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PHYSICS
SOLUTIONS
1. (a) The escape velocity on the earth is defined as r l
ve = 2g e R e
Where Re & ge are the radius & acceleration due to M m
gravity of earth.
Now for planet gP=2ge, RP=Re/4
ve x
So v P = 2g P R P = 2 ´ 2g e ´ R e / 4 =
2 r+l
r +l r +l
2. (c) Applying conservation of energy principle, we get GMm GMm 1
F= ò lx 2
dx =
l ò x2
dx
1 GMm GMm r r
mk 2 v e2 - =- r +l r +l
2 R r GMm GMm é x -2+1 ù
1 2GM GMm GMm
=
l ò x -2 dx =
l êë -2 + 1úû r
mk 2
r
Þ - =- r+l
2 R R r GMm -1 r + l GMm é 1 ù GMm
=- éx ù = - =
l ë ûr
2 2
Þ
k 1 1 1 1 k
- =- Þ = - l êë x úû r r (r + l )
R R r r R R
1 1 R k Mv 2
Þ = (1 - k 2 ) Þ r = 7. (b) F= = . Hence v µ R0
r R 1- k 2 R R
3. (d) The gravitational force due to the whole sphere at A mv 2 GmM GM
8. (d) = g= 2
point is (R + x ) (R + x ) 2 also R
GM e m o , where m is the assumed rest mass at
F1 = 0 mv 2 æ GM ö R 2
(2R ) 2 \ = mç
point A. (R + x) è R 2 ÷ø (R + x) 2
In the second case, when we made a cavity of radius mv 2 R2
(R/2), then gravitational force at point A is \ = mg
(R + x ) (R + x ) 2
GM e m o
F2 = \ F2/F1= 1/9 1/ 2
( R + R / 2) 2 gR 2 æ gR 2 ö
4. (c) According to Kepler’s law of period T2 µ R3 \ v2 = Þ v = çç ÷
÷
R+x èR+xø
T12 R13 (6 R )3 3
= = =8 9. (b) v= ve
T22 R23 (3 R )3 4
24 ´ 24 2
=8 1 1 æ3 ö 9
K.E. = mv 2 = m ç v e ÷ = mve2
T22 2 2 è4 ø 32
24 ´ 24 9 æ 2GM ö
T22 = = 72 = 36 × 2 = m
8 32 çè R ÷ø
T2 = 6 2 9 GMm GMm
K.E. = ; P.E. = -
16 R R
PE
5. (d) Total energy = – K E = 7 GMm
2 Total energy = K.E. + P.E. = -
1 16 R
K.E = mv 2 Let the height above the surface of earth be h, then
2
GMm
1 2 P.E. = -
\ Total energy = - mv h
2
6. (a) The force of attraction between sphere and shaded 7 GMm GMm 16R
- =- \ h=
æm ö 16 R h 7
ç dx ÷ 10. (a) When closer to the sun, velocity of planet will be greater.
l
position dF = GM è ø So time taken in covering a given area will be less.
x2
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11. (c) Applying conservation of total mechanical energy

