 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2015 (Offline)

A train is moving on a straight track with speed $20\,m{s^{ - 1}}.$ It is blowing its whistle at the frequency of $1000$ $Hz$. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound $= 320\,m{s^{ - 1}}$) close to :
A
$18\%$
B
$24\%$
C
$6\%$
D
$12\%$

Explanation

${f_1} = f\left[ {{v \over {v - {v_s}}}} \right] = f \times {{320} \over {300}}Hz$

${f_2} = f\left[ {{v \over {v + {v_s}}}} \right] = f \times {{320} \over {340}}Hz$

$\left( {{{{f_2}} \over {{f_1}}} - 1} \right) \times 100 = \left( {{{300} \over {340}} - 1} \right) \times 100 = 12\%$
2

JEE Main 2014 (Offline)

A pipe of length $85$ $cm$ is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below $1250$ $Hz$. The velocity of sound in air is $340$ $m/s$.
A
$12$
B
$8$
C
$6$
D
$4$

Explanation

Length of pipe $=85$ $cm$ $=0.85m$

Pipe is closed from one end so it behaves as a closed organ pipe

Frequency of oscillations of air column in closed organ pipe is given by,

$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}}$

$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}} \le 1250$

$\Rightarrow {{\left( {2n - 1} \right) \times 340} \over {0.85 \times 4}} \le 1250$

$\Rightarrow 2n - 1 \le 12.5 \approx 6$

Possible value of n = 1, 2, 3, 4, 5, 6

So, number of possible natural frequencies lie below 1250 Hz is 6.
3

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, $\,\,\,\,\,\,$ $f \approx 178.2\,Hz$
4

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