Heat And Thermodynamics Ques 217
- 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 frequency
(2013 Main)
(a) $\frac{1}{2 \pi} \frac{A _{\gamma} p _0}{V _0 M}$
(b) $\frac{1}{2 \pi} \frac{V _0 M p _0}{A^{2} \gamma}$
(c) $\frac{1}{2 \pi} \sqrt{\frac{A^{2} \gamma p _0}{M V _0}}$
(d) $\frac{1}{2 \pi} \sqrt{\frac{M V _0}{A _{\gamma} p _0}}$
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Answer:
Correct Answer: 217.(c)
Solution:
Formula:
- In equilibrium,
$$ p _0 A=M g $$
when slightly displaced downwards,
$d p=-\gamma\left(\frac{p _0}{V _0}\right) d V\left(\right.$ As in adiabatic process, $\left.\frac{d p}{d V}=-\gamma \frac{p}{V}\right)$
$\therefore$ Restoring force,
$$ \begin{aligned} F & =(d p) A=-\left(\frac{\gamma p _0}{V _0}\right)(A)(A x) \\ F & \propto-x \end{aligned} $$
Therefore, motion is simple harmonic comparing with
$$ \begin{aligned} F & =-k x \text { we have } \\ k & =\frac{\gamma p _0 A^{2}}{V _0} \\ \therefore \quad f & =\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{\gamma p _0 A^{2}}{M V _0}} \end{aligned} $$
BETA
