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Grade 11Thermal Physics

Plz answer the attached question.............................

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nagashravan hemadri

8 Years agoGrade 11

1 Answer

Eshan

8 Years ago

Under equilibrium, let the pressure be $P_0$ .
$\implies P_0V_0=Mg$
Let the piston be displaced downwards by $x$ .
The pressure would increase as

$P_0V_0^{\gamma}=P'(V_0-Ax)^{\gamma}$
$\implies P'=P_0(\dfrac{V_0}{V_0-Ax})^{\gamma}$
Since $Ax<<V_0$ ,

$V_0-Ax=V_0(1-\dfrac{Ax}{V_0})$
$\implies (\dfrac{1}{V_0-Ax})^{\gamma}=\dfrac{1}{V_0^{\gamma}}(1-\dfrac{Ax}{V_0})^{-\gamma}\approx \dfrac{1}{V_0^{\gamma}}(1+\dfrac{\gamma Ax}{V_0})$

Hence $P'=P_0(1+\dfrac{\gamma Ax}{V_0})$

Hence restoring force acting upwards= $P'A-Mg=P_0(1+\dfrac{\gamma Ax}{V_0})A-Mg=\dfrac{P_0\gamma A^2}{V_0}x=M\omega^2 x$

$\implies \omega=\sqrt{\dfrac{P_0\gamma A^2}{MV_0}}=2\pi f$
$\implies f=\dfrac{1}{2\pi}\sqrt{\dfrac{P_0\gamma A^2}{MV_0}}$

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