æ ö
principle GMm ÷ 1 8 2 GM
\ -4 ç + mv2 = 0 Þ v 2 =
ç a ÷ 2 a
1 2 è 2ø
mv = mg A h A = mg B hB
2
18. (c) The potential energy for a conservative force is defined
Þ g A hA = g B hB as
r r r
- dU
æg ö F= or U = - ò F.dr …… (i)
Þ hB = ç A ÷ hA = 9 × 2 = 18 m dr
èg ø ¥
B
r
GM1M 2 -GM1M 2
12. (b) Due to inertia of motion it will move tangentially to the or U r = ò r 2
dr =
r
…… (ii)
original orbit with same velocity. ¥
(Q U¥ = 0)
13. (a) F µ xM ´ (1 – x )M = xM 2 (1 - x ) If we bring the mass from the infinity to the centre of
dF earth, then we obtain work, ‘so it has negative
For maximum force, =0 (gravitational force do work on the object) sign &
dx
potential energy decreases. But if we bring the mass
dF
Þ = M 2 - 2 xM 2 = 0 Þ x = 1/ 2 from the surface of earth to infinite, then we must do
dx work against gravitational force & potential energy of
14. (c) Mass of the satellite = m and height of satellite from the mass increases.
earth (h) = 6.4 × 106 m. GM M
We know that gravitational potential energy of the Now in equation (i) if F = 51/ 2 2 instead of
r
satellite at height
GM1M 2
F= then
GM e m gR 2 m = - gR e m = -0.5 mgR r2
h =- =- e e
Re + h 2R e 2 r
GM1M 2 -2 GM1M 2
(where, GMe = gRe2 and h= Re)
Ur = ò r 5/2
dr =
3 r 3/2
¥
15. (d) Acceleration due to gravity on earth's surface 1
Þ Ur µ
M r +3/ 2
g=G
R2 19. (b) As we know, the minimum speed with which a body is
This implies that as radius decreases, the acceleration projected so that it does not return back is called escape
due to gravity increases. speed.
Dg DR DR 2GM 2GM 2GM
= -2 But = -1% Ve = = =
g R R r R+h 4R
('–' sign is due to shrinking of earth) 1

Dg æ GM ö 2
\ = -2 ´ (-1%) = 2% =ç ÷ (Q h = 3R)
g è 2R ø
16. (a) According to kepler's law of area 20. (a) Acceleration due to gravity at a height h above the
earth’s surface is
dA L
= æ 2h ö
dt 2m g h = g ç1 - ÷
è Rø
For central forces, torque = 0 Acceleration due to gravity at a depth d below the
\ L = constant earth’s surface is
dA æ dö
\ = constant g d = g ç1 - ÷
dt è Rø

17. (b) Potential energy of particle at the centre of square æ 2h ö

1- ÷
g h çè R ø (R - 2h)
Now, = =
æ ö gd æ dö (R - d)
çè 1 - ÷
ç GMm ÷ Rø
= -4 ç
a ÷ As h = 1 km, d = 1 km
ç ÷ gh R - 2
è 2 ø \ =
gd R - 1
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21. (a) At the surface of earth, the value of g = 9.8m/sec2. If 28. (c) Applying the properties of ellipse, we have
we go towards the centre of earth or we go above the 2 1 1 r1 + r2
surface of earth, then in both the cases the value of g = + =
R r1 r2 r1 r2
decreases.
Instant position
Hence W1=mgmine, W2=mgsea level, W3=mgmoun of satellite
So W1< W2 > W3 (g at the sea level = g at the suface Sun
of earth) R
22. (d) Time period does not depend upon the mass of satellite major axis
2 pr 2pr 2 pr 3 / 2 2p
23. (a) T = = = = r1 r2
v0 ( gR 2 / r )1/ 2 gR 2 w
2 r1 r2
gR2 gR 2 R=
Hence, r 3/ 2
= or r 3 = 2 r1 + r2
w w 29. (c) In a circular or elliptical orbital motion, torque is always
or, r = (gR2 / w2)1/3
acting parallel to displacement or velocity. So, angular
24. (b) g ' = g - w2 R cos 2 l momentum is conserved. In attractive field, potential
energy is negative. Kinetic energy changes as velocity
To make effective acceleration due to gravity zero at
increase when distance is less. So, option (c) is correct.
equator l = 0 and g ' = 0
2GM 2GM
30. (a) Here, v = and kv = .
g 1 rad R R+R
\ 0 = g - w2 R Þ w = =
R 800 s 1
Solving k =
25. (a) mg = 72 N (body weight on the surface) 2
GM G (2M) GM
g= 31. (a) g = =
R2 (2R)2 2R 2
R 1 2
At a height H = , From h = gt [ Q U= 0]
2 2
GM 2h hR 2
g¢ = 2 4 GM t= =2
æ Rö = g GM
çè R + ÷ø 9 R2
2 R GMm é1 1 ù
32. (b) P.E. = ò dr = -GMm ê - ú
R 2
Body weight at height H = ,
R0 r ë R R0 û
2 1
The K.E. acquired by the body at the surface = m v2
4 GM 2
mg ¢ = m ´
9 R2 1 2 é1 1 ù
\ mv = -GMm ê - ú
4 4 2 ë R R0 û
= m´
´ g = mg
9 9
æ 1 1ö
4 v = 2G M ç - ÷
= ´ 72 = 32 N è R0 R ø
9
26. (c) At a height h, mv 2 k 2 k
2 33. (b) = or mv =
2
R2 æ R ö R R R
g' =g Þ mg ' = mg ç
(R + h)2 è R + h ÷ø 1 k
Kinetic energy = mv 2 =
2 2 2R
æ R ö
Þ W' = Wç In case of satellites P.E = – 2 K.E
è R + h ÷ø
Here, h = R/2 and T.E = P. E + K. E

