### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

MCQ (Single Correct Answer)
A horizontal force of $10$ $N$ is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is $0.2$. The weight of the block is
A
$20N$
B
$50N$
C
$100N$
D
$2N$

## Explanation

Assume weight of the block = $W$ and friction force = $f$ = $\mu N$

For the block to remain stationary with the wall, weight of the block should balanced by the force of friction.

$\therefore$ $f=W$

$\Rightarrow$ $\mu N = W$

$\Rightarrow$ $W$ = 0.2 $\times$ 10 = 2 N
2

### AIEEE 2003

MCQ (Single Correct Answer)
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.
3

### AIEEE 2003

MCQ (Single Correct Answer)
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
4

### AIEEE 2003

MCQ (Single Correct Answer)
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 $\times$ $a$

$T=Ma$

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

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