SATHEE: Heat and Thermodynamics 6 Question 4

Heat and Thermodynamics 6 Question 4

8. 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) $\frac{1}{2 π} \frac{A_{γ} p_{0}}{V_{0} M}$

(b) $\frac{1}{2 π} \frac{V_{0} M p_{0}}{A^{2} γ}$

(d) $\frac{1}{2 π} \sqrt{\frac{M V_{0}}{A_{γ} p_{0}}}$

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Answer:

Correct Answer: 8. (c)

Solution:

In equilibrium,

$p_{0} A = M g$

when slightly displaced downwards,

$d p = - γ (\frac{p_{0}}{V_{0}}) d V ($ As in adiabatic process, $\frac{d p}{d V} = - γ \frac{p}{V})$

$∴$ Restoring force,

$\begin{aligned} F & = (d p) A = - (\frac{γ p_{0}}{V_{0}}) (A) (A x) \\ F & \propto - x \end{aligned}$

Therefore, motion is simple harmonic comparing with

$\begin{aligned} F & = - k x we have \\ k & = \frac{γ p_{0} A^{2}}{V_{0}} \\ ∴ f & = \frac{1}{2 π} \sqrt{\frac{k}{m}} = \frac{1}{2 π} \sqrt{\frac{γ p_{0} A^{2}}{M V_{0}}} \end{aligned}$

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Heat and Thermodynamics 6 Question 4

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Heat and Thermodynamics 6 Question 4

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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