33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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GravitationTelegram @unacademyplusdiscounts71(a) 1/2 (b) 1/2(c) 2 (d) 2 (2005)62. The satellite of mass m is orbiting around the earthin a circular orbit with a velocity v. What will be itstotal energy ?(a) (3/4)mv 2 (b) (1/2)mv 2(c) mv 2 (d) –(1/2)mv 2 (1991)8.11 Geostationary and Polar Satellites63. The mean radius of earth is R, its angular speed onits own axis is w and the acceleration due to gravityat earth’s surface is g. What will be the radius of theorbit of a geostationary satellite ?(a) (R 2 g/w 2 ) 1/3 (b) (Rg/w 2 ) 1/3(c) (R 2 w 2 /g) 1/3 (d) (R 2 g/w) 1/3 (1992)ANSWER KEY1. (b) 2. (b) 3. (b) 4. (a) 5. (b) 6. (c) 7. (c) 8. (a) 9. (d) 10. (a)11. (b) 12. (b) 13. (b) 14. (b) 15. (a) 16. (a) 17. (b) 18. (d) 19. (a) 20. (d)21. (d) 22. (b) 23. (b) 24. (a) 25. (c) 26. (b) 27. (a) 28. (c) 29. (b) 30. (c)31. (a) 32. (d) 33. (c) 34. (b) 35. (d) 36. (c) 37. (c) 38. (d) 39. (c) 40. (b)41. (a) 42. (b) 43. (c) 44. (a) 45. (b) 46. (b) 47. (a) 48. (b) 49. (a) 50. (a)51. (c) 52. (b) 53. (c) 54. (d) 55. (b) 56. (c) 57. (d) 58. (b) 59. (d) 60. (b)61. (a) 62. (d) 63. (a)Hints & Explanations1. (b) :BPerihelionAv ASv CCAphelionPoint A is perihelion and C is aphelion.So, v A> v B> v CAs kinetic energy K = (1/2) mv 2 or K ∝ v 2So, K A> K B> K C.2. (b) : According to the law of conservation of angularmomentum L 1= L 2vmv 1r 1= mv 2r 2⇒ v 1r 1= v 2r 2or 1 r2=v r3. (b) : Equal areas are swept in equal time.As it is given that area SCD = 2 × area of SABThe time taken to go from C to D, t 1= 2t 2where t 2is the time taken to go from A to B.4. (a) : Period of revolution of planet A, (T A) = 8T B.According to Kepler’s III law of planetary motion T 2 ∝ R 3 .3 2 2⎛ r TTherefore =A ⎞ A TB⎝⎜ rB⎠⎟ = ⎛ ⎞⎝ ⎜ TB⎠⎟ = ⎛ 8 ⎞⎝ ⎜ TB⎠⎟ = 64rAor = 4 or rrA= 4r BB5. (b) : Distance of two planets from sun, r 1= 10 13 mand r 2= 10 12 mRelation between time period (T) and distance of theplanet from the sun is T 2 ∝ r 3 or T ∝ r 3/2 .21Therefore, T 32 32131 r110= ⎛ /⎞T2⎝ ⎜ r122 ⎠⎟ = ⎛ /⎞⎝ ⎜ ⎟ = 10 3/2 =10 1010 ⎠6. (c) : In a circular or elliptical orbital motion aplanet, angular momentum is conserved. In attractivefield, potential energy and the total energy is negative.Kinetic energy increases with increase is velocity. If themotion is in a plane, the direction of L does not change.7. (c) :Applying the properties of ellipse, we have2 1 1 r1 + r22rr= + = ; R = 12R r1 r2rr 12r1 + r28. (a) : Since two astronauts are floating in gravitationalfree space. The only force acting on the two astronauts isGmthe gravitational pull of their masses, 1m2F = ,2rwhich is attractive in nature.Hence they move towards each other.9. (d) : Gravitational force of attraction between sunand planet provides centripetal force for the orbit ofplanet.
Telegram @unacademyplusdiscounts72 NEET-AIPMT Chapterwise Topicwise Solutions PhysicsMm 4FRGFe = = ×(R e+R42e / 2)9 = ×9 72 23. (b)= 32 N15. (a) : Force between the two masses, F = −G mm 24. (a) : Acceleration due to gravity4R2 g = GM 343πR × ρ 4× = G= G × πR× ρR 2 2R32This force will provide the necessary centripetal force forGMm mv 2 GM∴ = ; v =…(i) the masses to go around a circle, then2r r r2Time period of the planet is given byGmm mv 2 Gm 1 Gm= ⇒ v = ⇒ v =22 2 2 24RR 4R2 Rrr rT 2π 2Tv4π v4π,2 ⎛ GM(Using (i)) 16. (a) : Given : mass (m) = 6 × 10 24 kg;2 3r ⎠⎟2 4 rT = π angular velocity (w) = 2 × 10 –7 rad/s and…(ii) radius (r) = 1.5 × 10 8 km = 1.5 × 10 11 mGMAccording to question, T 2 = Kr 3Force exerted on the earth = mRw 2…(iii)= (6 × 10 24 ) × (1.5 × 10 11 ) × (2 × 10 –7 ) 2Comparing equations (ii) and (iii), we get= 36 × 10 21 N24πK = , ∴ GMK = 4π217. (b) : Centripetal force (F) = mv 2and the gravitationalGM Rforce (F) = GMm GMm= (where R 2 → R). Since210. (a) : For a point inside the earth i.e. r < RR R2mv GMmGME 3Rr= therefore v= GM . Thus velocity v isR Rindependent of R.where M and R be mass and radius of the earthrespectively.At the centre, r = 018. (d) : If universal gravitational constant becomes tentimes, then G′ = 10 G.So, acceleration due