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(C) Let the total length of the cylinder be $L = 84 	ext{ cm}$. Initially,the piston is at the center,so each part has a length $l = 42 	ext{ cm}$.L

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$2$

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(C) Let the total length of the cylinder be $L = 84 	ext{ cm}$. Initially,the piston is at the center,so each part has a length $l = 42 	ext{ cm}$.Let the final pressure in both parts be $P$ (since the piston is in equilibrium) and the displacement of the piston be $x$.The final lengths of the two parts are $(42 - x)$ and $(42 + x)$.For the first part (temperature remains $T_1 = 300 	ext{ K}$):$P_0 V_0 = P V_1 \Rightarrow 1 	imes 42 = P(42 - x) \Rightarrow P(42 - x) = 42 \quad ...(i)$For the second part (temperature increases to $T_2 = 57 + 273 = 330 	ext{ K}$):$\frac{P_0 V_0}{T_1} = \frac{P V_2}{T_2} \Rightarrow \frac{1 	imes 42}{300} = \frac{P(42 + x)}{330} \Rightarrow P(42 + x) = 42 	imes \frac{330}{300} = 46.2 \quad ...(ii)$Dividing $(ii)$ by $(i)$:$\frac{42 + x}{42 - x} = \frac{46.2}{42} = 1.1$$42 + x = 1.1(42 - x) = 46.2 - 1.1x$$2.1x = 4.2 \Rightarrow x = 2 	ext{ cm}$.

Column $I$	Column $II$
$(p)$ isobaric	$(x)$ $\frac{PV(1 - 2^{1 - \gamma})}{\gamma - 1}$
$(q)$ isothermal	$(y)$ $PV$
$(r)$ adiabatic	$(z)$ $PV \ln 2$

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