### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

Three forces start acting simultaneously on a particle moving with velocity, $\overrightarrow v \,\,.$ These forces are represented in magnitude and direction by the three sides of a triangle $ABC$. The particle will now move with velocity
A
less than $\overrightarrow v \,$
B
greater than $\overrightarrow v \,$
C
$\left| v \right|$ in the direction of the largest force $BC$
D
$\overrightarrow v \,\,,$ remaining unchanged

## Explanation

As shown in the figure, the three forces are represented by the sides of a triangle taken in the same order. So the particle will be in equilibrium under the three forces.

Therefore the resultant force is zero.

We know, ${\overrightarrow F_{net}} = m\overrightarrow a .$

$\therefore$ 0 = $m\overrightarrow a .$

$\Rightarrow \overrightarrow a = 0$

Therefore acceleration will be zero.

Hence the particle velocity remains unchanged.
2

### AIEEE 2003

A rocket with a lift-off mass $3.5 \times {10^4}\,\,kg$ is blasted upwards with an initial acceleration of $10m/{s^2}.$ Then the initial thrust of the blast is
A
$3.5 \times {10^5}N$
B
$7.0 \times {10^5}N$
C
$14.0 \times {10^5}N$
D
$1.75 \times {10^5}N$

## Explanation

Here, thrust force is responsible to accelerate the rocket,

So initial thrust of the blast

= (Lift-off mass) × acceleration

= (3.5 × 104) × (10)

= 3.5 × 105 N
3

### 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 $\times$ $a$

$T=Ma$

$\therefore$ $T = {{MP} \over {m + M}}$
4

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

$v = u + at$

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

$\therefore$ $\mu = 0.06$