### JEE Mains Previous Years Questions with Solutions

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1

### 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.
2

### 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 $\,\,\,\,$ $T = {{2\pi } \over \omega } = {{2\pi } \over {\sqrt \alpha }}$
3

### 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.

4

### AIEEE 2005

Two simple harmonic motions are represented by the equations ${y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right)$ and ${y_2} = 0.1\,\cos \,\pi t.$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is
A
${\pi \over 3}$
B
${{ - \pi } \over 6}$
C
${\pi \over 6}$
D
${{ - \pi } \over 3}$

## Explanation

${v_1} = {{d{y_1}} \over {dt}} = 0.1 \times 100\pi \cos \left( {100\pi t + {\pi \over 3}} \right)$

${v_2} = {{d{y_2}} \over {dt}} = - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi \over 2}} \right)$

$\therefore$ Phase diff. $= {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\pi - 3\pi } \over 6} = {\pi \over 6}$