### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :
A
$2f$
B
$f$
C
${f \over 2}$
D
${3f \over 4}$

## Explanation

The fundamental frequency in case $(a)$ is $f = {v \over {2\ell }}$

The fundamental frequency in case $(b)$ is

$f'{v \over {4\left( {\ell /2} \right)}} = {u \over {2\ell }} = f$
2

### JEE Main 2016 (Offline)

A uniform string of length $20$ $m$ is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the supports is :
(take ${\,\,g = 10m{s^{ - 2}}}$ )
A
$2\sqrt 2 s$
B
$2\pi \sqrt 2 s$
C
$2\pi \sqrt 2 s$
D
$2$ $s$

## Explanation

We know that velocity in string is given by

$v = \sqrt {{T \over \mu }} \,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$

where $\mu = {m \over {\rm I}} = {{mass\,\,\,of\,\,\,string} \over {length\,\,\,of\,\,\,string}}$

The tension $T = {m \over \ell } \times x \times g\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

From $(a)$ and $(b)$ ${{dx} \over {dt}} = \sqrt {gx}$

${x^{ - 1/2}}\,dx = \sqrt g \,dt$

$\therefore$ $\int\limits_0^\ell {{x^{ - 1/2}}} dx - \sqrt g \int\limits_0^\ell {dt}$

$2\sqrt \ell = \sqrt g \times t$

$\therefore$ $t = 2\sqrt {{\ell \over g}} = 2\sqrt {{{20} \over {10}}} = 2\sqrt 2$
3

### 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

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

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

$\left( {{{{f_2}} \over {{f_1}}} - 1} \right) \times 100 = \left( {{{300} \over {340}} - 1} \right) \times 100 = 12\%$
4

### 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.