SIR M.

V EDUCATION TRUST(R) ✖︎
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SIR M.V GROUP OF INSTITUTIONS
Near Government ITI College, Hadadi Road, Davangere

1. Assuming earth to be a sphere of radius , if is value of acceleration due to gravity at latitude of and at the equator,
∘
R g30∘ 30 g

the value of g − g30 is ∘

(A) 1

4
ω R
2
(B) 3

4
2
ω R

(C) ω R
2
(D) 1

2
mωR

Answer: 2
2 2 ∘
g30∘ = g − ω R cos 30

3 2
g − g30∘ = ω R
4

2. At what height from the surface of the earth, the total energy of the satellite is equal to its potential energy at a height of 2R from
the surface of the earth? (R = radius of earth )
(A) (B)
R R

4 2

(C) 2R (D) 4R

Answer: 2
GM m
Total energy of a satellite at a height h = −
2(R+h)

Potential energy at a height 2R from surface of earth

GM m GM m
= − = −
(2R+R) 3R

Given, − GM m
= −
GM m
or h =
R

2(R+h) 3R 2

3. Body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If R is the
radius of the earth, the maximum height attained by the body is:
(A) (B)
R R

6 3

(C)
2
R (D) R
3

Answer: 2
Let H be the maximum height By energy conservation,
K. E .i +P . E = P . EF
i

1 2 GM m GM m
mVe + (− ) = −
2 R R+H

2Gm
∵ Ve =
R

m(2GM ) GM m GM m
− = −
8R R R+H

4R
R + H =
3

R
H =
3

4. A remote-sensing satellite of earth revolves in a circular orbit at a height of above the surface of earth. If earth's
6
0.25 × 10 m

radius is 6.38 × 10 6
m and g = 9.8ms
−2
, then the orbital speed of the satellite is
(A) 19kms−1 (B) 6kms−1
(C) 7.76kms−1 (D) 12kms
−1

Answer: 3
−−−−−
g
The orbital speed of the satellite is vo = R√
(R+h)

where R is the earth's radius, g is the acceleration due to gravity on earth's surface and h is the height above the surface of earth.
6 −2
H ere, R = 6.38 × 10 m, g = 9.8ms

6
andh = 0.25 × 10 m
−−−−−−−−−−−−−−−
−2
(9.8ms )
6 3 −1 −1
∴ vo = (6.38 × 10 m) √ = 7.76 × 10 ms = 7.76kms
6 6
(6.38×10 m+0.25×10 m)

3
(∵ 1km = 10 m)
5. Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational attraction. The speed v
of each particle is
−−− −−−− Success
(A) (B)
1 Gm 4Gm
√ √
2 R R
−−− −−−
(C) (D)
1 1 Gm
√ √
2R Gm R

Answer: 1
The two masses, separated by a distance 2R are going round their common centre of mass, the centre of the circle.
Attractive force = −G mm . 2
R
mm
But the two masses are going round the centre of mass or the reduced mass μ =
m+m
is going round a circle of radius =
distance of separation
Centrifugal force =
m 2 m 2 1
∴ ω ⋅ 2R = v ⋅
2 2 2R
−−−
2 2

Now,
m v Gm Gm
× = ⇒ v = √
2
2 2R 4R R

6. An object of mass 100 kg falls from point A to B as shown in figure. The change in its weight, corrected to the nearest integer is (
RE is the radius of the earth)

(A) 49 N (B) 89 N
(C) 5 N (D) 10 N
Answer: 1
2
′ R
Mg = Mg 2
(R+h)
2
Mg
At A M g ′
= Mg
R
2
=
9
(R+2R)

R M g⋅4
At B Mg
′
= Mg 2
=
25
3R
(R+ )
2

Mg
Change in weight = Mg
4

25
−
9
= 49 N

7. A 400 kg satellite is in an circular orbit of radius 2 R about the earth, How much energy required to transfer it to a circular orbit
of radius 4 R
(A) 1.11 × 10
9
J (B) 2.12 × 10
9
J

(C) 4.14 × 10 9
J (D) 3.13 × 10
9
J

Answer: 4
GM m GM m
Ei = − , Ef = −
4R 8R
2
g = 9.8 m/s
6
R = 6.37 × 10 m
6
gmR 9.81×400×6.37×10
ΔE = =
8 8
6
= 3.37 × 10 m

3RE
8. A research satellite of mass 200 kg circles the earth in an orbit radius where RE is the radius of the earth. Assuming the
2

gravitational pull on a mass of 1 kg on the earth’s surface to be 10 N , the pull on the satellite will be
(A) 890 N (B) 889 N
(C) 885 N (D) 892 N
Answer: 2
Force between earth and mass of 1 kg is
GME ×1 GME
F = 2
= 2
(i)
R R
E E

where Mg is the mass of the earth and RE is the radius of the earth respectively.
GME ×200
(ii)
′
F = 2
3
( RE )
2

Divide (ii) by (i), we get

F 800
=
F 9
′ 800 F
F =
9

=
800
× 10 N (∵ F = 10 N (Given))
9
8000
= N = 889 N
9

9. The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is v . For a satellite orbiting at an
altitude of half of the earth's radius, the orbital velocity is
−
−
(A)
3
v 3
2 (B) √
2
v

−
−
(D)
2
(C) √
2
v 3
v
3

Answer: 3
10. Consider the following statements and select the correct statement(s) from the following.
I. When height of a satellite is increased, its potential energy increases while kinetic energy decreases
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II. The orbital velocity of satellite depends upon the density of planet around which it is revolving
III. For a satellite orbiting in circular orbit, the kinetic energy is always greater than potential energy.
(A) I & II (B) II & III
(C) III only (D) I, II and III
Answer: 4
−GM m
P E of satellite =
R+h

GM m
KE of satellite =
2(R+h)

−GM m
∴ T E of satellite =
2(R+h)

Hence on increasing height of satellite its potential energy increases but KE decreases and orbital velocity of satellite depends
upon the density of planet.

11. Gravitational potential differernce between a point on surface of planet and another point 10 m above is 4J /kg. Considering
gravitational field to be uniform, how much work is done in moving a mass of 2 kg from the surface to a point 5 m above the
surface :
(A) 4 J (B) 5 J
(C) 6 J (D) 7 J
Answer: 1
−4
Gravitational field g = −
ΔV

ΔX
= −(
10
) =
4

10
J /kgm

Work done in moving a mass of 2 kg from the surface to a point 5 m above the surface.
4 J
W = mgh = (2kg) ( ) (5m) = 4J
10 kgm

Prepared and generated by Tardigrade App