Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

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

$${1 \over {2\pi }}\,\sqrt {{{M{V_0}} \over {A\gamma {P_0}}}} $$

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

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

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

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

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 $$SHM.$$

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

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

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

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

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

MCQ (Single Correct Answer)

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

$${2 \over b}$$

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

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

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

MCQ (Single Correct Answer)

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

$${\pi \over 2}$$

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

MCQ (Single Correct Answer)

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

$${M \over {M + m}}$$

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

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

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

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

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

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

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

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

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

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

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

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