 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

A cylindrical tube, open at both ends, has a fundamental frequency, $f,$ in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now :
A
$f$
B
$f/2$
C
$3/4$
D
$2f$

Explanation

The fundamental frequency of open tube

${v_0} = {v \over {2{l_0}}}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

That of closed pipe

${v_c} = {\upsilon \over {4{l_c}}}\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$
According to the problem ${l_c} = {{{l_0}} \over 2}$

Thus ${v_c} = {\upsilon \over {{l_0}/2}} \Rightarrow {v_c}{\upsilon \over {2l}}\,\,\,\,...\left( {iii} \right)$

From equations $(i)$ and $(iii)$

${v_0} = {v_c}$

Thus, ${v_c} = f$ $\,\,\,\left( {\,\,} \right.$ as ${v_0} = f$ is given $\left. {\,\,} \right)$
2

AIEEE 2011

The transverse displacement $y(x, t)$ of a wave on a string is given by $y\left( {x,t} \right) = {e^{ - \left( {a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)}}.$ This represents $a:$
A
wave moving in $-x$ direction with speed $\sqrt {{b \over a}}$
B
standing wave of frequency $\sqrt b$
C
standing wave of frequency ${1 \over {\sqrt b }}$
D
wave moving in $+x$ direction speed $\sqrt {{a \over b}}$

Explanation

Given wave equation is

$y\left( {x,t} \right){ = _e}\left( { - a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)$

$= {e^{ - \left[ {{{\left( {\sqrt {ax} } \right)}^2} + {{\left( {\sqrt {bt} } \right)}^2} + 2\sqrt a x.\sqrt b t} \right]}}$

$= {e^{ - {{\left( {\sqrt a x + \sqrt b t} \right)}^2}}}$

$= {e^{ - {{\left( {x + \sqrt {{b \over a}} t} \right)}^2}}}$

It is a function of type $y = f\left( {x + vt} \right)$

$\Rightarrow$ Speed of wave $= \sqrt {{b \over a}}$
3

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

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

$\therefore$ $T = {v^2} \times \mu = {\left( {12.5} \right)^2} \times 0.04 = 6.25\,N$
4

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

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