 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

Starting from the origin a body oscillates simple harmonically with a period of $2$ $s.$ After what time will its kinetic energy be $75\%$ of the total energy?
A
${1 \over 6}s$
B
${1 \over 4}s$
C
${1 \over 3}s$
D
${1 \over 12}s$

Explanation

$K.E.\,$ of a body undergoing $SHM$ is given by,

$K.E. = {1 \over 2}m{a^2}{\omega ^2}{\cos ^2}\,\omega t,$

$T.E. = {1 \over 2}m{a^2}{\omega ^2}$

Given $K.E.=0.75T.E.$

$\Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = {\pi \over 6}$

$\Rightarrow t = {\pi \over {6 \times \omega }} \Rightarrow t = {{\pi \times 2} \over {6 \times 2\pi }} \Rightarrow t = {1 \over 6}s$
2

AIEEE 2006

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motoin of angular frequency $\omega .$ The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
A
at the mean position of the platform
B
for an amplitude of ${g \over {{\omega ^2}}}$
C
For an amplitude of ${{{g^2}} \over {{\omega ^2}}}$
D
at the height position of the platform

Explanation

Coin $A$ is moving in simple harmonic motion in vertical direction. Now we are assuming coin will leave contact with the platform when platform is at a distance of $x$ from the mean position which is also called amplitude. At distance $x$ the force acting on the coin is

$mg - N = m{\omega ^2}x$

For coin to leave contact $N=0$

$\Rightarrow mg = m{\omega ^2}x \Rightarrow x = {g \over {{\omega ^2}}}$

$\therefore$ Option (B) is correct.
3

AIEEE 2005

If a simple harmonic motion is represented by ${{{d^2}x} \over {d{t^2}}} + \alpha x = 0.$ its time period is
A
${{2\pi } \over {\sqrt \alpha }}$
B
${{2\pi } \over \alpha }$
C
$2\pi \sqrt \alpha$
D
$2\pi \alpha$

Explanation

${{{d^2}x} \over {d{t^2}}} = - \alpha x = - {\omega ^2}x$

$\Rightarrow \omega = \sqrt \alpha$ $\,\,\,\,$ or $\,\,\,\,$ $T = {{2\pi } \over \omega } = {{2\pi } \over {\sqrt \alpha }}$
4

AIEEE 2005

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would
A
first decrease and then increase to the original value
B
first increase and then decrease to the original value
C
increase towards a saturation value
D
remain unchanged

Explanation

Center of mass of combination of liquid and hollow portion (at position $\ell$ ), first goes down (to $\ell + \Delta \ell$) and when total water is drained out, center of mass regain its original position (to $\ell$), $$T = 2\pi \sqrt {{\ell \over g}}$$

$\therefore$ $'T'$ first increases and then decreases to original value. 