### JEE Mains Previous Years Questions with Solutions

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### JEE Main 2013 (Offline)

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M.$ The piston and the cylinder have equal cross sectional area $A$. When the piston is in equilibrium, the volume of the gas is ${V_0}$ and its pressure is ${P_0}.$ The piston is slightly displaced from the equilibrium position and released,. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frquency
A
${1 \over {2\pi }}\,{{A\gamma {P_0}} \over {{V_0}M}}$
B
${1 \over {2\pi }}\,{{{V_0}M{P_0}} \over {{A^2}\gamma }}$
C
${1 \over {2\pi }}\,\sqrt {{{A\gamma {P_0}} \over {{V_0}M}}}$
D
${1 \over {2\pi }}\,\sqrt {{{M{V_0}} \over {A\gamma {P_0}}}}$

## Explanation

${{Mg} \over A} = {P_0}$

$Mg = {P_0}A\,\,\,\,...\left( 1 \right)$

${P_0}V_0^\gamma = P{V^\gamma }$

$P = {{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^y}}}$

Let piston is displaced by distance $x$

$Mg - \left( {{{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right)A = {F_{restoring}}$

${P_0}A\left( {1 - {{x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right) = {F_{restoring}}$

$\left[ {{x_0} - x \approx {x_0}} \right]$

$F = - {{\gamma {P_0}Ax} \over {{x_0}}}$

$\therefore$ Frequency with which piston executes $SHM.$

$f = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}A} \over {{x_0}M}}} = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}{A^2}} \over {M{V_0}}}}$
2

### AIEEE 2012

If a simple pendulum has significant amplitude (up to a factor of $1/e$ of original ) only in the period between $t = 0s\,\,to\,\,t = \tau \,s,$ then $\tau \,$ may be called the average life of the pendulum When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $b$ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds :
A
${{0.693} \over b}$
B
$b$
C
${1 \over b}$
D
${2 \over b}$

## Explanation

The equation of motion for the pendulum, suffering retardation

$I\alpha = - mg\left( {\ell \sin \theta } \right) - mbv\left( \ell \right)$ where $I = m{\ell ^2}$

and $\alpha = {d^2}\theta /d{t^2}$

$\therefore$ ${{{d^2}\theta } \over {d{t^2}}} = - {g \over \ell }\tan \theta + {{bv} \over \ell }$

on solving we get $\theta = {\theta _0}\,{e^{{{bt} \over 2}\sin \left( {\omega t + \phi } \right)}}$

According to questions ${{{\theta _0}} \over e} = {\theta _0}{e^{{{ - b\tau } \over 2}}}$

$\therefore$ $\tau = {2 \over b}$
3

### AIEEE 2011

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance ${X_0}\left( {{X_0} > A} \right)$. If the maximum separation between them is $\left( {{X_0} + A} \right),$ the phase difference between their motion is:
A
${\pi \over 3}$
B
${\pi \over 4}$
C
${\pi \over 6}$
D
${\pi \over 2}$

## Explanation

For ${X_0} + A$ to be the maximum separation $y$ one body is at the mean position, the other should be at the extreme.
4

### AIEEE 2011

A mass $M,$ attached to a horizontal spring, executes $S.H.M.$ with amplitude ${A_1}.$ When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude ${A_2}.$ The ratio of $\left( {{{{A_1}} \over {{A_2}}}} \right)$ is :
A
${{M + m} \over M}$
B
${\left( {{M \over {M + m}}} \right)^{{1 \over 2}}}$
C
${\left( {{{M + m} \over M}} \right)^{{1 \over 2}}}$
D
${M \over {M + m}}$

## Explanation

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.

$\therefore$ $M{v_1} = \left( {M + m} \right){v_2}$

$M{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}}$

$\therefore$ $\left( {V = A\sqrt {{k \over M}} } \right)$

${A_1}\sqrt M = {A_2}\sqrt {M + m}$
$\therefore$ ${{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}}$