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

Correct answer: A. (1)/(2π) A ^ 2 γ P _ 0 MV _ 0 . For an adiabatic process, Pv ^ γ = constant Differentiating with respect to v , dP dv ( v ^ γ ) + Pγ v ^ γ - 1 = 0 dP = - γ P v dv = - γ P v Adx Thus, the acceleration, a = dF M = γ PA ^ 2 MV dx Comparing with , a = - ω ^ 2 dx , we get, ω = γ P _ 0 A ^ 2 M V _ 0 = 2 f = (1)/(2π) γ P _ 0 A ^ 2 M V _ 0

JEE Main 20137 April

Exam: JEE Main
Year: 2013
Date: 7 April
Subject: Physics
Chapter: Oscillations

JEE Main 2013 · 7 April · Physics · Oscillations

An ideal gas enclosed in a vertical cylindrical container supports...

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.]

Options

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

Correct answer

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

Step-by-step solution

For an adiabatic process,

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

Differentiating with respect to

v

\dfrac{\text{dP}}{\text{dv}}\left({\text{v}}^{\gamma}\right) + P\gamma{\text{v}}^{\gamma - 1} = 0

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

Thus, the acceleration,

\text{a} = \dfrac{\text{dF}}{\text{M}} = \dfrac{\gamma{\text{PA}}^{2}}{\text{MV}}\text{dx}

Comparing with ,

\text{a} = - {\omega}^{2}\text{dx}

, we get,

\omega = \sqrt{\dfrac{\gamma{P}_{0}{A}^{2}}{M{V}_{0}}} = 2\pif

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

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