 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2013 (Offline)

A sonometer wire of length $1.5$ $m$ is made of steel. The tension in it produces an elastic strain of $1\%$. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times {10^3}\,kg/{m^3}$ and $2.2 \times {10^{11}}\,N/{m^2}$ respectively ?
A
$188.5$ $Hz$
B
$178.2$ $Hz$
C
$200.5$ $Hz$
D
$770$ $Hz$

Explanation

Fundamental frequency,

$f = {v \over {2\ell }} = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2\ell }}\sqrt {{T \over {A\rho }}}$

$\left[ {\,\,} \right.$ as $v = \sqrt {{T \over \mu }}$ $\,\,\,\,\,\,$ and $\,\,\,\,\,\,$ $\left. {\mu = {m \over \ell }\,\,} \right]$

Also, $Y = {{T\ell } \over {A\Delta \ell }} \Rightarrow {T \over A} = {{Y\Delta \ell } \over \ell }$

$\Rightarrow f = {1 \over {2\ell }}\sqrt {{{\gamma \Delta \ell } \over {\ell \rho }}} ....\left( i \right)$

Putting the value of $\ell ,{{\Delta \ell } \over \ell },\rho$ $\,\,\,\,\,\,$ and

$\,\,\,\,\,\,$ $\gamma$ in $e{q^n}.\left( i \right)$ we get,

$f = \sqrt {{2 \over 7}} \times {{{{10}^3}} \over 3}$ $\,\,\,\,\,\,$ or, $\,\,\,\,\,\,$ $f \approx 178.2\,Hz$
2

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 $\left. {\,\,} \right)$
3

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 $= \sqrt {{b \over a}}$
4

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

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

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