### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

The displacement $y$ of a particle in a medium can be expressed as, $y = {10^{ - 6}}\,\sin$ $\left( {100t + 20x + {\pi \over 4}} \right)$ $m$ where $t$ is in second and $x$ in meter. The speed of the wave is
A
$20\,\,m/s$
B
$5\,m/s$
C
$2000\,m/s$
D
$5\,\pi \,m/s$

## Explanation

From equation given,

$\omega = 100$ and $k = 20,$ $v = {\omega \over k} = {{100} \over {20}} = 5m/s$
2

### AIEEE 2003

A tuning fork of known frequency $256$ $Hz$ makes $5$ beats per second with the vibrating string of a piano. The beat frequency decreases to $2$ beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
A
$256 + 2Hz$
B
$256 - 2Hz$
C
$256 - 5Hz$
D
$256 + 5Hz$

## Explanation

A tuning fork of frequency $256$ $Hz$ makes $5$ beats/ second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is $\left( {256 \pm 5} \right)$ $Hz$ ie either $261$$Hz or 251 Hz. When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is 251 Hz 3 ### AIEEE 2003 MCQ (Single Correct Answer) The displacement y of a wave travelling in the x-direction is given by$$y = {10^{ - 4}}\,\sin \left( {600t - 2x + {\pi \over 3}} \right)\,\,metres$\$
where $x$ is expressed in metres and $t$ in seconds. The speed of the wave - motion, in $m{s^{ - 1}}$, is
A
$300$
B
$600$
C
$1200$
D
$200$

## Explanation

$y = {10^{ - 4}}\sin \left( {600t - 2x + {\pi \over 3}} \right)$

But $y = A\sin \left( {\omega t - kx + \phi } \right)$

On comparing we get $\omega = 600;\,k = 2$

$v = {\omega \over k} = {{600} \over 2} = 300\,m{s^{ - 1}}$
4

### AIEEE 2003

A metal wire of linear mass density of $9.8$ $g/m$ is stretched with a tension of $10$ $kg$-$wt$ between two rigid supports $1$ metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency $n.$ The frequency $n$ of the alternating source is
A
$50$ $Hz$
B
$100$ $Hz$
C
$200$ $Hz$
D
$25$ $Hz$

## Explanation

KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by

$n = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2 \times }}\sqrt {{{10 \times 9.8} \over {9.8 \times {{10}^{ - 3}}}}} = 50Hz$

As the string is vibrating in resonance to a.c of frequency $n,$ therefore both the frequencies are same.