Class : XIth SUBJECT : PHYSICS
Date : Solutions DPP No. : 1
Topic :- GRAVITATION
1 (c)
It is self-evident that the orbit of the comet is elliptic with sun begin at one of the focus.
Now, as for elliptic orbits, according to kepler’s third law,
1/3
4𝜋2𝑎3 𝑇2𝐺𝑀
𝑇2 =
𝐺𝑀
⇒𝑎 =
4𝜋2( )
1/3
(76 × 3.14 × 107) × 6.67 × 10-11
𝑎=
[ 4𝜋2
× 2 × 1010
]
But in case of ellipse,
2𝑎 = 𝑟min + 𝑟max
∴ 𝑟max = 2𝑎 - 𝑟min = 2 × 2.7 × 1012 - 8.9 × 1010
≅5.3 × 1012m
2 (b)
𝐺𝑀
Acceleration due to gravity 𝑔 =
𝑅 2
, 𝑀= (43 𝜋𝑅 )ρ
3
4𝐺 𝜋𝑅 3
∴ 𝑔= ρ
3 𝑅2
4𝐺𝜋𝑅
⟹ 𝑔=
3 ( ρ ) (ρ = average density)
⟹𝑔 ∝ ρ or ρ ∝ g
3 (c)
𝐺𝑀 𝐿2
𝑔= 𝑅2
and 𝐾 = 2𝐼
1 1
If mass of the earth and its angular momentum remains constant then 𝑔 ∝ 𝑅2 and 𝐾 ∝ 𝑅2
𝑖.𝑒., if radius of earth decreases by 2% then 𝑔 and 𝐾 both increases by 4%
4 (a)
Acceleration due to gravity at a height above the earth surface
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nR
𝑅 2
𝑔' = 𝑔( )
𝑅+h
𝑔 𝑅+h 2
𝑔'
= ( )𝑅
𝑔 𝑅 + 𝑛𝑅 2
𝑔'
= ( ) 𝑅
𝑔
= (1 + 𝑛)2
𝑔'
5 (c)
Gravitational potential
1 1 1
𝑉 = 𝐺𝑀 (
+ + +…
𝑟1 𝑟2 𝑟3 )
1 1 1 1 1
=𝐺×1 + + + + (
1 2 4 8 16
+ …. )
𝑎
(1
= 𝐺 1 1/2
-
)
( ∴ sum of GP = 1-𝑟
)
= 2𝐺
6 (b)
g𝑒 𝑅𝑒ρ𝑒 2 4 g𝑒
= = × = 6 or g𝑚 =
g𝑚 𝑅𝑚ρ𝑚 3 1 6
For motion on earth, using the relation,
1
𝑠 = 𝑢𝑡 + 𝑎𝑡2
2
1 1
We have, 2 = 0 + 2 × 9.8𝑟2 or 𝑡 = 1/ 9.8𝑠
1
For motion on moon, 3 = 0 + 2(9.8/6)𝑡21
𝑡1
or 𝑡1 = 6 9.8𝑠 ∴ 𝑡 = 6 or 𝑡1 = 6𝑡
7 (c)
Escape velocity,
2 𝐺𝑀
𝑣ascape =
𝑅
8
=𝑅 𝜋𝐺ρ
3
∴ 𝑣𝑒 ∝ 𝑅 if ρ = constant.
Since the planet is having double radius in comparision to earth, therefore escape velocity
becomes twice 𝑖𝑒, 22 kms-1 .
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8 (a)
𝑣𝑝 𝑀𝑝 𝑅 𝑒 1
𝑣𝑒
= × = 6 × = 3 ∴ 𝑣𝑝 = 3𝑣𝑒
𝑀𝑒 𝑅 𝑝 2
12 (c)
𝑔𝑒 𝑅𝑒
𝑣𝑒
𝑣𝑚 =
𝑔𝑚 𝑅𝑚 = 6 × 10 = 60 = 8 (nearly)
13 (c)
-𝐺𝑀 𝑚
Gravitational potential energy of a body in the gravitational field, 𝐸 = 𝑟
. When 𝑟
decreases negative value of 𝐸 increase 𝑖𝑒, 𝐸 decreases
14 (b)
Actually gravitational force provides the centripetal force
15 (c)
The earth moves around the sun is elliptical path, so by using the properties of ellipse
𝑟1 = (1 + 𝑒)𝑎 and 𝑟2 = (1 - 𝑒)𝑎
𝑟1 + 𝑟2
⇒𝑎 = 2
and 𝑟1𝑟2 = (1 - 𝑒2)𝑎2
Where 𝑎 = semi major axis
𝑏 = semi minor axis
𝑒 = eccentricity
𝑏2
Now required distance = semi latusrectum = 𝑎
𝑎2(1 - 𝑒2) (𝑟1𝑟2) 2𝑟1𝑟2
= = =
𝑎 (𝑟1 + 𝑟2)/2 𝑟1 + 𝑟2
16 (c)
At a certain velocity of projection of the body will go out of the gravitational field of earth
and never to return to earth. The initial velocity is called escape velocity
𝑣𝑒 = 2𝑔𝑅
Where 𝑔 is acceleration due to gravity and 𝑅 the radius. As is clear from above formula,
that escape velocity dose not depends upon mass of body hence, it will be same for a body
of 100kg as for 1kg body.
17 (d)
Telecommunication satellites are geostationary satellite
18 (b)
Weight of body at height above the earth’s surface is
𝑤
𝑤' = 2
1+h (
𝑟 )
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80
⟹ 40 = 2
(1 + h𝑟)
⟹ h = 0.41𝑟
19 (d)
As we know gas molecules cannot escape from earth’s atmosphere because their root
mean square velocity is less than escape velocity at earth’s surface. If we fill this
requirement, then gas molecules can escape from earth’s atmosphere.
𝑖𝑒, 𝑣rms = 𝑣es
3𝑅𝑇
or = 2𝑔𝑅𝑒
𝑀
2𝑀𝑔𝑅𝑒
or 𝑇= ….(i)
3𝑅
Given, 𝑀 = 2 × 10-3kg, 𝑔 = 9.8 ms-2
𝑅𝑒 = 6.4 × 106 m, 𝑅 = 8.31 Jmol-1–K-1
Substituting in Eq. (i), we have
2 × 2 × 10-3 × 9.8 × 6.4 × 106
𝑇=
3 × 8.31
4
= 10 K
20 (d)
𝐺𝑀
= 𝑔𝑅
2
𝑣0 =
𝑟 𝑅+h
ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
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A. C B C A C B C A C A
Q. 11 12 13 14 15 16 17 18 19 20
A. A C C B C C D B D D
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