Gravitation The Law Of Falling(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

GRAVITATION THE LAW OF FALLING MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1. The gravitational potential energy of a body of mass 'm' at the earth's surface

- m g R_{e}

. Its gravitational potential energy at a height

R_{e}

from the earth's surface will be (Here

R_{e}

is the radius of the earth)

−2mgRe
2mgRe
12mgRe
−12mgRe

Discuss Question

Answer: Option D. -> −12mgRe
:
D

△ U = U_{2} - U_{1} = \frac{m g h}{1 + \frac{h}{R_{e}}} = \frac{m g R_{e}}{1 + \frac{R_{e}}{R_{e}}} = \frac{m g R_{e}}{2}

\Rightarrow U_{2} - (- m g R_{e}) = \frac{m g R_{e}}{2} \Rightarrow U_{2} = - \frac{1}{2} m g R_{e}

Question 2. In a double star system of two stars of masses m and 2m, rotating about their centre of mass. Their time period of rotation about their centre of mass will be proportional to

r3
r
m12
m−12

Discuss Question

Answer: Option D. -> m−12
:
D

r_{2} = \frac{2 m r}{m + 2 m} = \frac{2 r}{3}

In A Double Star System Of Two Stars Of Masses M And 2m, R...

T_{2}^{2} = \frac{4 π^{2} r_{2}^{2}}{G M}

T_{2}^{2} = \frac{32 π^{2} r_{2}^{2}}{27 G M}

T_{2} \propto r^{\frac{3}{2}}

;

T_{2} \propto m^{\frac{- 1}{2}}

Question 3. If the period of revolution of an artificial satellite just above the earth's surface is T and the density of earth is

ρ

, then

ρ T^{2}

A universal constant whose value is 3πG
A universal constant whose value is 3π2G
Proportional to radius of earth R
Proportional to square of the radius of earth R2 Here, G= universal gravitational constant

Discuss Question

Answer: Option A. -> A universal constant whose value is 3πG
:
A
Here, G = universal gravitation constant
Time period of a satellite above the earth's surface is :

T^{2} = \frac{4 π^{2} R^{3}}{G M} = \frac{3 π}{G ⎛ ⎜ ⎝ \frac{M}{\frac{4}{3} π R^{3}} ⎞ ⎟ ⎠} = \frac{3 π}{G ρ}

ρ T^{2} = \frac{3 π}{G}

= a universal constant.

Question 4. Select the correct statement from the following

The orbital velocity of a satellite increases with the radius of the orbit
Escape velocity of a particle from the surface of the earth depends on the speed with which it is fired
The time period of a satellite does not depend on the radius of the orbit
The orbital velocity is inversely proportional to the square root of the radius of the orbit

Discuss Question

Answer: Option D. -> The orbital velocity is inversely proportional to the square root of the radius of the orbit
:
D

v_{0} = \sqrt{\frac{G M}{r}}

Question 5. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between planet and star is proportional to

R^{\frac{- 5}{2}}

, then

T^{2}

is proportional to

Discuss Question

Answer: Option B. -> R72
:
B
For revolution of planet centripetal force is provided by gravitational force of attraction

m ω^{2} R \propto R^{\frac{- 5}{2}} \Rightarrow \frac{1}{T^{2}} \propto R^{\frac{7}{2}}

Question 6. A small body starts falling on to the earth from a distance equal to the radius of the earth's orbit. How long will the body take to reach the sun? Express the time in terms of T, period of revolution of the earth round the sun

T4√2
T2√2
T√2
T

Discuss Question

Answer: Option A. -> T4√2
:
A
Let x be the distance moved by the particle in time t.
Then by conservation of energy

- \frac{G M m}{d} + 0 = - \frac{G M m}{d - x} + \frac{1}{2} m v^{2}

Where M is mass of the sun and d is the initial separation.

v = \sqrt{\frac{2 G M}{d}} \sqrt{\frac{x}{d - x}}

\frac{d x}{d t} = \sqrt{\frac{2 G M}{d}} \sqrt{\frac{x}{d - x}}

d t = \sqrt{\frac{2 G M}{d}} \int_{0}^{d} \sqrt{\frac{x}{d - x}} d x

t = \sqrt{\frac{d}{2 G M}} {[\sqrt{x (d - x)} - d s i n^{- 1} \sqrt{\frac{d - x}{d}}]}_{0}^{- 1} = \sqrt{\frac{d}{2 G M}} (d s i n^{- 1} 1)

\sqrt{\frac{d}{2 G M}} \times d \times \frac{π}{2}

But,

\frac{G M m_{e}}{d^{2}} = m_{e} ω^{2} d

where

m_{e}

= mass of the earth and

ω

= angular speed of earth.

∴ G M = ω^{2} d^{2}

t = \sqrt{\frac{d}{2 ω^{2} d^{2}}} \times d \times \frac{π}{2} = \frac{π}{2 \sqrt{2} ω} = \frac{π}{2 \sqrt{2} \frac{2 π}{T_{e}}} = \frac{T_{e}}{4 \sqrt{2}}

Question 7. An earth satellite of mass m revolves in a circular orbit at a height h from the surface of the earth. R is the radius of the earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

gR2R+h
gR
gRR+h
√gR2R+h

Discuss Question

Answer: Option D. -> √gR2R+h
:
D

v_{0} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{g R^{2}}{R + h}}

Question 8. Two satellites of masses

m_{1}

and

m_{2} (m_{1} > m_{2})

are revolving round the earth in circular orbits of radius

r_{1}

and

r_{2} (r_{1} > r_{2})

respectively. Which of the following statements is true regarding their speeds

v_{1}

and

v_{2}

v1=v2
v1
v1>v2
v1r1=v2r2

Discuss Question

Answer: Option B. -> v1
:
B

v = \sqrt{\frac{G M}{r}} i f r_{1} > r_{2} t h e n v_{1} < v_{2}

Orbit speed of satellite does not depends upon the mass of satellite

Question 9. The potential at the surface of a planet of mass M and radius R is assumed to be zero. Choose the most appropriate option

The potential at infinity is GMR
The potential at the center of planet is −GMR
Both (a) and (b) are correct
Both (a) and (b) are wrong

Discuss Question

Answer: Option C. -> Both (a) and (b) are correct
:
C
At all places potential will increase by

\frac{G M}{R}

Question 10. A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface (

R_{e}

= 6400 km) will approximately be

Discuss Question

Answer: Option C. -> 2 h
:
C
Time period of a satellite very close to earth's surface is 84.6 min. Time period increases as the distance of the satellite from the surface of earth increase. So, time period of spy satellite orbiting a few hundred km above the earth's surface should be slightly greater than 84.6 min. Therefore, the more appropriate option is (c) or 2 h