### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2010

The equation of a wave on a string of linear mass density $0.04\,\,kg\,{m^{ - 1}}$ is given by $$y = 0.02\left( m \right)\,\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}} - {x \over {0.50\left( m \right)}}} \right)} \right].$$

The tension in the string is

A
$4.0N$
B
$12.5$ $N$
C
$0.5$ $N$
D
$6.25$ $N$

## Explanation

$y = 0.02\left( m \right)\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}}} \right) - {x \over {0.50\left( m \right)}}} \right]$

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

$\therefore$ $\omega = {{2\pi } \over {0.04}} \Rightarrow v = {1 \over {0.04}} = 25\,Hz$

$k = {{2\pi } \over {0.50}} \Rightarrow \lambda = 0.5m$

$\therefore$ velocity, $v = v\lambda = 25 \times 0.5\,m/s = 12.5\,m/s$

Velocity on a string is given by

$v = \sqrt {{T \over \mu }}$

$\therefore$ $T = {v^2} \times \mu = {\left( {12.5} \right)^2} \times 0.04 = 6.25\,N$
2

### AIEEE 2009

A motor cycle starts from rest and accelerates along a straight path at $2m/{s^2}.$ At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at $94\%$ of its value when the motor cycle was at rest? (Speed of sound $= 330\,m{s^{ - 1}}$)
A
$98$ $m$
B
$147$ $m$
C
$196\,m$
D
$49$ $m$

## Explanation

$v_m^2 - {u^2} = 2as \Rightarrow v_m^2 = 2 \times 2 \times s$

$\therefore$ ${v_m} = 2\sqrt s$

According to Doppler's effect

$0.94v = v\left[ {{{330 - 2\sqrt s } \over {330}}} \right] \Rightarrow s = 98.01\,m$
3

### AIEEE 2009

Three sound waves of equal amplitudes have frequencies $\left( {v - 1} \right),\,v,\,\left( {v + 1} \right).$ They superpose to give beats. The number of beats produced per second will be :
A
$3$
B
$2$
C
$1$
D
$4$

## Explanation

Maximum number of beats $= \left( {v + 1} \right) - \left( {v - 1} \right) = 2$
4

### AIEEE 2008

A wave travelling along the $x$-axis is described by the equation $y(x, t)=0.005$ $\cos \,\left( {\alpha \,x - \beta t} \right).$ If the wavelength and the time period of the wave are $0.08$ $m$ and $2.0s$, respectively, then $\alpha$ and $\beta$ in appropriate units are
A
$\alpha = 25.00\pi ,\,\beta = \pi$
B
$\alpha = {{0.08} \over \pi },\,\beta = {{2.0} \over \pi }$
C
$\alpha = {{0.04} \over \pi },\,\beta = {{1.0} \over \pi }$
D
$\alpha = 12.50\pi ,\,\beta = {\pi \over {2.0}}$

## Explanation

$y\left( {x,t} \right) = 0.005\,\cos \left( {\alpha x - \beta t} \right)$ (Given)

Comparing it with the standard equation of wave

$y\left( {x,t} \right) = a\cos \left( {kx - \omega t} \right)$ we get

$k = \alpha$ $\,\,\,\,\,$ and $\,\,\,\,\,$ $\omega = \beta$

$\therefore$ ${{2\pi } \over \gamma } = \alpha$ $\,\,\,\,\,$ and $\,\,\,\,\,$ ${{2\pi } \over T} = \beta$

$\therefore$ $\alpha = {{2\pi } \over {0.08}} = 25\pi$ $\,\,\,\,\,$ and $\,\,\,\,\,$ $\beta = {{2\pi } \over 2} = \pi$