### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

A satellite is revolving in a circular orbit at a height $'h'$ from the earth's surface (radius of earth $R;h < < R$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to : (Neglect the effect of atmosphere.)
A
$\sqrt {gR/2}$
B
$\sqrt {gR} \left( {\sqrt 2 - 1} \right)$
C
$\sqrt {2gR}$
D
$\sqrt {gR}$

## Explanation

Orbital velocity of satellite,

${v_0} = \sqrt {{{GM} \over {R + h}}}$

= $\sqrt {{{GM} \over R}}$ [ As $h < < R$ then R + h = R ]

= $\sqrt {gR}$ [ As $g = {{GM} \over {{R^2}}}$ ]

Escape velocity

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

= $\sqrt {2gR}$ [ As $g = {{GM} \over {{R^2}}}$ ]

$\therefore$ The minimum increase in its orbital velocity required to escape from the earth gravitational field

$= \sqrt {2gR} - \sqrt {gR} = \sqrt {gR} \left( {\sqrt 2 - 1} \right)$
2

### JEE Main 2015 (Offline)

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
3

### JEE Main 2014 (Offline)

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

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

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

$= {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } \right)}$
4

### JEE Main 2013 (Offline)

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

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

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