### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2006

A ball of mass $0.2$ $kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2$ $m$ while applying the force and the ball goes upto $2$ $m$ height further, find the magnitude of the force. (consider $g = 10\,m/{s^2}$).
A
$4N$
B
$16$ $N$
C
$20$ $N$
D
$22$ $N$

## Explanation

According to energy conservation law,

Work done by the hand and due to gravity = total change in the kinetic energy

Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $\Delta K$ = 0

${W_{hand}} + {W_{gravity}} = \Delta K$

[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.
And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]
$\Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)$ $= 0 \Rightarrow F = 22\,N$

$\therefore$ Option (D) is correct.
2

### 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. ]
$\Rightarrow F = Mg\left( {\sqrt 2 - 1} \right)$
3

### 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 $= M \times v = M \times \sqrt {{k \over M}} .L = \sqrt {kM} .L$
4

### 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 $= {{m{v^2}t} \over {{T^2}}}$