### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2002

A spring of force constant $800$ $N/m$ has an extension of $5$ $cm.$ The work done in extending it from $5$ $cm$ to $15$ $cm$ is
A
$16J$
B
$8J$
C
$32J$
D
$24J$

## Explanation

When we extend the spring by $dx$ then the work done

$dW = k\,x\,dx$

Applying integration both sides we get,

$\therefore$ $W = k\int\limits_{0.05}^{0.15} {x\,dx}$

$= {{800} \over 2}\left[ {{{\left( {0.15} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]$

$= 8\,J$
2

### AIEEE 2002

If a body looses half of its velocity on penetrating $3$ $cm$ in a wooden block, then how much will it penetrate more before coming to rest?
A
$1$ $cm$
B
$2$ $cm$
C
$3$ $cm$
D
$4$ $cm$

## Explanation

We know the work energy theorem, $W = \Delta K = FS$

For first penetration, by applying work energy theorem we get,

${1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2} = F \times 3\,\,...(i)$

For second penetration, by applying work energy theorem we get,

${1 \over 2}m{\left( {{v \over 2}} \right)^2} - 0 = F \times S\,...(ii)$

On dividing $(ii)$ by $(i)$

${{1/4} \over {3/4}} = S/3$

$\therefore$ $S = 1\,cm$