JEE Main 2013 — Physics Waves & Oscillations

2013mcqhard

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $\text{M}$ . The piston and the cylinder have equal cross-sectional area $\text{A}$ . When the piston is in equilibrium, the volume of the gas is ${V}_{0}$ and its pressure is ${M}_{0}$ . 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 frequency

[Assume the system is in space.]

Official previous-year question

Held on 7 Apr 2013 · Verified 6 Jul 2026.

Options

A
$\frac{1}{2\pi }\sqrt{\frac{{\text{A}}^{2}\gamma {\text{P}}_{0}}{{\text{MV}}_{0}}}$ .
B
$\frac{1}{2\pi }\sqrt{\frac{{\text{MV}}_{0}}{\text{A}\gamma {\text{P}}_{0}}}$ .
C
$\frac{1}{2\pi }\frac{\text{A}\gamma {\text{P}}_{0}}{{\text{V}}_{0}\text{M}}$
D
$\frac{1}{2\pi }\frac{{\text{V}}_{0}{\text{MP}}_{0}}{{\text{A}}^{2}\gamma }$ .

Did you get this right?

Solution

For an adiabatic process,

${\text{Pv}}^{\gamma }=\text{constant}$

Differentiating with respect to $v$ ,

$\frac{\text{dP}}{\text{dv}}({\text{v}}^{\gamma })+P\gamma {\text{v}}^{\gamma -1}=0$

$\text{dP}=\frac{-\gamma P}{\text{v}}\text{dv}=\frac{-\gamma P}{\text{v}}\text{Adx}$

JEE Main 2013 Physics, Oscillations — solution figure

Thus, the acceleration, $\text{a}=\frac{\text{dF}}{\text{M}}=\frac{\gamma {\text{PA}}^{2}}{\text{MV}}\text{dx}$

Comparing with , $\text{a}=-{\omega }^{2}\text{dx}$ , we get,

$\omega =\sqrt{\frac{\gamma {P}_{0}{A}^{2}}{M{V}_{0}}}=2\pi f$

$f=\frac{1}{2\pi }\sqrt{\frac{\gamma {P}_{0}{A}^{2}}{M{V}_{0}}}$

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Options

$\frac{1}{2\pi }\sqrt{\frac{{\text{A}}^{2}\gamma {\text{P}}_{0}}{{\text{MV}}_{0}}}$ .

$\frac{1}{2\pi }\sqrt{\frac{{\text{MV}}_{0}}{\text{A}\gamma {\text{P}}_{0}}}$ .

$\frac{1}{2\pi }\frac{\text{A}\gamma {\text{P}}_{0}}{{\text{V}}_{0}\text{M}}$

$\frac{1}{2\pi }\frac{{\text{V}}_{0}{\text{MP}}_{0}}{{\text{A}}^{2}\gamma }$ .

Solution

For an adiabatic process,

${\text{Pv}}^{\gamma }=\text{constant}$

Differentiating with respect to $v$ ,

$\frac{\text{dP}}{\text{dv}}({\text{v}}^{\gamma })+P\gamma {\text{v}}^{\gamma -1}=0$

$\text{dP}=\frac{-\gamma P}{\text{v}}\text{dv}=\frac{-\gamma P}{\text{v}}\text{Adx}$

JEE Main 2013 Physics, Oscillations — solution figure

Thus, the acceleration, $\text{a}=\frac{\text{dF}}{\text{M}}=\frac{\gamma {\text{PA}}^{2}}{\text{MV}}\text{dx}$

Comparing with , $\text{a}=-{\omega }^{2}\text{dx}$ , we get,

$\omega =\sqrt{\frac{\gamma {P}_{0}{A}^{2}}{M{V}_{0}}}=2\pi f$

$f=\frac{1}{2\pi }\sqrt{\frac{\gamma {P}_{0}{A}^{2}}{M{V}_{0}}}$

JEE Main 2013 — Physics Waves & Oscillations

Options

Did you get this right?

Solution

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JEE Main 2013 — Physics Waves & Oscillations

Options

Did you get this right?

Solution

Your note

JEE Main Waves & Oscillations in other years

More JEE Main Waves & Oscillations questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Thermodynamics

Keep practicing Waves & Oscillations