### JEE Mains Previous Years Questions with Solutions

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1

### 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.

2

### 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. $= {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\pi - 3\pi } \over 6} = {\pi \over 6}$
3

### AIEEE 2005

The function ${\sin ^2}\left( {\omega t} \right)$ represents
A
a periodic, but not $SHM$ with a period ${\pi \over \omega }$
B
a periodic, but not $SHM$ with a period ${{2\pi } \over \omega }$
C
a $SHM$ with a period ${\pi \over \omega }$
D
a $SHM$ with a period ${{2\pi } \over \omega }$

## Explanation

y = sin2$\omega$t

= ${{1 - \cos 2\omega t} \over 2}$

$= {1 \over 2} - {1 \over 2}\cos \,2\omega t$

$\therefore$ Angular speed = 2$\omega$

$\therefore$ Period (T) = ${{2\pi } \over {angular\,speed}}$ = ${{2\pi } \over {2\omega }}$ = ${\pi \over \omega }$

So it is a periodic function.

As y = sin2$\omega$t

${{dy} \over {dt}}$ = 2$\omega$sin$\omega$t cos$\omega$t = $\omega$ sin2$\omega$t

${{{d^2}y} \over {d{t^2}}}$ = $2{\omega ^2}$ cos2$\omega$t which is not proportional to -y.

Hence it is is not SHM.
4

### AIEEE 2004

In forced oscillation of a particle the amplitude is maximum for a frequency ${\omega _1}$ of the force while the energy is maximum for a frequency ${\omega _2}$ of the force; then
A
${\omega _1} < {\omega _2}$ when damping is small and ${\omega _1} > {\omega _2}$ when damping is large
B
${\omega _1} > {\omega _2}$
C
${\omega _1} = {\omega _2}$
D
${\omega _1} < {\omega _2}$

## Explanation

The maximum of amplitude and energy is obtained when the frequency is equal to the natural frequency (resonance condition)

$\therefore$ ${\omega _1} = {\omega _2}$