 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

AIEEE 2003

A light spring balance hangs from the hook of the other light spring balance and a block of mass $M$ $kg$ hangs from the former one. Then the true statement about the scale reading is
A
Both the scales read $M$ $kg$ each
B
The scale of the lower one reads $M$ $kg$ and of the upper one zero
C
The reading of the two scales can be anything but the sum of the reading will be $M$ $kg$
D
Both the scales read $M/2$ $kg$ each

Explanation

Question says, both spring balance are light that means springs are massless. The Earth pulls the block by a force $Mg.$ The block in turn exerts a force $Mg$ on the spring of spring balance ${S_1}$ which therefore shows a reading of $Mkgf.$

The spring ${S_1}$ is massless. Therefore it exerts a force of $Mg$ on the spring of spring balance ${S_2}$ which shows the reading of $Mkgf.$

2

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

$mg-T=ma$

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

$= 5\left( {9.8 - 5} \right) = 24\,N$
3

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.
4

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

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