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 the universal constant and $g$ is the acceleration due to gravity)

\frac{R T}{n g} [\frac{l_{1} - 3 l_{2}}{l_{1} l_{2}}]

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\frac{R T}{g} [\frac{2 l_{1} + l_{2}}{l_{1} l_{2}}]

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\frac{n R T}{g} [\frac{l_{1} - l_{2}}{l_{1} l_{2}}]

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\frac{n R T}{g} [\frac{1}{l_{2}} + \frac{1}{l_{1}}]

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Solution

The correct option is C $\frac{n R T}{g} [\frac{l_{1} - l_{2}}{l_{1} l_{2}}]$

As the piston is stationary, so from FBD of the piston,

$P_{2} A = P_{1} A + m g$

Using, $V = A l$ and ideal gas equation

$\frac{n R T \cdot A}{A l_{2}} = \frac{n R T \cdot A}{A l_{1}} + m g$

$\Rightarrow n R T (\frac{1}{l_{2}} - \frac{1}{l_{1}}) = m g$

$∴ m = \frac{n R T}{g} (\frac{l_{1} - l_{2}}{l_{1} \cdot l_{2}})$

Hence, option $(D)$ is correct.

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The correct option is C nRTg[l1−l2l1l2] As the piston is stationary, so from FBD of the piston, P2A=P1A+mg Using, V=Al and ideal gas equation nRT⋅AAl2=nRT⋅AAl1+mg ⇒nRT(1l2−1l1)=mg ∴m=nRTg(l1−l2l1⋅l2) Hence, option (D) is correct.