33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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GravitationFrom equation (i) and (ii), we get f = 1 244. (a) : Use2 2ghv = given h = R.h1+RGM∴ v= gR =R2GM45. (b) : ve= 2gRe=ReRP= 1 Re446. (b)v P= 2v e= 2 × 11.2 = 22.4 km/s.47. (a) : Escape velocity of a body (v e) = 11.2 km/s; Newmass of the earth M′ e= 2M eand new radius of the earthR′ e= 0.5 R e.Escape velocity (v e) =2GMe∝ReMeReTherefore v e MeReve′ = R× 05 .e M= 12= 1e 4 2or, v′ e= 2v e= 22.4 km/s48. (b) : Escape velocity does not depend on the angleof projection.49. (a)50. (a) : Time period of Geostationary satellite is,\3aT = 2π⇒ T ∝ aGM2 3T1 2 a1 3 2( 24) ( 7RE)T2 2 =a2 3 ⇒ =T (. 35R)32 2 3E2 322 ( 24) × (.) 35( 24)⇒ T 2 =⇒ T32 = = 6 2h.() 7851. (c) : The orbital speed of the satellite is,gv o = R( R + h)where R is the earth’s radius, g is the acceleration due togravity on earth’s surface and h is the height above thesurface of earth.Here, R = 6.38 × 10 6 m, g = 9.8 m s –2 , h = 0.25 × 10 6 m6∴ v o = (. 638×10 m)−2(. 98ms)6 6(. 638× 10 m+ 025 . × 10 m)= 7.76 × 10 3 m s –1 = 7.76 km s –1Telegram @unacademyplusdiscounts52. (b) : The gravitational force on the satellite S actstowards the centre of the earth, so the acceleration ofthe satellite S is always directed towards the centre of theearth.53. (c) : According to Kepler’s third law T ∝ r 3/2∴ = ⎛ 32 / 32 /T2r ⎞ ⎛ + ⎞⎝ ⎜ 2 R 2R1⎠⎟=⎜⎝ + ⎟=T1r1R 5R⎠32 /2Since T 1= 24 hours so,T21 24 24= or T32 2 = = =6 2 hours24 / 32 /22 2 275GM54. (d) : Escape velocity, ve = 2 ...(i)Rwhere M and R be the mass and radius of the earthrespectively.The orbital velocity of a satellite close to the earth’ssurface isvo =GMRFrom (i) and (ii), we get ve= 2vo...(ii)55. (b) : Orbital speed of the satellite around the earth isv =GMrFor satellite A, r A= 4R, v A= 3VGMvA= r...(i)AFor satellite B, r B= R, v B= ?GMvB= r...(ii)BDividing equation (ii) by equation (i), we getvBrArA= or vB= vAvArBrBSubstituting the given values, we getRvB = 3 4VRor v B= 6V56. (c) : Since no external torque is applied therefore,according to law of conservation of angular momentum,the ball will continue to move with the same angularvelocity along the original orbit of the spacecraft.57. (d) : Time period of satellite does not depend on itsmass.As T 2 ∝ r 3T32 /A r 1= =T 32 / 32 /B 2 r 2 258. (b) : Total energy of satellite at height h from theearth's surface,GMm 1 2E = PE + KE =− + mv...(i)( R+h)2
Telegram @unacademyplusdiscounts76 NEET-AIPMT Chapterwise Topicwise Solutions Physics2mv GMmAlso, =( R+h) ( R+ h) 2 or, v 2 GM = ...(ii)R + hFrom eqns. (i) and (ii),GMm 1 GMm 1 GMmE =− + =−( R+h) 2 ( R+h) 2 ( R+h)21 GM mR=− ×2 2R ( R+h) =− mg0R22( R+h)59. (d)60. (b) : The satellite of mass m is moving in a circularorbit of radius r.\ Kinetic energy of the satellite, K = GMm ... (i)2 rGMmPotential energy of the satellite, U = − ... (ii)rGMOrbital speed of satellite, v = ... (iii)rTime-period of satellite,T = ⎡ ⎛ ⎝ ⎜ 4 2 12 /π ⎞⎢ ⎟GM ⎠r⎤3⎥... (iv)⎣⎢⎦⎥Given m S1= 4m S2Since M, r is same for both the satellites S 1and S 2.\ From equation (ii), we get U ∝ m\USUS12mS1= = 4 or, UmS1= 4U S2.S2Option (a) is wrong.From (iii), since v is independent of the mass of a satellite,the orbital speed is same for both satellites S 1and S 2.Hence option (b) is correct.From (i), we get K ∝ m\ K S m1 S1= = 4 or, KK mS1= 4K S2.S2S2Hence option (c) is wrong.From (iv), since T is independent of the mass of asatellite, time period is same for both the satellites S 1andS 2. Hence option (d) is wrong.GMm GMm61. (a) : KE . . = ; PE . . =−2RR| PE . .| KE . .∴ KE . . = or,=2 | PE . .|62. (d) : Total energy = –K.E. =− 1 2mv263. (a) : GMm22= mω r ⇒ 3 GM gR2r = =r2 2ω ω\ r = (gR 2 /w 2 ) 1/3 .12vvv
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76 NEET-AIPMT Chapterwise Topicwise Solutions Physics
2
mv GMm
Also, =
( R+
h) ( R+ h) 2 or, v 2 GM = ...(ii)
R + h
From eqns. (i) and (ii),
GMm 1 GMm 1 GMm
E =− + =−
( R+
h) 2 ( R+
h) 2 ( R+
h)
2
1 GM mR
=− ×
2 2
R ( R+
h) =− mg0R
2
2( R+
h)
59. (d)
60. (b) : The satellite of mass m is moving in a circular
orbit of radius r.
\ Kinetic energy of the satellite, K = GMm ... (i)
2 r
GMm
Potential energy of the satellite, U = − ... (ii)
r
GM
Orbital speed of satellite, v = ... (iii)
r
Time-period of satellite,
T = ⎡ ⎛ ⎝ ⎜ 4 2 12 /
π ⎞
⎢ ⎟
GM ⎠
r
⎤
3
⎥
... (iv)
⎣⎢
⎦⎥
Given m S1
= 4m S2
Since M, r is same for both the satellites S 1
and S 2
.
\ From equation (ii), we get U ∝ m
\
US
US
1
2
mS
1
= = 4 or, U
m
S1
= 4U S2
.
S
2
Option (a) is wrong.
From (iii), since v is independent of the mass of a satellite,
the orbital speed is same for both satellites S 1
and S 2
.
Hence option (b) is correct.
From (i), we get K ∝ m
\ K S m
1 S1
= = 4 or, K
K m
S1
= 4K S2
.
S2
S2
Hence option (c) is wrong.
From (iv), since T is independent of the mass of a
satellite, time period is same for both the satellites S 1
and
S 2
. Hence option (d) is wrong.
GMm GMm
61. (a) : KE . . = ; PE . . =−
2R
R
| PE . .| KE . .
∴ KE . . = or,
=
2 | PE . .|
62. (d) : Total energy = –K.E. =− 1 2
mv
2
63. (a) : GMm
2
2
= mω r ⇒ 3 GM gR
2
r = =
r
2 2
ω ω
\ r = (gR 2 /w 2 ) 1/3 .
1
2
vvv
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