### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2014 (Offline)

A block of mass $m$ is placed on a surface with a vertical cross section given by $y = {{{x^3}} \over 6}.$ If the coefficient of friction is $0.5,$ the maximum height above the ground at which the block can be placed without slipping is:
A
${1 \over 6}m$
B
${2 \over 3}m$
C
${1 \over 3}m$
D
${1 \over 2}m$

## Explanation

At limiting equilibrium, $\mu = \tan \theta$

Equation of the surface,

$y = {{{x^3}} \over 6}$

Slope, $\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}$

Given that, Coefficient of friction $\mu = 0.5$

$\therefore$ $\,\,\,\,0.5 = {{{x^2}} \over 2}$

$\Rightarrow \,\,\,x = \pm \,1$

Now, $y = {{{x^3}} \over 6} = {1 \over 6}m$
2

### AIEEE 2010

Two fixed frictionless inclined planes making an angle ${30^ \circ }$ and ${60^ \circ }$ with the vertical are shown in the figure. Two blocks $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$ ?
A
$4.9m{s^{ - 2}}$ in horizontal direction
B
$9.8m{s^{ - 2}}$ in vertical direction
C
Zero
D
$4.9m{s^{ - 2}}$ in vertical direction

## Explanation

Along inclined plane the equation of motion of the body

$mg\,\sin \,\theta = ma$ $\,\,\,\,\,\,\,\,$ $\therefore$ $a = g\,\sin \,\theta$

where $a$ is along the inclined plane.

$\therefore$ vertical component of acceleration is $\left( {g\sin \theta } \right)\sin \theta$ = $g\,{\sin ^2}\theta$ (Along vertical)

For block A,
${a_{A\left( {along\,vertical} \right)}} = g{\sin ^2}60^\circ$

For block B,
${a_{B\left( {along\,vertical} \right)}} = g{\sin ^2}30^\circ$

$\therefore$ relative vertical acceleration of $A$ with respect to $B$ is

$g{\sin ^2}60^\circ$ - $g{\sin ^2}30^\circ$

=$g\left( {{{\sin }^2}60 - {{\sin }^2}\left. {30} \right]} \right.$

$= 4.9$ $\,\,m/{s^2}$ in vertical direction
3

### AIEEE 2007

A block of mass $m$ is connected to another block of $mass$ $M$ by a spring (massless) of spring constant $k.$ The block are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force $F$ starts acting on the block of mass $M$ to pull it. Find the force of the block of mass $m.$
A
${{MF} \over {\left( {m + M} \right)}}$
B
${{mF} \over M}$
C
${{\left( {M + m} \right)F} \over m}$
D
${{mF} \over {\left( {m + M} \right)}}$

## Explanation

From free body-diagram of $m$

we get $T = ma$

From free body-diagram of $M$

we get $F-T=Ma$

where $T$ is force due to spring

$\Rightarrow F - ma = Ma$

$\Rightarrow$ $F=Ma+ma$

$\therefore$ $a = {F \over {M + m}}$

Now, force acting on the block of mass $m$ is

$ma = m\left( {{F \over {M + m}}} \right) = {{mF} \over {m + M}}.$
4

### AIEEE 2006

A player caught a cricket ball of mass $150$ $g$ moving at a rate of $20$ $m/s.$ If the catching process is completed in $0.1s,$ the force of the blow exerted by the ball on the hand of the player is equal to
A
$150$ $N$
B
$3$ $N$
C
$30$ $N$
D
$300$ $N$

## Explanation

We know, Force$\times$ time = Impulse = Change in momentum

$\therefore$ $F \times t = m\left( {v - u} \right)$

$\Rightarrow$ $F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$