### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2006

A mass of $M$ $kg$ is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of ${45^ \circ }$ with the initial vertical direction is
A
$Mg\left( {\sqrt 2 + 1} \right)$
B
$Mg\sqrt 2$
C
${{Mg} \over {\sqrt 2 }}$
D
$Mg\left( {\sqrt 2 - 1} \right)$

## Explanation

From work energy theorem we can say,

Work done by tension $+$ work done by force (applied) $+$ Work done by gravitational force $=$ change in kinetic energy

Here Work done by tension is zero

$\Rightarrow 0 + F \times AB - Mg \times AC = 0$
$\Rightarrow F = Mg\left( {{{AC} \over {AB}}} \right) = Mg\left[ {{{1 - {1 \over {\sqrt 2 }}} \over {{1 \over 2}}}} \right]$
[ as $AB = \ell \sin {45^ \circ } = {\ell \over {\sqrt 2 }}$
and $AC = OC - OA = \ell - \ell \,\cos \,{45^ \circ } = \ell \left( {1 - {1 \over {\sqrt 2 }}} \right)$
where $\ell =$ length of the string. ]
$\Rightarrow F = Mg\left( {\sqrt 2 - 1} \right)$
2

### AIEEE 2005

The block of mass $M$ moving on the frictionless horizontal surface collides with the spring of spring constant $k$ and compresses it by length $L.$ The maximum momentum of the block after collision is
A
${{k{L^2}} \over {2M}}$
B
$\sqrt {Mk} \,\,L$
C
${{M{L^2}} \over k}$
D
Zero

## Explanation

Elastic energy stored in the spring = ${1 \over 2}k{L^2}$

And kinetic energy of the block = ${1 \over 2}M{v^2}$

$\therefore$ ${1 \over 2}M{v^2} = {1 \over 2}k{L^2}$

$\Rightarrow v = \sqrt {{k \over M}} .L$

$\therefore$ Momentum $= M \times v = M \times \sqrt {{k \over M}} .L = \sqrt {kM} .L$
3

### AIEEE 2005

A body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T.$ The instantaneous power delivered to the body as a function of time is given by
A
${{m{v^2}} \over {{T^2}}}.{t^2}$
B
${{m{v^2}} \over {{T^2}}}.t$
C
${1 \over 2}{{m{v^2}} \over {{T^2}}}.{t^2}$
D
${1 \over 2}{{m{v^2}} \over {{T^2}}}.t$

## Explanation

$u = 0;v = u + aT;v = aT$

Instantaneous power $= F \times v = m.\,a.\,at = m.{a^2}.t$

$\therefore$ Instantaneous power $= {{m{v^2}t} \over {{T^2}}}$
4

### AIEEE 2005

A spherical ball of mass $20$ $kg$ is stationary at the top of a hill of height $100$ $m$. It rolls down a smooth surface to the ground, then climbs up another hill of height $30$ $m$ and finally rolls down to a horizontal base at a height of $20$ $m$ above the ground. The velocity attained by the ball is
A
$20$ $m/s$
B
$40$ $m/s$
C
$10\sqrt {30} \,\,\,m/s$
D
$10\,\,m/s$

## Explanation

Loss in potential energy $=$ gain in kinetic energy

$m \times g \times 80 = {1 \over 2}m{v^2}$

$\Rightarrow$ $10 \times 80 = {1 \over 2}{v^2}$

$\Rightarrow$${v^2} = 1600$ or $v = 40\,m/s$