to gravity increases. i.e, (d) is the\ E = 0wrong option.For a point outside the earth i.e. r > R,19. (a) : Gravitational force acting on particle of mass m isGMGMpmE =− F =r 2( Dp/ 2)2On the surface of the earth i.e. r = R,Acceleration due to gravity experienced by the particle isF GMp4GMpGMg = = =E =−R 2m2 2( Dp/ 2)DpThe variation of E with distance r from the centre is asGM G ⎛ 4 3 ⎞shown in the figure.20. (d) : g = = × rGrr r ⎝⎜⎠⎟ = 4π ρ × πρ2 2 3 311. (b) : Gravitational field due to the thin spherical shell g ′ 3R=Inside the shell, (for r < R) F = 0g R ⇒ g ′ = 3 g.On the surface of the shell, (for r = R)21. (d) : From equation of acceleration due to gravity.GM3F =GMeG( 4/ 3)πRe2ge= =ρR22 eReReOutside the shell, (for r > R)ge ∝ReρeGMF =r 2Acceleration due to gravity of planet gp ∝RpρpThe variation of F with distance r from the centre is as R er e= R pr p⇒ R er e= R p2r eshown in the figure.⇒ R p= 112. (b) : The gravitational force does not depend upon2 R (Q R e = R)the medium in which objects are placed.22. (b) : Initial velocity of the mass on both the planetsis same.13. (b)ie .., 2g′ h′ = 2gh14. (b) : F G Mmsurface = 22× g ′ × h′ = 2× 9g ′ × 2 ⇒ 2h′ = 36R e ⇒ h′ = 18 m
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Gravitation
Telegram @unacademyplusdiscounts
71
(a) 1/2 (b) 1/
2
(c) 2 (d) 2 (2005)
62. The satellite of mass m is orbiting around the earth
in a circular orbit with a velocity v. What will be its
total energy ?
(a) (3/4)mv 2 (b) (1/2)mv 2
(c) mv 2 (d) –(1/2)mv 2 (1991)
8.11 Geostationary and Polar Satellites
63. The mean radius of earth is R, its angular speed on
its own axis is w and the acceleration due to gravity
at earth’s surface is g. What will be the radius of the
orbit of a geostationary satellite ?
(a) (R 2 g/w 2 ) 1/3 (b) (Rg/w 2 ) 1/3
(c) (R 2 w 2 /g) 1/3 (d) (R 2 g/w) 1/3 (1992)
ANSWER KEY
1. (b) 2. (b) 3. (b) 4. (a) 5. (b) 6. (c) 7. (c) 8. (a) 9. (d) 10. (a)
11. (b) 12. (b) 13. (b) 14. (b) 15. (a) 16. (a) 17. (b) 18. (d) 19. (a) 20. (d)
21. (d) 22. (b) 23. (b) 24. (a) 25. (c) 26. (b) 27. (a) 28. (c) 29. (b) 30. (c)
31. (a) 32. (d) 33. (c) 34. (b) 35. (d) 36. (c) 37. (c) 38. (d) 39. (c) 40. (b)
41. (a) 42. (b) 43. (c) 44. (a) 45. (b) 46. (b) 47. (a) 48. (b) 49. (a) 50. (a)
51. (c) 52. (b) 53. (c) 54. (d) 55. (b) 56. (c) 57. (d) 58. (b) 59. (d) 60. (b)
61. (a) 62. (d) 63. (a)
Hints & Explanations
1. (b) :
B
Perihelion
A
v A
S
v C
C
Aphelion
Point A is perihelion and C is aphelion.
So, v A
> v B
> v C
As kinetic energy K = (1/2) mv 2 or K ∝ v 2
So, K A
> K B
> K C
.
2. (b) : According to the law of conservation of angular
momentum L 1
= L 2
v
mv 1
r 1
= mv 2
r 2
⇒ v 1
r 1
= v 2
r 2
or 1 r2
=
v r
3. (b) : Equal areas are swept in equal time.
As it is given that area SCD = 2 × area of SAB
The time taken to go from C to D, t 1
= 2t 2
where t 2
is the time taken to go from A to B.
4. (a) : Period of revolution of planet A, (T A
) = 8T B
.
According to Kepler’s III law of planetary motion T 2 ∝ R 3 .
3 2 2
⎛ r T
Therefore =
A ⎞ A TB
⎝
⎜ rB
⎠
⎟ = ⎛ ⎞
⎝ ⎜ TB
⎠
⎟ = ⎛ 8 ⎞
⎝ ⎜ TB
⎠
⎟ = 64
rA
or = 4 or r
r
A
= 4r B
B
5. (b) : Distance of two planets from sun, r 1
= 10 13 m
and r 2
= 10 12 m
Relation between time period (T) and distance of the
planet from the sun is T 2 ∝ r 3 or T ∝ r 3/2 .
2
1
Therefore, T 32 32
13
1 r1
10
= ⎛ /
⎞
T2
⎝ ⎜ r
12
2 ⎠
⎟ = ⎛ /
⎞
⎝ ⎜ ⎟ = 10 3/2 =10 10
10 ⎠
6. (c) : In a circular or elliptical orbital motion a
planet, angular momentum is conserved. In attractive
field, potential energy and the total energy is negative.
Kinetic energy increases with increase is velocity. If the
motion is in a plane, the direction of L does not change.
7. (c) :
Applying the properties of ellipse, we have
2 1 1 r1 + r2
2rr
= + = ; R = 12
R r1 r2
rr 12
r1 + r2
8. (a) : Since two astronauts are floating in gravitational
free space. The only force acting on the two astronauts is
Gm
the gravitational pull of their masses, 1m2
F = ,
2
r
which is attractive in nature.
Hence they move towards each other.
9. (d) : Gravitational force of attraction between sun
and planet provides centripetal force for the orbit of
planet.