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### JEE Mains Previous Years Questions with Solutions

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### AIEEE 2007

MCQ (Single Correct Answer)
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

$ma = m\left( {{F \over {M + m}}} \right) = {{mF} \over {m + M}}.$
2

### AIEEE 2006

MCQ (Single Correct Answer)
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

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

$\Rightarrow$ $F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$
3

### AIEEE 2005

MCQ (Single Correct Answer)
A block is kept on a frictionless inclined surface with angle of inclination $'\,\alpha \,'.$ The incline is given an acceleration $a$ to keep the block stationary. Then $a$ is equal to
A
$g$ $cosec$ $\alpha$
B
$g/tan$ $\alpha$
C
$g$ $tan$ $\alpha$
D
$g$

## Explanation

Acceleration of the block is to the right. Pseudo force acting on the block to the left.

From diagram we can say,

m$a$cos$\alpha$ = m$g$sin$\alpha$

$\Rightarrow a = g\tan \alpha$
4

### AIEEE 2005

MCQ (Single Correct Answer)
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, ${v^2} = {u^2} + 2as$

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

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

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

$\Rightarrow s = 1000\,m$

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