Verified Solution • 2025–26 Session moderate

QJEEM13Phf6ad: JEE Main 2013 Physics - Class 11 JEEM13Phf6ad Solution: An ideal gas enclosed in a vertical cylindrical container…

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M.M. The piston and the cylinder have equal cross sectional area AA . When the piston is in equilibrium, the volume of the gas is V0{V_0} and its pressure is P0.{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

Multiple Choice Options

  • A 12π AγP0V0M{1 \over {2\pi }}\,{{A\gamma {P_0}} \over {{V_0}M}}
  • B 12π V0MP0A2γ{1 \over {2\pi }}\,{{{V_0}M{P_0}} \over {{A^2}\gamma }}
  • C 12π AγP0V0M{1 \over {2\pi }}\,\sqrt {{{A\gamma {P_0}} \over {{V_0}M}}}
  • D 12π MV0AγP0{1 \over {2\pi }}\,\sqrt {{{M{V_0}} \over {A\gamma {P_0}}}}

Step-by-Step Verified Solution

MgA=P0{{Mg} \over A} = {P_0}

Mg=P0A    ...(1)Mg = {P_0}A\,\,\,\,...\left( 1 \right)

P0V0γ=PVγ{P_0}V_0^\gamma = P{V^\gamma }

P=P0x0γ(x0−x)yP = {{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^y}}}

Let piston is displaced by distance xx

Mg−(P0x0γ(x0−x)γ)A=FrestoringMg - \left( {{{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right)A = {F_{restoring}}

JEE Main 2013 (Offline) Physics - Simple Harmonic Motion Question 133 English Explanation

P0A(1−x0γ(x0−x)γ)=Frestoring{P_0}A\left( {1 - {{x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right) = {F_{restoring}}

[x0−x≈x0]\left[ {{x_0} - x \approx {x_0}} \right]

F=−γP0Axx0F = - {{\gamma {P_0}Ax} \over {{x_0}}}

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

f=12πγP0Ax0M=12πγP0A2MV0f = {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}}}}

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