### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2008

An athlete in the olympic games covers a distance of $100$ $m$ in $10$ $s.$ His kinetic energy can be estimated to be in the range
A
$200J-500J$
B
$2 \times {10^5}J - 3 \times {10^5}J$
C
$20,000J - 50,000J$
D
$2,000J - 5,000J$

## Explanation

The average speed of the athelete

$v = {{100} \over {10}} = 10m/s\,\,\,\,$ $\therefore$ $K.E. = {1 \over 2}m{v^2}$

If mass of athlete is $40$ $kg$ then, $K.E.$ $= {1 \over 2} \times 40 \times {\left( {10} \right)^2} = 2000J$

If mass of athlete is $100$ $kg$ then, $K.E.$ $= {1 \over 2} \times 100 \times {\left( {10} \right)^2} = 5000J$

His kinetic energy can be in the range = 2000 J to 5000 J.
2

### AIEEE 2007

A $2$ $kg$ block slides on a horizontal floor with a speed of $4m/s.$ It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is $15N$ and spring constant is $10, 000$ $N/m.$ The spring compresses by
A
$8.5cm$
B
$5.5cm$
C
$2.5cm$
D
$11.0cm$

## Explanation

Let the block compress the spring by $x$ before coming to rest.

Initial kinetic energy of the block $=$ (potential energy of compressed spring) $+$ work done due to friction.

${1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x$

$10,000{x^2} + 30x - 32 = 0$

$\Rightarrow 5000{x^2} + 15x - 16 = 0$

$\therefore$ $x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}$

$\,\,\,\,\, = 0.055m = 5.5cm.$
3

### AIEEE 2006

The potential energy of a $1$ $kg$ particle free to move along the $x$-axis is given by $V\left( x \right) = \left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right)J$.

The total mechanical energy of the particle is $2J.$ Then, the maximum speed (in $m/s$) is

A
${3 \over {\sqrt 2 }}$
B
${\sqrt 2 }$
C
${1 \over {\sqrt 2 }}$
D
$2$

## Explanation

Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimum

For minimum potential energy,

${{dV} \over {dx}} = 0$

$\Rightarrow {x^3} - x = 0$

$\Rightarrow x = \pm 1$

$\Rightarrow$ Min. Potential energy (P.E.) =${1 \over 4} - {1 \over 2} = - {1 \over 4}J$

$K.E{._{\left( {\max .} \right)}} + P.E{._{\left( {\min .} \right)}} = 2\,$ (Given)

$\therefore$ $K.E{._{\left( {\max .} \right)}} = 2 + {1 \over 4} = {9 \over 4}$

$\therefore$ ${1 \over 2}mv_{\max }^2$ = ${9 \over 4}$

$\Rightarrow {1 \over 2} \times 1 \times {v^2}_{\max .} = {9 \over 4}$

$\Rightarrow {v_{\max }} = {3 \over {\sqrt 2 }}$ m/s
4

### AIEEE 2006

A particle of mass $100g$ is thrown vertically upwards with a speed of $5$ $m/s$. The work done by the force of gravity during the time the particle goes up is
A
$-0.5J$
B
$-1.25J$
C
$1.25J$
D
$0.5J$

## Explanation

Kinetic energy at point of throwing is converted into potential energy of the particle during rise.

$K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J$

$W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J$

$\left[ \, \right.$ As we know, $mgh = {1 \over 2}m{v^2}$ by energy conservation $\left. \, \right]$