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 2 of 3 pages

Question 11. An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential)energy

E_{o}

. Its potential energy is

−Eo
1.5 Eo
2 Eo
Eo

Discuss Question

Answer: Option C. -> 2 Eo
:
C
Potential energy =2

\times

(Total energy)=

2 E_{o}

Becuase we know =

U = \frac{- G M m}{r}

and

E_{o} = \frac{- G M m}{2 r}

Question 12. The earth revolves about the sun in an elliptical orbit with mean radius

9.3 \times 10^{7} m

in a period of 1 year. Assuming that there are no outside influences

The earth's kinetic energy remains constant
The earth's angular momentum remains constant
The earth's potential energy remains constant
All are correct

Discuss Question

Answer: Option B. -> The earth's angular momentum remains constant
:
B
Kinetic and potential energies varies with position of earth w.r.t. sun. Angular momentum remains constant every where.

Question 13. The masses and radii of the earth and moon are

M_{1}, R_{1} a n d M_{2}, R_{2}

respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escapes to infinity is

2√Gd(M1+M2)
2√2Gd(M1+M2)
2√Gmd(M1+M2)
2√Gm(M1+M2)d(R1+R2)

Discuss Question

Answer: Option A. -> 2√Gd(M1+M2)
:
A
Gravitational potential at mid point

V = \frac{- G M_{1}}{\frac{d}{2}} + \frac{- G M_{2}}{\frac{d}{2}}

Now,

PE=m×V=frac−2Gmd(M1+M2)

[m = mass of particle]
So, for projecting particle from mid point to infinity
KE=|PE|

\Rightarrow \frac{1}{2} m v^{2} = \frac{2 G M}{d} (M_{1} + M_{2}) \Rightarrow v = 2 \sqrt{\frac{G (M_{1} + M_{2})}{d}}

Question 14. Escape velocity of a body on a planet is

100 m s^{- 1}

. What is the gravitational potential energy of the body at the surface, if the mass of the body is

1 k g

−5000J
−1000J
−2400J
5000J

Discuss Question

Answer: Option A. -> −5000J
:
A
Given, escape velocity,

v_{e} = 100 m s^{- 1}

, mass of the body,

m = 1 k g

We know universal gravitational constant,

G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}

.
Let

M

be the mass of the planet and

R

be the radius of the planet and

U

be the gravitational potential energy.
We know,

v_{e} = \sqrt{\frac{2 G M}{R}} = 100

\Rightarrow 10000 = \frac{2 G M}{R} \Rightarrow \frac{G M}{R} = 5000

Potential energy,

U = - \frac{G M m}{R} = 5000 \times 1 = - 5000 J

Question 15. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of earth

The acceleration of S is always directed towards the centre of the earth
The angular momentum of S about the centre of the earth changes in direction but its magnitude remains constant
The total mechanical energy of S varies periodically with time
The linear momentum of S remains constant in magnitude

Discuss Question

Answer: Option A. -> The acceleration of S is always directed towards the centre of the earth
:
A
The satellite revolves around the earth due to gravitational force that acts towards the centre of the earth. And according to the Kepler's laws the angular momentum of the satellite reamains constant but not the linear momentum

Question 16. A particle is projected upward from the surface of earth (radius = R) with a speed equal to the orbital speed of a satellite near the earth's surface. The height to which it would rise is

√R
R√2
R
2R

Discuss Question

Answer: Option C. -> R
:
C

v = v_{0} = \frac{v_{e}}{\sqrt{2}}

(orbital speed

v_{0}

of a satellit near the earth's surface is equal to

\frac{1}{\sqrt{2}}

times the escape velocity of a particle on earth's surface)
Now from conservetion of mechanical energy:
Decrease in kinetic energy = increase in potential energy
or

\frac{1}{2} m \frac{v_{e}^{2}}{2} = \frac{m g h}{(1 + \frac{h}{R})}

\frac{1}{2} m (\frac{2 g R}{2}) = \frac{m g h}{(1 + \frac{h}{R})}

or h = R

Question 17. A clock S is based on oscillation of a spring and a clock P is based on pendulum motion. Both clocks run at the same rate on earth. On a planet having the same density as earth but twice the radius

S will run faster than P
P will run faster than S
They will both run at the same rate as on the earth
None of these

Discuss Question

Answer: Option B. -> P will run faster than S
:
B

g = \frac{4}{3} π ρ G R

. If density is same then

g \propto R

According to problem

R_{p} = 2 R_{e} ∴ g_{p} = 2 g_{e}

For clock P (based on pendulum motion)

T = 2 π \sqrt{\frac{l}{g}}

Time period decreases on planet so it will run faster because

g_{p} > g_{e}

For clock S (based on oscillation of spring)

T = 2 π \sqrt{\frac{m}{k}}

So it does not change.

Question 18. Time period of revolution of a nearest satellite around a planet of radius R is T. Period of revolution around another planet, whose radius is 3R but having same density is

T
3 T
9 T
3√3T

Discuss Question

Answer: Option A. -> T
:
A
Time period of satellite which is very near to planet

T = 2 π \sqrt{\frac{R^{3}}{G M}} = 2 π \sqrt{\frac{R^{3}}{G \frac{4}{3} π R^{3} ρ}} ∴ T \propto \sqrt{\frac{1}{ρ}}

i.e. time period of nearest satellite does not depends upon the radius of planet, it only depends upon the density of the planet. In the problem, density is same so time period will be same.

Question 19. If the value of ‘g’ at a height h above the surface of the earth is the same as at a depth x below it, then (both x and h being much smaller than the radius of the earth)

x=h
x=2h
x=h2
x=h2

Discuss Question

Answer: Option B. -> x=2h
:
B
The value of g at the height h from the surface of earth

g^{^{'}} = g (1 - \frac{2 h}{R})

The value of g at depth x below the surface of earth

g^{^{'}} = g (1 - \frac{x}{R})

These two are given equal, hence

g^{^{'}} = g (1 - \frac{2 h}{R}) = g^{^{'}} = g (1 - \frac{x}{R})

On solving, we get

x = 2 h

Question 20. The density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be

Discuss Question

Answer: Option D. -> R2
:
D