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(B,D,C) $1.$ Since the cylinder is thermally conducting and the process is slow,it is isothermal. Initial state: $P_1 = P_0$,$V_1 = \pi R^2 L$. Final

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$A, D, C$

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(B,D,C) $1.$ Since the cylinder is thermally conducting and the process is slow,it is isothermal. Initial state: $P_1 = P_0$,$V_1 = \pi R^2 L$. Final state: $V_2 = \pi R^2 (2L)$. By Boyle's Law,$P_1 V_1 = P_2 V_2 \implies P_0 (\pi R^2 L) = P_2 (\pi R^2 2L) \implies P_2 = \frac{P_0}{2}$. Correct option is $(B)$.$2.$ Let the new equilibrium distance be $x$. The pressure inside is $P$. Equilibrium condition: $P_0 \pi R^2 = P \pi R^2 + Mg \implies P = P_0 - \frac{Mg}{\pi R^2}$. Using isothermal process from the state at $2L$: $P_{initial} V_{initial} = P_{final} V_{final} \implies P_0 (\pi R^2 2L) = (P_0 - \frac{Mg}{\pi R^2}) (\pi R^2 x) \implies x = \frac{P_0 (2L)}{P_0 - \frac{Mg}{\pi R^2}} = \left(\frac{P_0 \pi R^2}{\pi R^2 P_0 - Mg}\right)(2L)$. Correct option is $(D)$.$3.$ Initial state of air: $P_0, V_0 = \pi R^2 L_0$. Final state: $P, V = \pi R^2 (L_0 - H)$. By Boyle's Law: $P_0 (\pi R^2 L_0) = P (\pi R^2 (L_0 - H)) \implies P = \frac{P_0 L_0}{L_0 - H}$. From hydrostatic equilibrium: $P + \rho g (L_0 - H) = P_0 \implies \frac{P_0 L_0}{L_0 - H} + \rho g (L_0 - H) = P_0$. Multiplying by $(L_0 - H)$: $P_0 L_0 + \rho g (L_0 - H)^2 = P_0 (L_0 - H) \implies \rho g (L_0 - H)^2 - P_0 (L_0 - H) + P_0 L_0 = 0$. Correct option is $(C)$.

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