### JEE Mains Previous Years Questions with Solutions

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1

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

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

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

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

### 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

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

= ${{5GMm} \over {6R}}$
3

### AIEEE 2012

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
4

### AIEEE 2012

Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is
A
m1r1 : m2r2
B
m1 : m2
C
r1 : r2
D
1 : 1

## Explanation

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

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

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

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

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