Gravitation — Page 10 | NEET Physics

Q91

Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be :

A

{2 \over 9}

F

B

{16 \over 9}

F

C

{8 \over 9}

F

D F

Correct Answer

Option C

Solution

Let the masses are m and distance between them is l, then

F = {{G{m^2}} \over {{l^2}}}

. When 1/3rd mass is transferred to the other then masses will be

{{4m} \over 3}

and

{{2m} \over 3}

. so new force will be

F' = {{G{{4m} \over 3} \times {{2m} \over 3}} \over {{l^2}}} = {8 \over 9}{{G{m^2}} \over {{l^2}}} = {8 \over 9}F

Q92

Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth

d=\dfrac{R}{2}

from the surface of earth, if its weight on the surface of earth is 200 N, will be: (Given R = radius of earth)

A 100 N

B 400 N

C 300 N

D 500 N

Correct Answer

Option A

Solution

The gravitational field strength, or equivalently, the weight of an object, decreases linearly from its surface value to zero at the center of a sphere of uniform mass density.

This is because only the mass inside the radius at which the object is located contributes to the gravitational force at that location.

If ( $d = \dfrac{R}{2}$ ) is the depth below the surface of the Earth, then the radius of the sphere contributing to the gravitational force at that depth is ( $R - d = R - \dfrac{R}{2} = \dfrac{R}{2}$ ).

Since the gravitational force (or weight) decreases linearly with the radius in a sphere of uniform density, the weight of the object at depth ( $d = \dfrac{R}{2}$ ) is half its weight at the surface of the Earth.

So, if the weight of the body on the surface of the Earth is 200 N, its weight at depth ( $d = \dfrac{R}{2}$ ) is half of that, or 100 N.

Therefore, the correct answer is 100 N.

Q93

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : A pendulum clock when taken to Mount Everest becomes fast. Reason R : The value of g (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth. In the light of the above statements, choose the most appropriate answer from the options given below

A Both A and R are correct but R is NOT the correct explanation of A

B A is correct but R is not correct

C Both A and R are correct and R is the correct explanation of A

D A is not correct but R is correct

Correct Answer

Option D

Solution

When we go on the Mount Everest the value of gravitational acceleration decreases $\left(g=\dfrac{g_{0}}{\left(1+\dfrac{h}{R_{e}}\right)^{2}}\right)$ .

Therefore, the time period of oscillation $\left(T=2 \pi \sqrt{\dfrac{I}{g}}\right)$ increases and the pendulum clock becomes slow thus the assertion is wrong but reason is correct.

Q94

The escape velocity of a body depends upon mass as

A

{m^0}

B

{m^1}

C

{m^2}

D

{m^3}

Correct Answer

Option A

Solution

Escape velocity,

{v_e} = \sqrt {2gR} = \sqrt {{{2GM} \over R}} \Rightarrow {V_e}\, \propto \,{m^0}

Where

M,R

are the mass and radius of the planet respectively. In this expression the mass of the body

(m)

is not present. The escape velocity is independent of the mass m or it depends on m0.

Q95

The height at which the acceleration due to gravity becomes

{g \over 9}

(where

g=

the acceleration due to gravity on the surface of the earth) in terms of

R,

the radius of the earth, is:

A

{R \over {\sqrt 2 }}

B

R/2

C

\sqrt 2 \,\,R

D

2\,R

Correct Answer

Option D

Solution

Given that, at height h from ground the acceleration due to gravity becomes

{g \over 9}

. We know acceleration at earth surface due to gravity g =

{{GM} \over {{R^2}}}

and acceleration at height h due to gravity g' =

{{GM} \over {{{\left( {R + h} \right)}^2}}}

So

{{g} \over 9}

=

{{GM} \over {{{\left( {R + h} \right)}^2}}}

$\Rightarrow$

{{g} \over 9}

=

{{GM} \over {{R^2}}}.{{{R^2}} \over {{{\left( {R + h} \right)}^2}}}

=

g.{\left( {{R \over {R + h}}} \right)^2}

\Rightarrow {1 \over 9} = {\left( {{R \over {R + h}}} \right)^2}

\Rightarrow {R \over {R + h}} = {1 \over 3}

\Rightarrow 3R = R + h

$\therefore$

h = 2R

Q96

The mass of the moon is

\dfrac{1}{144}

times the mass of a planet and its diameter is

\dfrac{1}{16}

times the diameter of a planet. If the escape velocity on the planet is

v

, the escape velocity on the moon will be :

A

\dfrac{\mathrm{v}}{4}

B

\dfrac{\mathrm{v}}{6}

C

\dfrac{\mathrm{V}}{12}

D

\dfrac{\mathrm{v}}{3}

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{V}_{\text{escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ & \mathrm{V}_{\text{planet }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\mathrm{V} \\ & \mathrm{V}_{\text{Moon }}=\sqrt{\frac{2 \mathrm{GM} \times 16}{144 \mathrm{R}}}=\frac{1}{3} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ & \mathrm{V}_{\text{Moon }}=\frac{\mathrm{V}_{\text{Planet }}}{3}=\frac{\mathrm{V}}{3} \end{aligned}