### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### 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.
2

### AIEEE 2002

The minimum velocity (in $m{s^{ - 1}}$) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction $0.6$ to avoid skidding is
A
$60$
B
$30$
C
$15$
D
$25$

## Explanation

For no skidding along curved track,

The maximum velocity possible ${v_{\max }} = \sqrt {\mu rg}$

Here $\mu = 0.6,\,r = 150m,\,g = 9.8$

$\therefore$ ${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$
3

### AIEEE 2002

Three identical blocks of masses $m=2$ $kg$ are drawn by a force $F=10.2$ $N$ with an acceleration of $0.6$ $m{s^{ - 2}}$ on a frictionless surface, then what is the tension (in $N$) in the string between the blocks $B$ and $C$?
A
$9.2$
B
$3.4$
C
$4$
D
$7.8$

## Explanation

Assume tension between block A and B is T1 and block B and C is T2. Acceleration a = 0.6 m/s2.

For block A :

F - T1 = ma

For block B :

T1 - T2 = ma

By adding both those equation we get,

F - T2 = 2ma

T2 = F - 2ma

T2 = 10.2 - 2$\times$2$\times$0.6 = 10.2 - 2.4 = 7.8 N
4

### AIEEE 2002

A light string passing over a smooth light pulley connects two blocks of masses ${m_1}$ and ${m_2}$ (vertically). If the acceleration of the system is $g/8$, then the ratio of the masses is
A
$8:1$
B
$9:7$
C
$4:3$
D
$5:3$

## Explanation

Assume that, mass m1 is greater than mass m2, so the heavier mass m1 is accelerating downward and the lighter mass m2 is accelerating upwards.

For mass ${m_1}$ the equation will be

${m_1}$$g-T=$${m_1}$$a For mass {m_2} the equation will be T-$${m_2}$$g=$${m_2}$$a$

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

$\therefore$ ${g \over 8} = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}$

$\Rightarrow {1 \over 8} = {{{{{m_1}} \over {{m_2}}} - 1} \over {{{{m_1}} \over {{m_2}}} + 1}}$

$\Rightarrow$ ${{{m_1}} \over {{m_2}}} + 1 =$ ${8\left( {{{{m_1}} \over {{m_2}}} - 1} \right)}$

$\Rightarrow$ ${{{m_1}} \over {{m_2}}} = {9 \over 7}$