### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

A block rests on a rough inclined plane `making an angle of ${30^ \circ }$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8.$ If the frictionless force on the block is $10$ $N,$ the mass of the block (in $kg$) is $\left( {take\,\,\,g\, = \,10\,\,m/{s^2}} \right)$
A
$1.6$
B
$4.0$
C
$2.0$
D
$2.5$

## Explanation

For equilibrum of block,

$mg\,\,\sin \theta = {f_s}\,\,$

$\Rightarrow m \times 10 \times \sin {30^ \circ } = 10$

$\Rightarrow m \times 5 = 10$

$\Rightarrow m = 2.0\,\,kg$
2

### AIEEE 2004

Two masses ${m_1} = 5kg$ and ${m_2} = 4.8kg$ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when left free to move? $\left( {g = 9.8m/{s^2}} \right)$
A
$5\,\,m/{s^2}$
B
$9.8\,\,m/{s^2}$
C
$0.2\,\,m/{s^2}$
D
$4.8\,\,m/{s^2}$

## Explanation

For m1 mass :

m1g - T = m1a .........(1)

For m2 mass :

T - m2g = m2a ....... (2)

By solving (1) and (2) we get,

Acceleration a = $\left( {{{{m_1} - {m_2}} \over {{m_1} + {m_2}}}} \right)g$

$= {{\left( {5 - 4.8} \right) \times 9.8} \over {\left( {5 + 4.8} \right)}}m/{s^2}$ = 0.2 m/s2
3

### AIEEE 2003

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
4

### 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.