Question

A fixed thermally conducting cylinder has a radius R and height $L_0$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown. The atmospheric pressure is $P_0$.

While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

• A. $=\left(\frac{2P_0\pi R^2}{\pi R^2{P}_0+Mg}\right)\left(2L\right)$
• B. $=\left(\frac{P_0\pi R^2-Mg}{\pi R^2{P}_0}\right)\left(2L\right)$
• C. $=\left(\frac{P_0\pi R^2+Mg}{\pi R^2{P}_0}\right)\left(2L\right)$
• D. $=\left(\frac{P_0\pi R^2}{\pi R^2{P}_0-Mg}\right)\left(2L\right)$

Answer 1

The correct option is D

$=\left(\frac{P_0\pi R^2}{\pi R^2{P}_0-Mg}\right)\left(2L\right)$

Net P = Pressure in equilibrium,

Then $PA=P_{0}A-Mg$

$P=P_0-\frac{Mg}{A}=P_0-\frac{Mg}{\pi R^2}$

Applying, $P_1{V}_1=P_2{V}_2$

$\therefore P_0\left(2AL\right)=\left(P\right)\left(A{L}^{\prime }\right)$

$\therefore L^{\prime }=\frac{2P_{0}L}{P}=\left(\frac{P_0}{P_0-\frac{Mg}{\pi R^2}}\right)\left(2L\right)$

$=\left(\frac{P_0\pi R^2}{\pi R^2{P}_0-Mg}\right)\left(2L\right)$