### JEE Mains Previous Years Questions with Solutions

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1

### 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

$mg-T=ma$

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

$= 5\left( {9.8 - 5} \right) = 24\,N$
2

### 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.
3

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

$\therefore$ ${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$
4

### 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