Gravitation Test

Result

Gravitation Test - 12

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Two masses m₁ & m₂ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently, due to gravitational attraction between them, their relative velocity of approach at a separation distance of d is

A

B

C

D
(m₁ + m₂)^1/2 2Gd
Question 2
1 / -0

A man of mass m starts falling towards a planet of mass M and radius R. As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass 2M/3 and a point mass M/3 at the centre. Change in the force of gravity experienced by the man is

A

B
0

C

D

Solution

Gravity does not depend upon shape and size of the celestial body. It depends only on the mass of the two bodies and distance between them.
Fold = GmM/R² and Fnew = GmM/R²
Question 3
1 / -0

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 of g at poles. The escape velocity of a particle on the pole of planet in terms of V.

A
V_e = 2V

B
V_e= V

C
V_e= V/2

D
V_e =

Solution
Question 4
1 / -0

The escape velocity for a planet is v_e. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

A
v_e

B

C

D
Zero

Solution
Question 5
1 / -0

If a tunnel is cut at any orientation through earth, then a ball released from one end will reach the other end in time (neglect earth rotation)

A
84.6 minutes

B
42.3 minutes

C
8 minutes

D
Depends on orientation

Solution

Total time period will be 84.6 min when ball released from one end and it come backs to same point as in oscillation.
When ball is released from one end then time taken to reach other end will be half of total time period then that will be 42.3 min.
Question 6
1 / -0

Two uniform spherical stars made of same material have radii R and 2R. Mass of the smaller planet is m. They start moving from rest towards each other from a large distance under mutual force of gravity. The collision between the stars is inelastic with coefficient of restitution 1/2.
Kinetic energy of the system just after the collision is

A

B

C

D
Cannot be determined

Solution
Question 7
1 / -0

Two uniform spherical stars made of same material have radii R and 2R. Mass of the smaller planet is m. They start moving from rest towards each other from a large distance under mutual force of gravity. The collision between the stars is inelastic with coefficient of restitution 1/2.
The maximum separation between their centres after their first collision

A
4R

B
6R

C
8R

D
12R
Question 8
1 / -0

A mass is at the center of a square, with four masses at the corners as shown.
(A) (B)
(C) (D)
Rank the choices according to the magnitude of the gravitational force on the center mass.

A
F_A = F_B < F_C = F_D

B
F_A > F_B < F_D < F_C

C
F_A = F_B > F_C = F_D

D
None

Solution

Gravitational force by mass M on another body of m at a distance of r is given by F=GMmr²
For given figure Right diagonal is x axis and left y axis in figure A forces due to 5M will cancel out each other so effective force will be due to only 3M an M. In figure B 2M forces will cancel out each other and effective force will be due to 3M and M.
Hence total force on M in both the cases will be the same.
Similarly, in figure C, 5M mass will cancel each other out and in Figure D, 2M will cancel each other out and effective force will be due to 3M and M on the mass of 2M. So, force on c and d will be the same.
Force on C will be more than B.
Hence answer is F_A=F_BC=F_D
Question 9
1 / -0

A satellite of mass m, initially at rest on the earth, is launched into a circular orbit at a height equal to the radius of the earth. The minimum energy required is

A

B

C

D
Question 10
1 / -0

A satellite of mass 5M orbits the earth in a circular orbit. At one point in its orbit, the satellite explodes into two pieces, one of mass M and the other of mass 4M. After the explosion the mass M ends up travelling in the same circular orbit, but in opposite direction. After explosion the mass 4M is in

A
Bound orbit

B
Unbound orbit

C
Partially bound orbit

D
Data is insufficient to determine the nature of the orbit

Solution

Applying the conservation of momentum, we get:
V = velocity of 5M, V1 = velocity of 4M, V2 = velocity of M and ATQ V2 = -V
5MV = 4MV1 + M(-V)
6MV = 4MV1
V1 = 3/2 V
Now, V_o = orbital velocity and V_e = escape velocity
V^e = √2 V_o
In a bound orbit the object is gravitationally bound to the body that is the source of gravity (like the Earth is bound to the Sun, or the Moon to the Earth). An unbound orbit is typically hyperbolic, and the object will escape from the source of gravity
Question 11
1 / -0

The orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse.

If t₁ is the time taken by the planet to travel along ACB and t₂ the time along BDA, then

A
t₁ = t₂

B
t₁ > t₂

C
t₁ < t₂

D
Nothing can be concluded

Solution

Here, the angular momentum is conserved i.e. mvr is constant. Where r is the distance from centre of the sun to centre of the planet.
Now, consider one point each from both the mentioned paths. (Shown in fig.)
Applying conservation of angular momentum for these points we get
mv₁r₁=mv₂r₂. Simplifying this, v₁/v₂=r₂/r₁<1
Time period are given by
t₁=L/ v₁, t₂= L/ v₂
Comparing them by taking the ratio
t₁/t₂=(L/ v₁)×(v₂/L)
So, t₁/t₂=v₂/v₁>1.Thus t₁>t₂
Question 12
1 / -0

The orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse.

If U is the potential energy and K kinetic energy then |U| > |K| at

A
Only D

B
Only C

C
Both D & C

D
Neither D nor C

Solution

If a planet is orbiting the sun then |U|> |K| else it will leave the system because it won’t be under the influence of the sun’s gravity.
Question 13
1 / -0

Satellites A and B are orbiting around the earth in orbits of ratio R and 4R respectively. The ratio of their areal velocities is

A
1 : 2

B
1 : 4

C
1 : 8

D
1 : 16

Solution

Answer :- a
Solution :- L = mvr
⇒L=m√GMr/r
L = m√GMr−−−−.....(1)
L = 2mdA/dt....(2)
From (1) and (2)
dA/dt ∝ (r)^1/2
= √(dA/dt)1/(dA/dt)2
= √4/1 = 2/1
Question 14
1 / -0

Figure shows the variation of energy with the orbit radius r of a satellite in a circular motion. Select the correct statement.

A
Z is total energy, Y is kinetic energy and X is potential energy

B
X is kinetic energy, Y is total energy and Z is potential energy

C
X is kinetic energy, Y is potential energy and Z is total energy

D
Z is kinetic energy, X is potential energy and Y is total energy

Solution

Kinetic Energy of the Satellite =GMm/2r
Potential Energy of the satellite =−GMm/r
Total Energy of the Satellite =−GMm/2r
Hence, Kinetic Energy Cure will be above 0 , hence X
Potential energy will be the lowest curve , hence Y
Total Energy is the curve is below 0 , but higher magnitude than Potential Energy . Hence, Z