### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2005

Consider a car moving on a straight road with a speed of $100$ $m/s$. The distance at which car can be stopped is $\left[ {{\mu _k} = 0.5} \right]$
A
$1000$ $m$
B
$800$ $m$
C
$400$ $m$
D
$100$ $m$

## Explanation

Acceleration due to friction = $\left( { - {\mu _k}g} \right)$

We know, ${v^2} = {u^2} + 2as$

$\Rightarrow$ ${0^2} = {u^2} + 2\left( { - {\mu _k}g} \right)s$

$\Rightarrow$ $2 { {\mu _k}g}s$ = ${u^2}$

$\Rightarrow s = {{{{100}^2}} \over {2 \times 0.5 \times 10}}$

$\Rightarrow s = 1000\,m$
2

### AIEEE 2005

A smooth block is released at rest on a ${45^ \circ }$ incline and then slides a distance $'d'$. The time taken to slide is $'n'$ times as much to slide on rough incline than on a smooth incline. The coefficient of friction is
A
${\mu _k} = \sqrt {1 - {1 \over {{n^2}}}}$
B
${\mu _k} = 1 - {1 \over {{n^2}}}$
C
${\mu _k} = \sqrt {1 - {1 \over {{n^2}}}}$
D
${\mu _s} = 1 - {1 \over {{n^2}}}$

## Explanation

For smooth surface,

$d = {1 \over 2}\left( {g\,\sin \,\theta } \right)t_1^2,$

${t_1} = \sqrt {{{2d} \over {g\,\sin \,\theta }}} ,$

When surface is rough

$d = {1 \over 2}\left( {g\,\sin \,\theta - \mu g\,\cos \theta } \right)t_2^2$

${t_2} = \sqrt {{{2d} \over {g\,\sin \,\theta - \mu g\,\cos \theta }}}$

According to question, ${t_2} = n{t_1}$

$n\sqrt {{{2d} \over {g\,\sin \,\theta }}} = \sqrt {{{2d} \over {g\,\sin \,\theta - \mu g\,\cos \theta }}}$

$n = {1 \over {\sqrt {1 - {\mu _k}} }}$ ( as $\cos \,{45^ \circ } = \sin \,{45^ \circ } = {1 \over {\sqrt 2 }}$ )

${n^2} = {1 \over {1 - {\mu _k}}}$

or $1 - {\mu _k} = {1 \over {{n^2}}}$

or ${\mu _k} = 1 - {1 \over {{n^2}}}$

3

### AIEEE 2005

A parachutist after bailing out falls $50$ $m$ without friction. When parachute opens, it decelerates at $2\,\,m/{s^2}.$ He reaches the ground with a speed of $3$ $m/s$. At what height, did he bail out?
A
$182$ $m$
B
$91$ $m$
C
$111$ $m$
D
$293$ $m$

## Explanation

The velocity of parachutist when parachute opens at 50 m is

$u = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 50} = \sqrt {980}$

The velocity at ground, $v=3m/s$

$\therefore$ $S = {{{v^2} - {u^2}} \over {2 \times \left( { - 2} \right)}} = {{{3^2} - 980} \over { - 4}} \approx 243\,m$

Initially he has fallen $50$ $m.$

$\therefore$ Total height from where

He bailed out $=243+50=293m$
4

### AIEEE 2004

A block rests on a rough inclined plane `making an angle of ${30^ \circ }$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8.$ If the frictionless force on the block is $10$ $N,$ the mass of the block (in $kg$) is $\left( {take\,\,\,g\, = \,10\,\,m/{s^2}} \right)$
A
$1.6$
B
$4.0$
C
$2.0$
D
$2.5$

## Explanation

For equilibrum of block,

$mg\,\,\sin \theta = {f_s}\,\,$

$\Rightarrow m \times 10 \times \sin {30^ \circ } = 10$

$\Rightarrow m \times 5 = 10$

$\Rightarrow m = 2.0\,\,kg$