### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

A uniform chain of length $2$ $m$ is kept on a table such that a length of $60$ $cm$ hangs freely from the edge of the table. The total mass of the chain is $4$ $kg.$ What is the work done in pulling the entire chain on the table?
A
$12$ $J$
B
$3.6$ $J$
C
$7.2$ $J$
D
$1200$ $J$

## Explanation

Mass of hanging part $(m') = {4 \over 2} \times \left( {0.6} \right)kg$ = 1.2 kg

Let at the surface $PE=0$

Center of mass of hanging part $=0.3$ $m$ below the surface of the table

${U_i} = - m'gx = - 1.2 \times 10 \times 0.30$ = - 3.6 J

$\Delta U = m'gx = 3.6 J =$ Work done in putting the entire chain on the table.
2

### AIEEE 2004

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to
A
$x$
B
${e^x}$
C
${x^2}$
D
${\log _e}x$

## Explanation

Given that, retardation $\propto$ displacement

$\Rightarrow$ $a=-kx$

But we know $a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$

$\therefore$ ${{vdv} \over {dx}} = - kx$

$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx}$

$\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}$

$\Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} \right) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} \right)$

$\therefore$ Loss in kinetic energy is proportional to ${x^2}$.

$\therefore$ $\Delta K \propto {x^2}$
3

### AIEEE 2003

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $'t'$ is proportional to
A
${t^{3/4}}$
B
${t^{3/2}}$
C
${t^{1/4}}$
D
${t^{1/2}}$

## Explanation

We know that $F \times v =$ Power

According to the question, power is constant.

$\therefore$ $F \times v = c\,\,\,\,$ where $c=$ constant

$\therefore$ $m{{dv} \over {dt}} \times v = c$ $\,\,\,\,\left( \, \right.$ $\therefore$ $\left. {F = ma = {{mdv} \over {dt}}\,\,} \right)$

$\therefore$ $m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$ $\therefore$ ${1 \over 2}m{v^2} = ct$

$\therefore$ $v = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$

${{dx} \over {dt}} = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}\,\,\,\,$ where $v = {{dx} \over {dt}}$

$\therefore$ $\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}} dt$
$x = \sqrt {{{2c} \over m}} \times {{2{t^{{\raise0.5ex\hbox{\scriptstyle 3} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}} \over 3} \Rightarrow x \propto {t^{{\raise0.5ex\hbox{\scriptstyle 3} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$
4

### AIEEE 2003

A wire suspended vertically from one of its ends is stretched by attaching a weight of $200N$ to the lower end. The weight stretches the wire by $1$ $mm.$ Then the elastic energy stored in the wire is
A
$0.2$ $J$
B
$10$ $J$
C
$20$ $J$
D
$0.1$ $J$

## Explanation

The elastic potential energy

$= {1 \over 2} \times$ Force $\times$ extension

$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$