### JEE Mains Previous Years Questions with Solutions

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1

### 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 $v = {{dx} \over {dt}}$

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

### AIEEE 2003

A spring of spring constant $5 \times {10^3}\,N/m$ is stretched initially by $5$ $cm$ from the unstretched position. Then the work required to stretch it further by another $5$ $cm$ is
A
$12.50$ $N$-$m$
B
$18.75$ $N$-$m$
C
$25.00$ $N$-$m$
D
$625$ $N$-$m$

## Explanation

Given $k = 5 \times {10^3}N/m$

Work done when a spring stretched from x1 cm to x2 cm,

$W = {1 \over 2}k\left( {x_2^2 - x_1^2} \right)$

$= {1 \over 2} \times 5 \times {10^3}\left[ {{{\left( {0.1} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]$

$= {{5000} \over 2} \times 0.15 \times 0.05 = 18.75\,\,Nm$
3

### 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

$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$
4

### AIEEE 2003

Consider the following two statements:
$A.$$\,\,\,\,\,\, Linear momentum of a system of particles is zero B.$$\,\,\,\,\,\,$ Kinetic energy of a system of particles is zero.
then
A
$A$ does not imply $B$ and $B$ does not imply $A$
B
$A$ implies $B$ but $B$ does not imply $A$
C
$A$ does not imply $B$ but $B$ implies $A$
D
$A$ implies $B$ and $B$ implies $A$

## Explanation

Kinetic energy of a system of particle is zero this is possible only when the speed of each particles is zero. And if speed of each particle is zero, the linear momentum of the system of particle has to be zero.

Also the linear momentum of the system may be zero even when the particles are moving in different direction. This is because linear momentum is a vector quantity. In this case the kinetic energy of the system of particles will not be zero.

$\therefore$ $A$ does not implies $B$ but $B$ implies $A.$