 400 m (d) 150 m AB, BC, CD ds vuqfn”k 15ms–1 ds osx ls 99. Two balls of same size but the density of one is greater xfr dj jgk gSA bldk vkSlr osx gS & than that of the other are dropped from the same height, then which ball will reach the earth first (air resistance is negligible) – leku vkdkj dh nks xsans ftlesa ls ,d dk ?kuRo nwljs ls vf/kd gS] leku ÅWpkbZ ls fxjk;h tkrh gS] rks buesa ls dkSu igys i`Foh ry ij (a) 15 ms–1 (b) 10 ms–1 igqWpsxh] ¼ok;q izfrjks/k ux.; ekuk tk;s½ & (c) 7.5 ms–1 (d) 5 ms–1 (a) Heavy ball / Hkkjh xsan 109. A body has speed V, 2V and 3V in first 1/3 of distance S, (b) Light ball / gYdh xsan seconds 1/3 of S and third 1/3 respectively. Its average (c) Both simultaneously / nksuksa lkFk&lkFk speed will be : (d) Will depend upon the density of the balls fdlh oLrq ds }kjk r; dh xbZ dqy nwjh S dh izFke ,d frgkbZ f}rh; xsan dk ?kuRo ij fuHkZj djsxkA ,d frgkbZ rFkk r`rh; ,d frgkbZ nwjh esa pysa Øe”k% V, 2V vkSj 3V 100. A packet is dropped from a balloon which is going gSA bldh vkSlr pky gksxhA upwards with the velocity 12 m/s, the velocity of the (a) V (b) 2 V packet after 2 seconds will be – (c) 18/11 V (d) 11/18 V ,d xqCckjk 12 eh0@lS0 ds osx ls Åij mBk jgk gSA xqCckjs ls ,d 110. A particle is projected upwards. The times corresponding iSdsV NksM+k tkrk gS 2 lSd.M Ik”pkr~ iSdsV dk osx gksxk & to height h white ascending and while descending and t1 (a) - 12 m/s (b) 12 m/s and t2 respectively. The velocity of projection will be: (c) - 7.6 m/s (d) 7.6 m/s ,d d.k Å/okZ/kj Åij dh vksj iz{ksfir fd;k tkrk gSA ;fn h Å¡pkbZ 101. If a freely falling body travels in the last second a Åij tkus esa rFkk uhps vkus esa yxs le; Øe”k% t1 rFkk t2 gSa] rks distance equal to the distance travelled by it in the first iz{ksi.k osx gSA three second, the time of the travel is – (a) gt1 (b) gt2 ;fn Lora= :Ik ls fxjrh oLrq }kjk vafre lSd.M esa r; dh x;h nwjh dk eku izFke rhu lSd.M esa r; dh x;h nwjh ds cjkcj gks rks xfr esa (c) g(t1 + t2) (d) g(t1 + t2) /2 yxk le; gS & 111. The displacement of the particle varies with time (a) 6 sec (b) 5 sec according to the relation x = k/b [1 – e−bt] Then the (c) 4 sec (d) 3 sec velocity of the particle is: 102. The effective acceleration of a body, when thrown fdlh d.k dk foLFkkiu] le; ds lkFk laca/k x = k/b [1 – e−bt] ds upwards with acceleration a will be – vuqlkj ifjorhZ gS] rks d.k dk osx gSA ,d leku Roj.k a ls Å/okZ/kj Åij Qsadh x;h oLrq dk izHkkoh Roj.k gksxk (a) k(e–bt) (b) k/ b2 e−bt (a) (b) (c) k b e–bt (d) None of these a−g a −g 112. A body falls freely from the top of a tower. It covers 36% (c) (a – g ) (d) ( a+ g ) of the total height in the last second before strinking the 103. A body