4 k k k
\ W' = W Total energy = - =-
9 2R R 2R
27. (c) Gravitational P.E. = m × gravitational potential 34. (d) Variation of g with altitude is,
U = mV, so the graph of U will be same as that of V for é 2h ù
g h = g ê1 - ú ;
a spherical shell. ë Rû
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variation of g with depth is, Now, if the shell shrinks then its radius decrease then
é dù density increases, but mass is constant. so from above
g d = g ê1 - ú expression if a decreases, then V increases.
ë Rû
Equating gh and gd, we get d = 2h æ dö g æ dö
42. (b) g ' = g ç1 - ÷ Þ = g ç1 - ÷
35. (a) The total momentum will be zero and hence velocity è Rø n è Rø
will be zero just after collisiion. The pull of earth will æ n -1 ö
Þd =ç ÷R R/4 Moon
make it fall down. è n ø
36. (b) Loss in potential energy = Gain in kinetic energy
43. (d) E earth = E moon
GMm æ 3 GMm ö 1 2
- -ç- ÷ = mv
R è 2 R ø 2 GM GM / 81
Þ = 60 R
2
GMm 1 GM x (60R - x) 2 x
Þ = mv 2 Þ v = = gR
2R 2 R 1 1
37. (d) Þ x = 9 (60R - x) R Earth

1 Þ x = 54 R from centre of earth.

38. (b) gµ
R2 44. (b) Acceleration due to gravity at lattitude’ l ’ is given by
R decreasing g increase hence, curve b represents
correct variation. g l = g e - R e w 2 cos 2 l
39. (d) Angular momentum, L = Iw; moment of inertia of sphere At equator, l = 90° Þ cos l = cos 90° = 0
along the axis passing through centre of mass, or g l = g e = g (as given in question)
2 2 2p 3
I = MR and w = . 2 2
Rw 2
5 T At 30°, g 30 = g - Rw cos 30 = g -
4
4p MR 2 3
Putting these values, L = Rw 2
or, g - g 30 =
5T 4
45. (a) As we know,
(R + h) 3 -GMm
40. (c) T = 2p Gravitational potential energy =
GM r
and orbital velocity, v0 = GM / R + h
R3 (1.01R )3 1 GMm 1 GM GMm
T1 = 2 p , T2 = 2p Ef = mv02 - = m -
GM GM 2 3R 2 3R 3R
GMm æ 1 ö - GMm
T2 - T1 = ç - 1÷ =
´ 100 = 1.5% 3R è 2 ø 6R
T1
-GMm
41. (a) The gravitational potential at the centre of uniform Ei = +K
R
spherical shell is equal to the gravitational potential at
the surface of shell i.e., Ei = E f

-GM 5GMm
V= , where a is radius of spherical shell Therefore minimum required energy, K =
a 6R