Gravitation — Page 6 | NEET Physics

Q51

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

A 1/2

B 1/

\sqrt 2

C 2

D

\sqrt 2

Correct Answer

Option A

Solution

K.E. of satellite moving in an orbit around the earth is

K = {1 \over 2}m{v^2} = {1 \over 2}m{\left( {\sqrt {{{GM} \over r}} } \right)^2} = {{GMm} \over {2r}}

P.E. of satellite and earth system is

U = {{GMm} \over r}

\Rightarrow {K \over U} = {{{{GMm} \over {2r}}} \over {{{GMm} \over r}}} = {1 \over 2}

Q52

Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then

A g' = g/9

B g' = 27g

C g' = 9g

D g' = 3g

Correct Answer

Option D

Solution

We know that

g = {{GM} \over {{R^2}}} = {{G\left( {{4 \over 3}\pi {R^3}} \right)} \over {{R^2}}} = {4 \over 3}\pi GR\rho

{{g'} \over g} = {{R'} \over R} = {{3R} \over R} = 3

$\therefore$ g' = 3g

Q53

The density of a newly discovered planet is twice that of 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

A 2R

B 4R

C

{1 \over 4}

R

D

{1 \over 2}

R

Correct Answer

Option D

Solution

From equation of acceleration due to gravity.

{g_e} = {{G{M_e}} \over {R_e^2}} = {{G\left( {4/3} \right)\pi R_e^3} \over {R_e^2}}{\rho _e}

{g_e} \propto {R_e}{\rho _e}

Acceleration due to gravity of planet

{g_p} \propto {R_p}{\rho _p}

$\therefore$

{R_e}{\rho _e} = {R_p}{\rho _p} \Rightarrow {R_e}{\rho _e} = {R_p}2{\rho _e}

\Rightarrow {R_e} = {1 \over 2}R

( $\because$ R e = R)

Q54

Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be

A 3 F

B F

C F/3

D F/9

Correct Answer

Option B

Solution

Gravitational force is independent of medium, Hence, this will remain same.

Q55

The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2 m on the surface of A. What is the height of jump by the same person on the planet B?

A (2/9) m

B 18 m

C 6 m

D (2/3) m

Correct Answer

Option B

Solution

The velocity of the mass while reaching the surface of both the planets will be same. i.e.,

\sqrt {2g'h'} = \sqrt {2gh}

\sqrt {2 \times g \times h'} = \sqrt {2 \times 9g \times 2}

\Rightarrow 2h' = 36

\Rightarrow h' = 18\,m

Q56

A body of mass m is placed on earth's surface which is taken from earth surface to a height of h = 3R, then change in gravitational potential energy is

A

{{mgR} \over 4}

B

{2 \over 3}mgR

C

{3 \over 4}mgR

D

{{mgR} \over 2}

Correct Answer

Option C

Solution

Gravitational potential energy on earth’s surface =

- {{GMm} \over R}

, where M and R are the mass and radius of the earth respectively, m is the mass of the body and G is the universal gravitational constant.

Gravitational potential energy at a height h = 3R

= - {{GMm} \over {R + h}} = - {{GMm} \over {R + 3R}} = - {{GMm} \over {4R}}

$\therefore$ Change in potential energy

= - {{GMm} \over {4R}} - \left( { - {{GMm} \over R}} \right)

= - {{GMm} \over {4R}} + {{GMm} \over R}

= {3 \over 4}{{GMm} \over R}

= {3 \over 4}mgR

Q57

With what velocity should a particle be projected so that its height becomes equal to radius of earth?

A

{\left( {{{GM} \over R}} \right)^{1/2}}

B

{\left( {{{8GM} \over R}} \right)^{1/2}}

C

{\left( {{{2GM} \over R}} \right)^{1/2}}

D

{\left( {{{4GM} \over R}} \right)^{1/2}}

Correct Answer

Option A

Solution

Using

{v^2} = {{2gh} \over {1 + {h \over R}}}

and given h = R $\therefore$

v = \sqrt {gR} = \sqrt {{{GM} \over R}}

Q58

A body of weight 72 N moves from the surface of earth at a height half of the radius of earth, then gravitational force exerted on it will be

A 36 N

B 32 N

C 144 N

D 50 N

Correct Answer

Option B

Solution

{F_{surface}} = G{{Mm} \over {R_e^2}}

{F_{{R_e}/2}} = G{{Mm} \over {{{\left( {{R_e} + {R_e}/2} \right)}^2}}} = {4 \over 9} \times {F_{surface}}

= {4 \over 9} \times 72 = 32\,N

Q59

For a planet having mass equal to mass of the earth but radius is one fourth of radius of the earth. Then escape velocity for this planet will be

A 11.2 km/sec

B 22.4 km/sec

C 5.6 km/sec

D 44.8 km/sec.

Correct Answer

Option B

Solution

{v_{earth}} = \sqrt {{{2G{M_e}} \over {{R_e}}}}

{v_{planet}} = \sqrt {{{2G{M_p}} \over {{R_p}}}} = \sqrt {{{2G{M_e}} \over {{R_e}/4}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}}

{{{v_{planet}}} \over {{v_{earth}}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}} \times \sqrt {{{{R_e}} \over {2G{M_e}}}} = 2

$\therefore$

{v_{planet}} = 2 \times {v_{earth}}

= 2 $\times$ 11.2 = 22.4 km/s

Q60

Gravitational force is required for

A stirring of liquid

B convection

C conduction

D radiation.

Correct Answer

Option B

Solution

Gravitational force is required for convection