 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

A block of mass $M$ is pulled along a horizontal frictionless surface by a rope of mass $m.$ If a force $P$ is applied at the free end of the rope, the force exerted by the rope on the block is
A
${{Pm} \over {M + m}}$
B
${{Pm} \over {M - m}}$
C
$P$
D
${{PM} \over {M + m}}$

Explanation

Taking the rope and the block as a system Acceleration of block($a$) = ${{Force\,applied} \over {total\,mass}}$

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

Force on block(T) = Mass of block $\times$ $a$

$T=Ma$

$\therefore$ $T = {{MP} \over {m + M}}$
2

AIEEE 2003

A marble block of mass $2$ $kg$ lying on ice when given a velocity of $6$ $m/s$ is stopped by friction in $10$ $s.$ Then the coefficient of friction is
A
$0.02$
B
$0.03$
C
$0.04$
D
$0.06$

Explanation

The retarding force is created by the frictional force.

Given, $u = 6m/s,\,v = 0,\,t = 10s,$

Retardation$(a) = - {f \over m} = {{ - umg} \over m} = - \mu g = - 10\mu$

$v = u + at$

$0 = 6 - 10\mu \times 10$

$\therefore$ $\mu = 0.06$
3

AIEEE 2003

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $49$ $N,$ when the lift is stationary. If the lift moves downward with an acceleration of $5 m/{s^2}$, the reading of the spring balance will be
A
$24$ $N$
B
$74$ $N$
C
$15$ $N$
D
$49$ $N$

Explanation When lift is stationary then,

T1 = $mg$ = 49 N

$m$ = 5

For the bag accelerating down

$mg-T=ma$

$\therefore$ $T=m(g-a)$

$= 5\left( {9.8 - 5} \right) = 24\,N$
4

AIEEE 2002

One end of a mass-less rope, which passes over a mass-less and friction-less pulley $P$ is tied to a hook $C$ while the other end is free. Maximum tension that the rope can bear is $360$ $N.$ With what value of maximum safe acceleration (in $m{s^{ - 2}}$) can a man of $60$ $kg$ climb on the rope? A
$16$
B
$6$
C
$4$
D
$8$

Explanation Assuming acceleration $a$ of the man is downwards. So the equation will be

$mg - T = ma$

$\therefore$ $a = g - {T \over m} = 10 - {{360} \over {60}} = 4$ $m/{s^2}$

So the maximum acceleration of man is 4 m/s2 downwards.