is thrown vertically upwards with velocity u. The ground level. The height of the tower is – distance travelled by it in the fifth and the sixth seconds एक िपंड एक टावर के शीष से तं प से िगरता है। यह जमीनी र पर are equal. The velocity u is given by (g = 9.8 m/s2) – हमला करने से पहले अंितम से कंड म कुल ऊंचाई का 36% कवर करता है । टावर ,d oLrq dks Å/okZ/kj Åij dh vksj u osx ls Qsadk tkrk gSA blds की ऊंचाई है - }kjk xfr ds ikWporsa ,oa NBs lSd.M esa r; dh x;h nwjh dk eku leku (a) 50 m (b) 75 m gSA izkjfEHkd osx u dk eku gS & (g = 9.8 m/s2) (c) 100 m (d) 125 m (a) 24.5 m/s (b) 49.0 m/s 113. If the velocity of a particle is (10 + 2t2) m/s, then the (c) 73.5 m/s (d) 98.0 m/s average acceleration of the particle between 2s and 5s is. 104. A ball is thrown up under gravity (g = 10 m / sec2). ;fn fdlh d.k dk foLFkkiu (10 + 2t2) m/s, gS rks 2s rFkk 5s ds Find its velocity after 1.0 sec at a height of 10 m : chp d.k dk vkSlr Roj.k gSA ,d xsan xq:Roh; izHkko esa Åij Qsadh tkrh gS (g = 10 m / sec2). (a) 2 m/s2 (b) 4 m/s2 dh 10 m Å¡pkbZ ij 1.0 sec ckn osx Kkr dhft, A (c) 12 m/s2 (d) 14 m/s2 (a) 5 m/ sec2 (b) 5 m/ sec 114. A bullet moving with a velocity of 200 cm/s penetrates a (c) 10 m/sec (d) 15 m /sec : wooden block and comes to rest after traversing 4 cm 105. Free fall of an object (in vacuum) is a case of motion with inside it. What velocity is needed for travelling distance fuokZr~ esa eqDr :i ls fxjrh gqbZ oLrq dh xfr gSA of 9 cm in same block. (a) Uniform velocity / ,d leku osx cUnwd dh ,d xksyh 200 lseh@ls ds osx ls ydM+h ds r[rsa esa izos”k (b) Uniform acceleration / ,d leku Roj.k djrh gS rFkk blds vUnj 4 lseh0 rd /kaldj :d tkrh gSA mlh (c) Variable acceleration / ifjorhZ Roj.k r[rs esa 9 lseh rd izos”k djus ds fy, vko”;d osx gksxkA (d) Constant momentum / fu;r laosx (a) 100 cm/s (b) 136.2 cm/s 106. A boy standing at the top of a tower of 20m height drops (c) 300 cm/s (d) 250 cm/s a stone. Assuming g = 10 ms-2, the velocity with which it 115. The speed of a body moving with uniform acceleration is hits the ground is : u. This speed is oubled while covering a distance S. ,d ckyd 20m Å¡ph ehukj ds “kh’kZ ij [kM+k gS vkSj og ,d iRFkj When it covers an additional distance S, its speed would fxjkrk gSA ;fn g = 10 ms-2gks rks iRFkj dk i`Foh ry ij Vdjkrs become – le; osx gksxkA एकसमान रण से गितमान िकसी िपंड की गित u है । दू री S तय करते समय यह गित दोगुनी हो जाती है । जब यह अित र दू री S तय करती है , तो इसकी गित हो (a) 5.0 m/s (b) 10.0 m/s जाएगी - (c) 20.0 m/s (d) 40.0 m/s (a) √3 u (b) √5 u 107. A stone falls freely rest from a height h and it travels a (c) √11 u (d) √7 u distance in the last second. The value of h is : 116. A bird flies for 4 