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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
From a solid sphere of mass $$M$$ and radius $$R,$$ a spherical portion of radius $$R/2$$ is removed, as shown in the figure. Taking gravitational potential $$V=0$$ at $$r = \infty ,$$ the potential at the center of the cavity thus formed is:
($$G=gravitational $$ $$constant$$)
A
$${{ - 2GM} \over {3R}}$$
B
$${{ - 2GM} \over R}$$
C
$${{ - GM} \over {2R}}$$
D
$${{ - GM} \over R}$$

Explanation



Before removing the spherical portion, potential at point $$P$$(Center of cavity)

$${V_{sphere}}\,\, = {{ - GM} \over {2{R^3}}}\left[ {3{R^2} - {{\left( {{R \over 2}} \right)}^2}} \right]$$

$$ = {{ - GM} \over {2{R^3}}}\left( {{{11{R^2}} \over 4}} \right) = - 11{{GM} \over {8R}}$$

Mass of removed part = $${M \over {{4 \over 3}\pi {R^3}}} \times {4 \over 3}\pi {\left( {{R \over 2}} \right)^2}$$ = $${M \over 8}$$

Due to cavity part potential at point $$P$$

$${V_{cavity}}\,\,$$ = $$ - {{G{M \over 8}} \over {2{{\left( {{R \over 2}} \right)}^3}}}\left[ {3{{\left( {{R \over 2}} \right)}^2} - {0^2}} \right]$$

           = $$ - {{3GM} \over {8R}}$$

So potential at P due to remaining part,

$$ = {V_{sphere}}\,\, - \,\,{V_{cavity}}\,\,$$

$$ = \,\, - {{11GM} \over {8R}} - \left( { - {3 \over 8}{{GM} \over R}} \right)$$

$$ = {{ - GM} \over R}$$



Note : Potential inside the sphere of radius R and at a distance r from the center,

V = $$ - {{GM} \over {2{R^3}}}\left[ {3{R^2} - {r^2}} \right]$$

Here M = mass of the sphere, r = distance from the center of the sphere
2

JEE Main 2014 (Offline)

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

Explanation

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 Fcp = $${{M{v^2}} \over R}$$

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

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

$$ = {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } \right)} $$
3

JEE Main 2013 (Offline)

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

Explanation

Energy of the satellite on the surface of the planet

Ei = 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

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

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

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

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

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

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

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

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

= Ef - Ei

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

= $${{5GMm} \over {6R}}$$
4

AIEEE 2012

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

Explanation

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

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