Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Four particles, each of mass $$M$$ and equidistant from each other, move along a circle of radius $$R$$ under the action of their mutual gravitational attraction. The speed of each particle is :

A

$$\sqrt {{{GM} \over R}} $$

B

$$\sqrt {2\sqrt 2 {{GM} \over R}} $$

C

$$\sqrt {{{GM} \over R}\left( {1 + 2\sqrt 2 } \right)} $$

D

$${1 \over 2}\sqrt {{{GM} \over R}\left( {1 + 2\sqrt 2 } \right)} $$

All those particles are moving due to their mutual gravitational attraction.

The force between each masses are repulsive force.

On mass M at C, due to mass at D the repulsive force is F in the vertical direction.

On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.

On mass M at C, due to mass at A the repulsive force is F'

Net force acting on particle at C,

= $$2F\,\cos \,{45^ \circ } + F'$$

Where $$F = {{G{M^2}} \over {{{\left( {\sqrt 2 R} \right)}^2}}}$$ and $$F' = {{G{M^2}} \over {4{R^2}}}$$

$$ \Rightarrow {{2 \times G{M^2}} \over {\sqrt 2 {{\left( {R\sqrt 2 } \right)}^2}}} + {{G{M^2}} \over {4{R^2}}}$$

$$ \Rightarrow {{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] $$

This net force will balance by the centripetal force F_{cp} = $${{M{v^2}} \over R}$$

$$\therefore$$ $${{M{v^2}} \over R} = $$ $${{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] $$

$$ \Rightarrow $$ $$v = \sqrt {{{Gm} \over R}\left( {{{\sqrt 2 + 4} \over {4\sqrt 2 }}} \right)} $$

$$ = {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } \right)} $$

The force between each masses are repulsive force.

On mass M at C, due to mass at D the repulsive force is F in the vertical direction.

On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.

On mass M at C, due to mass at A the repulsive force is F'

Net force acting on particle at C,

= $$2F\,\cos \,{45^ \circ } + F'$$

Where $$F = {{G{M^2}} \over {{{\left( {\sqrt 2 R} \right)}^2}}}$$ and $$F' = {{G{M^2}} \over {4{R^2}}}$$

$$ \Rightarrow {{2 \times G{M^2}} \over {\sqrt 2 {{\left( {R\sqrt 2 } \right)}^2}}} + {{G{M^2}} \over {4{R^2}}}$$

$$ \Rightarrow {{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] $$

This net force will balance by the centripetal force F

$$\therefore$$ $${{M{v^2}} \over R} = $$ $${{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] $$

$$ \Rightarrow $$ $$v = \sqrt {{{Gm} \over R}\left( {{{\sqrt 2 + 4} \over {4\sqrt 2 }}} \right)} $$

$$ = {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } \right)} $$

2

MCQ (Single Correct Answer)

What is the minimum energy required to launch a satellite of mass $$m$$ from the surface of a planet of mass $$M$$ and radius $$R$$ in a circular orbit at an altitude of $$2R$$?

A

$${{5GmM} \over {6R}}$$

B

$${{2GmM} \over {3R}}$$

C

$${{GmM} \over {2R}}$$

D

$${{GmM} \over {3R}}$$

Energy of the satellite on the surface of the planet

E_{i} = K.E + P.E = 0 + $$\left( { - {{GMm} \over R}} \right)$$ = $${ - {{GMm} \over R}}$$

Energy of the satellite at 2R distance from the surface of the planet while moving with velocity v

E_{f} = $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} \right)$$

In the orbital of planet, the centripetal force is provided by the gravitational force

$$\therefore$$ $${{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} \right)}^2}}}$$

$$ \Rightarrow {v^2} = {{GM} \over {3R}}$$

$$\therefore$$ E_{f} = $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} \right)$$

$$ = {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}$$

= $$ - {{GMm} \over {6R}}$$

$$\therefore$$ Minimum energy required required to launch the satellite

= E_{f} - E_{i}

= $$ - {{GMm} \over {6R}}$$ - $$\left( { - {{GMm} \over R}} \right)$$

= $${{5GMm} \over {6R}}$$

E

Energy of the satellite at 2R distance from the surface of the planet while moving with velocity v

E

In the orbital of planet, the centripetal force is provided by the gravitational force

$$\therefore$$ $${{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} \right)}^2}}}$$

$$ \Rightarrow {v^2} = {{GM} \over {3R}}$$

$$\therefore$$ E

$$ = {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}$$

= $$ - {{GMm} \over {6R}}$$

$$\therefore$$ Minimum energy required required to launch the satellite

= E

= $$ - {{GMm} \over {6R}}$$ - $$\left( { - {{GMm} \over R}} \right)$$

= $${{5GMm} \over {6R}}$$

3

MCQ (Single Correct Answer)

The mass of a spaceship is $$1000$$ $$kg.$$ It is to be launched from the earth's surface out into free space. The value of $$g$$ and $$R$$ (radius of earth ) are $$10\,m/{s^2}$$ and $$6400$$ $$km$$ respectively. The required energy for this work will be:

A

$$6.4 \times {10^{11}}\,$$ Joules

B

$$6.4 \times {10^8}\,$$ Joules

C

$$6.4 \times {10^9}\,$$ Joules

D

$$6.4 \times {10^{10}}\,$$ Joules

Potential energy at earth surface = $$ - {{GMm} \over R}$$

and at free space potential energy = 0

Work done for this = $$0 - \left( { - {{GMm} \over R}} \right)$$ = $${{{GMm} \over R}}$$

So the required energy for this work is

= $${{GMm} \over R}$$

=$${{{g{R^2}m} \over R}}$$ [ as $$g = {{GM} \over {{R^2}}}$$ ]

= $$mgR$$

$$ = 1000 \times 10 \times 6400 \times {10^3}$$

$$ = 6.4 \times {10^{10}}\,\,$$ Joules

and at free space potential energy = 0

Work done for this = $$0 - \left( { - {{GMm} \over R}} \right)$$ = $${{{GMm} \over R}}$$

So the required energy for this work is

= $${{GMm} \over R}$$

=$${{{g{R^2}m} \over R}}$$ [ as $$g = {{GM} \over {{R^2}}}$$ ]

= $$mgR$$

$$ = 1000 \times 10 \times 6400 \times {10^3}$$

$$ = 6.4 \times {10^{10}}\,\,$$ Joules

4

MCQ (Single Correct Answer)

Two cars of masses m_{1} and m_{2} are moving in circles of radii r_{1} and r_{2}, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is

A

m_{1}r_{1} : m_{2}r_{2}

B

m_{1} : m_{2}

C

r_{1} : r_{2}

D

1 : 1

We know, $$a = r\,{w^2} = r \times {\left( {{{2\pi } \over T}} \right)^2}$$

Given, $${T_1} = {T_2} = T$$

$${a_1} = {r_1} \times {\left( {{{2\pi } \over T}} \right)^2}$$

$${a_2} = {r_2} \times {\left( {{{2\pi } \over T}} \right)^2}$$

$$\therefore$$ $${{{a_1}} \over {{a_2}}} = {{{r_1}} \over {{r_2}}}$$

Given, $${T_1} = {T_2} = T$$

$${a_1} = {r_1} \times {\left( {{{2\pi } \over T}} \right)^2}$$

$${a_2} = {r_2} \times {\left( {{{2\pi } \over T}} \right)^2}$$

$$\therefore$$ $${{{a_1}} \over {{a_2}}} = {{{r_1}} \over {{r_2}}}$$

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