s with a velocity of |t -2| m/s in a straight ,d fi.M h Å¡pkbZ ls fole voLFkk ls eqDr :i esa fxjrk gSA rFkk ;g line, where t is time in seconds. It covers a distance of – एक प ी |t -2| के वेग से 4 s तक उड़ता है एक सीधी रे खा म m/s, जहाँ t सेकंड म vafre lsd.M esa nwjh r; djrk gSA h dk eku gksxkA समय है । की दू री तय करती है - (a) 145 m (b) 100 m (a) 2m (b) 4m (c) 122.5 m (d) 200 m (c) 6m (d) 8m 7 117. If a ball is thrown vertically with speed u, the distance covered during the last t seconds of its ascent is – ANsWERs यिद एक गद को u गित से लंबवत फका जाता है , तो उसके चढ़ने के अंितम t से कंड के दौरान तय की गई दू री है - (a) 1/2 gt2 (b) ut – 1/2 gt2 (c) (u – gt)t (d) ut 1 (a) 21 (a) 41 (a) 61 (b) 81 (a) 101 (b) 118. A small block slides without friction down an inclined 2 (a) 22 (b) 42 (a) 62 (a) 82 (b) 102 (c) plane starting from rest. Let Sn be the distance travelled from time t = n – 1 to t = n. Then Sn / Sn + 1. 3 (c) 23 (c) 43 (b) 63 (b) 83 (c) 103 (b) एक छोटा सा गुटका आराम से शु करके एक झुके ए तल पर िबना घषण के 4 (a) 24 (c) 44 (d) 64 (d) 84 (b) 104 (b) िफसलता है । मान लीिजए िक समय t = n – 1 से t = n तक तय की गई दू री Sn है । िफर Sn / Sn + 1. 5 (b) 25 (b) 45 (c) 65 (b) 85 (a) 105 (b) (a) (b) 6 (d) 26 (a) 46 (b) 66 (b) 86 (a) 106 (c) (c) (d) 7 (a) 27 (b) 47 (c) 67 (d) 87 (b) 107 (c) 8 (b) 28 (b) 48 (d) 68 (c) 88 (b) 108 (d) 119. A point initially at rest moves along x-axis. Its acceleration varies with time as a = (6t + 5) m/s2. If it 9 (a) 29 (b) 49 (a) 69 (d) 89 (d) 109 (c) starts from origin, the distance covered in 2s is – 10 (d) 30 (a) 50 (c) 70 (b) 90 (b) 110 (d) ारं भ म िवरामाव ा म एक िबंदु x-अ के अनुिदश गित करता है । इसका रण समय के साथ a = (6t + 5) m/s2 के प म बदलता रहता है । यिद यह मूल िबंदु 11 (c) 31 (b) 51 (a) 71 (b) 91 (b) 111 (a) से शु होता है , तो 2s म तय की गई दू री है - 12 (b) 32 (c) 52 (d) 72 (c) 92 (c) 112 (d) (a) 20 m (b) 18 m 13 (d) 33 (a) 53 (c) 73 (b) 93 (b) 113 (d) (c) 16 m (d) 25 m 120. A parachutist after bailing out falls 50 m without friction. 14 (c) 34 (c) 54 (d) 74 (a) 94 (a) 114 (c) When parachute opens, it decelerates at 2 m/s2. He 15 (d) 35 (a) 55 (a) 75 (d) 95 (a) 115 (d) reaches the ground with a speed of 3 m/s. At what height, 16 (b) 36 (c) 56 (a) 76 (a) 96 (b) 116 (b) did he bail out – 17 (b) 37 (b) 57 (c) 77 (b) 97 (a) 117 (a) एक पैराशू िट बाहर िनकलने के बाद िबना िकसी घषण के 50 मीटर तक िगरता है । जब पैराशू ट खु लता है, तो इसकी गित 2 m/s2 हो जाती है । वह 3 मीटर/से कड 18 (d) 38 (d) 58 (c) 78 (b) 98 (b) 118 (c) की गित से जमीन पर प ं चता है । वह िकस ऊंचाई पर जमानत पर छूटा - 19 (c) 39 (c) 59 (d) 79 (a) 99 (c) 119 (b) (a) 293 m (b) 111 m (c) 91 m (d) 182 m 20 (d) 40 (d) 60 (b) 80 (a) 100 (c) 120 (a) 8