 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

JEE Main 2013 (Offline)

The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5s$. In another $10s$ it will decrease to $\alpha$ times its original magnitude, where $\alpha$ equals
A
$0.7$
B
$0.81$
C
$0.729$
D
$0.6$

Explanation

as $\,\,A = {A_0}{e^{{{bt} \over {2m}}}}$ (where, ${A_0} =$ maximum amplitude)

According to the questions, after $5$ second,

$0.9{A_0} = {A_0}{e^{ - {{b\left( 5 \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

After $10$ more second,

$A = {A_0}{e^{ - {{b\left( {15} \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

From eqns $(i)$ and $(ii)$

$A = 0.729\,{A_0}$

$\therefore$ $\alpha = 0.729$
2

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 $SHM.$

$f = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}A} \over {{x_0}M}}} = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}{A^2}} \over {M{V_0}}}}$
3

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

$\therefore$ $\tau = {2 \over b}$
4

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.