33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Gravitation
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73
or ρ = 3 g
4π
GR
25. (c) : Given : radius of earth (R e
) = 6400 km; radius
of mars (R m
) = 3200 km; mass of earth (M e
) = 10 M m
and
weight of the object on earth (W e
) = 200 N.
Wm
mgm
Mm
Re
= = × ⎛ 2
⎞
We
mge
Me
⎝ ⎜ Rm
⎠
⎟ = 1
× 2
2
10
= 2
()
5
or W m
= W e × 2 = 200 × 0.4 = 80 N
5
26. (b) : Gravitational force at a height h,
mg 0 72
mg h
= =
2 2 or mg
⎛ h ⎞ ⎛ R/
2 ⎞
h
= 32 N
1+
1
⎝
⎜
r ⎠
⎟ +
⎝
⎜
R ⎠
⎟
or F g
= 32 N
27. (a) : Acceleration due to gravity at a depth d,
⎛ d ⎞
g d
= g 1−
⎝
⎜
R ⎠
⎟
⎛ R / 2 ⎞ g
For d = R/2 ⇒ g d
= g 1−
⎝
⎜
R ⎠
⎟ =
2
Required weight W′ = mg d
= mg = W 200
= = 100 N
2 2 2
28. (c) : The acceleration due to gravity at a height h is
given as
⎛ 2h
⎞
gh
= g −
⎝
⎜1
Re
⎠
⎟
where R e
is radius of earth.
The acceleration due to gravity at a depth d is given as
⎛ d ⎞
gd
= g −
⎝
⎜1
Re
⎠
⎟
Given, g h
= g d
⎛ h ⎞
∴ −
⎝
⎜
⎠
⎟ = ⎛
⎝
⎜ − d ⎞
g 1 2 g 1
R R ⎠
⎟ or, d = 2h = 2 × 1 = 2 km
e
e
(Q h = 1 km)
29. (b) : Acceleration due to gravity is given by
⎧4
πρGr ; r ≤ R
⎪3
g = ⎨ 3
⎪4
πρRG
; r > R
⎩⎪
3 2
r
30. (c) : Acceleration due to gravity at a height h from
the surface of earth is
g
g ′ =
... (i)
2
⎛ h ⎞
+
⎝
⎜1
⎠
⎟
R
where g is the acceleration due to gravity at the surface of
earth and R is the radius of earth.
Multiplying m (mass of the body) on both sides in (i),
we get
mg
mg ′ =
2
⎛ h ⎞
+
⎝
⎜1
R ⎠
⎟
\ Weight of body at height h, W′ = mg′
Weight of body at surface of earth, W = mg
Q W′ =
1
1 1
W ∴ =
16
16
2
⎛ h ⎞
+
⎝
⎜1
⎠
⎟
R
2
⎛ h ⎞
h
1+
16 1 4
⎝
⎜
⎠
⎟ = or + =
R
R
h
or = 3 or h=
3R
R
31. (a)
32. (d) : Work done = Change in potential energy
= u f
– u i = − GMm
− ⎛ − GMm ⎞
+ ⎝
⎜
⎠
⎟
( R h)
R
where M is the mass of earth and R is the radius of earth.
⎡ 1 1 ⎤
∴ W = GMm⎢
−
⎣ R ( R + h )
⎥
⎦
Now, h = R
∴ W ⎡
= GMm 1 1 ⎤
⎢ −
⎣ R R⎦
⎥ = GMm
2 2R
⎡ ⎤
⇒ W = mgR GM
⎢Q
g =
2 2 ⎥
⎣ R ⎦
33. (c) : Gravitation potential at a height h from the
surface of earth, V h
= –5.4 × 10 7 J kg –1
At the same point acceleration due to gravity, g h
= 6 m s –2
R = 6400 km = 6.4 × 10 6 m
GM
We know, V h =− ( R + h) ,
GM Vh
Vh
gh
= =− ⇒ R+ h=−
( R+
h) 2 R + h
g
V
∴ =− − =− − × 7
h ( 54 . 10 ) 6
h R
− 64 . × 10
gh
6
= 9 × 10 6 – 6.4 × 10 6 = 2600 km
34. (b) : The resulting gravitational potential at the
origin O due to each of mass 2 kg located at positions as
shown in figure is
G G G G
V =− × 2
− × 2
− × 2
− × 2
− ...
1 2 4 8
⎡ ⎤
⎡
⎤
=− ⎢ + + + + ⎥
⎣
⎦
=− ⎢ ⎥
2 1 1 1
1 1
G
...... 2G⎢
⎥
2 4 8
⎢
1−
1 2 ⎥
⎣ ⎦
⎡
=− 2 2
G
⎣
⎢ ⎤ 1 ⎦ ⎥ =− 4G
h