### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

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 $\omega = 600;\,k = 2$

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

### 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.
3

### AIEEE 2002

When temperature increases, the frequency of a tuning fork
A
increases
B
decreases
C
remains same
D
increases or decreases depending on the material

## Explanation

KEY CONCEPT : The frequency of a tuning fork is given by the expression

$f = {{{m^2}k} \over {4\sqrt 3 \pi {\ell ^2}}}\sqrt {{Y \over \rho }}$

As temperature increases, $\ell$ increases and therefore $f$ decreases.
4

### AIEEE 2002

A wave $y=a$ $\sin \left( {\omega t - kx} \right)$ on a string meets with another wave producing a node at $x=0.$ Then the equation of the unknown wave is
A
$y = \alpha \,\sin \,\left( {\omega t + kx} \right)$
B
$y = - \alpha \,\sin \,\left( {\omega t + kx} \right)$
C
$y = \alpha \,\sin \,\left( {\omega t - kx} \right)$
D
$y = - \alpha \,\sin \,\left( {\omega t - kx} \right)$

## Explanation

To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $\pi$. The equation of the reflected wave will be

$y = a\sin \left( {\omega t + kx + \pi } \right)$

$\Rightarrow y = - a\sin \left( {\omega t + kx} \right)$