A vertical closed cylinder is separated into two parts by a frictionless piston of mass $m$ and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is $l_1$, and that below the piston is $l_2$, such that $l_1 > l_2$. Each part of the cylinder contains $n$ moles of an ideal gas at equal temperature $T$. If the piston is stationary, its mass $m$ will be given by: ($R$ is universal gas constant and $g$ is the acceleration due to gravity)
Answer: (B) $\dfrac{nRT}{g}\left[\dfrac{l_1 - l_2}{l_1l_2}\right]$
With cross-section $A$: $P_1 = \dfrac{nRT}{Al_1}$ (above), $P_2 = \dfrac{nRT}{Al_2}$ (below).
Equilibrium of the piston: $P_2A = P_1A + mg$
$$mg = nRT\left(\frac1{l_2} - \frac1{l_1}\right) \Rightarrow m = \frac{nRT}{g}\left[\frac{l_1 - l_2}{l_1l_2}\right]$$
Solution by Sreeraj P, M.Sc Physics