Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Two bodies of masses $$m$$ and $$4$$ $$m$$ are placed at a distance $$r.$$ The gravitational potential at a point on the line joining them where the gravitational field is zero is:

A

$$ - {{4Gm} \over r}$$

B

$$ - {{6Gm} \over r}$$

C

$$ - {{9Gm} \over r}$$

D

zero

Let the gravitational field at $$P,$$ distant $$x$$ from mass $$m,$$ be zero.

$$\therefore$$ $${{Gm} \over {{x^2}}} = {{4Gm} \over {{{\left( {r - x} \right)}^2}}}$$

$$ \Rightarrow 4{x^2} = {\left( {r - x} \right)^2}$$

$$ \Rightarrow 2x = r - x$$

$$ \Rightarrow x = {r \over 3}$$

Gravitational potential at point $$P,$$

$$V = - {{Gm} \over {{r \over 3}}} - {{4Gm} \over {{{2r} \over 3}}} = -{{9Gm} \over r}$$

$$\therefore$$ $${{Gm} \over {{x^2}}} = {{4Gm} \over {{{\left( {r - x} \right)}^2}}}$$

$$ \Rightarrow 4{x^2} = {\left( {r - x} \right)^2}$$

$$ \Rightarrow 2x = r - x$$

$$ \Rightarrow x = {r \over 3}$$

Gravitational potential at point $$P,$$

$$V = - {{Gm} \over {{r \over 3}}} - {{4Gm} \over {{{2r} \over 3}}} = -{{9Gm} \over r}$$

2

MCQ (Single Correct Answer)

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

$$2\,R$$

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 $${{g} \over 9} $$ = $${{GM} \over {{{\left( {R + h} \right)}^2}}}$$

$$ \Rightarrow $$ $${{g} \over 9} $$ = $${{GM} \over {{R^2}}}.{{{R^2}} \over {{{\left( {R + h} \right)}^2}}}$$

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

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

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

$$ \Rightarrow 3R = R + h$$

$$\therefore$$ $$h = 2R$$

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 $${{g} \over 9} $$ = $${{GM} \over {{{\left( {R + h} \right)}^2}}}$$

$$ \Rightarrow $$ $${{g} \over 9} $$ = $${{GM} \over {{R^2}}}.{{{R^2}} \over {{{\left( {R + h} \right)}^2}}}$$

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

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

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

$$ \Rightarrow 3R = R + h$$

$$\therefore$$ $$h = 2R$$

3

MCQ (Single Correct Answer)

This question contains Statement - $$1$$ and Statement - $$2$$. of the four choices given after the statements, choose the one that best describes the two statements.

**Statement - $$1$$:**

For a mass $$M$$ kept at the center of a cube of side $$'a'$$, the flux of gravitational field passing through its sides $$4\,\pi \,GM.$$

**Statement - 2:**

If the direction of a field due to a point source is radial and its dependence on the distance $$'r'$$ from the source is given as $${1 \over {{r^2}}},$$ its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.

For a mass $$M$$ kept at the center of a cube of side $$'a'$$, the flux of gravitational field passing through its sides $$4\,\pi \,GM.$$

If the direction of a field due to a point source is radial and its dependence on the distance $$'r'$$ from the source is given as $${1 \over {{r^2}}},$$ its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.

A

Statement - $$1$$ is false, Statement - $$2$$ is true

B

Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is a correct explanation for Statement - $$1$$

C

Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is not a correct explanation for Statement - $$1$$

D

Statement - $$1$$ is true, Statement - $$2$$ is false

Gravitational field $$\overrightarrow g $$ = $$ - {{GM} \over {{r^2}}}$$

where, $$M=$$ mass enclosed in the closed surface

Gravitational flux through a closed surface is given by

$${\left| {\overrightarrow g .d\overrightarrow S } \right|}$$ = $$4\pi {r^2}.{{GM} \over {{r^2}}}$$ = $$4\pi GM$$

So Statement - 1 is correct.

Statement - 2 is also correct because when the shape of the earth is spherical, area of the Gaussian surface is $$4\pi {r^2}$$. This proves inverse square law.

where, $$M=$$ mass enclosed in the closed surface

Gravitational flux through a closed surface is given by

$${\left| {\overrightarrow g .d\overrightarrow S } \right|}$$ = $$4\pi {r^2}.{{GM} \over {{r^2}}}$$ = $$4\pi GM$$

So Statement - 1 is correct.

Statement - 2 is also correct because when the shape of the earth is spherical, area of the Gaussian surface is $$4\pi {r^2}$$. This proves inverse square law.

4

MCQ (Single Correct Answer)

A planet in a density solar system is $$10$$ times more massive than the earth and its radius is $$10$$ times smaller. Given that the escape velocity from the earth is $$11\,\,km\,{s^{ - 1}},$$ the escape velocity from the surface of the planet would be

A

$$1.1\,\,km\,{s^{ - 1}}$$

B

$$100\,\,km\,{s^{ - 1}}$$

C

$$110\,\,km\,{s^{ - 1}}$$

D

$$0.11\,\,km\,{s^{ - 1}}$$

Let M_{e} is mass of earth then mass of planet M_{p} = 10M_{e}.

And let R_{e} is radius of earth then radius of planet R_{p} = $${{{R_e}} \over {10}}$$

Escape velocity of earth, $${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} $$

Escape velocity of planet, $${v_p} = \sqrt {{{2G{M_p}} \over {{R_p}}}} $$

$$\therefore$$ $${{{{ {{v_p}}}}} \over {{{ {{v_e}} }}}} = {{\sqrt {{{2G{M_p}} \over {{R_p}}}} } \over {\sqrt {{{2G{M_e}} \over {{R_e}}}} }}$$

$$ = \sqrt {{{{M_p}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} $$

$$ = \sqrt {{{10{M_e}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{{\mathop{\rm R}\nolimits} _e}/10}}} = 10$$

$$\therefore$$ $${{v_p}} = 10 \times {{v_e}}$$

$$ = 10 \times 11 = 110\,km/s$$

And let R

Escape velocity of earth, $${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} $$

Escape velocity of planet, $${v_p} = \sqrt {{{2G{M_p}} \over {{R_p}}}} $$

$$\therefore$$ $${{{{ {{v_p}}}}} \over {{{ {{v_e}} }}}} = {{\sqrt {{{2G{M_p}} \over {{R_p}}}} } \over {\sqrt {{{2G{M_e}} \over {{R_e}}}} }}$$

$$ = \sqrt {{{{M_p}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} $$

$$ = \sqrt {{{10{M_e}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{{\mathop{\rm R}\nolimits} _e}/10}}} = 10$$

$$\therefore$$ $${{v_p}} = 10 \times {{v_e}}$$

$$ = 10 \times 11 = 110\,km/s$$

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