0.5 mole of an ideal gas at constant temperature $27^{o} C$ kept inside a cylinder of length L. and cross-sectional area. A closed by a massless piston. The cylinder is attached with a conducting rod of length L cross-section area $(1 / 9) m^{2}$ and thermal conductivity k whose other end is maintained at $0^{o} C$ . If piston is moved such that rate of heat flow through the conducting rod is constant then the velocity of piston when it is at height L/2 from the bottom of cylinder is [Neglect any kind of heat loss from system]

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$\frac{k}{90 R}$

$\frac{k}{110 R}$

$\frac{k}{100 R}$

$\frac{k}{50 R}$

answer is B.

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Detailed Solution

$\frac{Δ Q}{Δ t} = \frac{Δ w}{Δ t}$ = work done per unit time $= \frac{k a θ}{L}$
Power = F x Velocity = $P A V' = \frac{n R T}{V} A V'$
where $V \to v o l u m e, V' \to v e l o c i t y$
$\begin{array}{l} \Rightarrow \frac{0.5 R (300)}{V} A V' = \frac{k a θ}{L} \\ \Rightarrow \frac{0.5 R (300)}{A . \frac{L}{2}} A V' = \frac{k a θ}{L} \\ \Rightarrow V' = \frac{k a}{R} (\frac{27}{300}) = \frac{k}{100 R} \end{array}$