Step-by-step Solution:

1. Analyze the initial state before sealing: The problem states that the cylinder has a small hole at the top. While the piston is held at a distance of $2L$ from the top, the hole is open to the atmosphere. This means the air trapped in the volume above the piston is at atmospheric pressure, $P_0$.

   • Initial pressure of the trapped gas, $P_i = P_0$.
   • The volume of the trapped gas is the volume of the cylinder above the piston. Let the cross-sectional area of the cylinder be $A = \pi R^2$. The initial volume is $V_i = A \times (2L) = \pi R^2 (2L)$.

2. Analyze the final state at equilibrium: After the hole at the top is sealed, the piston is released. It moves to a new position and stays in equilibrium. Let the final distance of the piston from the top be $x$. The piston is in mechanical equilibrium, which means the net force on it is zero.

   Let's consider the forces acting on the piston in the vertical direction:

   • Force due to the trapped gas above the piston: This force acts downwards. If the final pressure of the trapped gas is $P_f$, this force is $F_{gas} = P_f A = P_f (\pi R^2)$.
   • Weight of the piston: This force acts downwards. $W = Mg$.
   • Force due to the atmospheric pressure from below the piston: The cylinder is open at its bottom, so the pressure below the piston is the atmospheric pressure $P_0$. This force acts upwards. $F_{atm} = P_0 A = P_0 (\pi R^2)$.

   For equilibrium, the total upward force must equal the total downward force: $F_{atm} = F_{gas} + W$ $P_0 A = P_f A + Mg$ We can solve for the final pressure $P_f$ of the trapped gas: $P_f A = P_0 A - Mg$ $P_f = P_0 - {{Mg} \over A} = P_0 - {{Mg} \over {\pi R^2}}$

   The final volume of the trapped gas is $V_f = A \times x = (\pi R^2) x$.

3. Apply the ideal gas law for the process: The cylinder is described as "thermally conducting." This implies that any temperature change due to compression or expansion of the gas is quickly equalized with the surroundings. Therefore, the process is isothermal, meaning the temperature $T$ of the trapped gas remains constant.

   For an isothermal process, Boyle's Law applies: $P_i V_i = P_f V_f$

4. Solve for the final distance x: Substitute the expressions for the initial and final pressures and volumes into Boyle's Law: $P_0 \times (A \cdot 2L) = \left( P_0 - {{Mg} \over A} \right) \times (A \cdot x)$ The area $A$ cancels out from both sides of the inner terms: $P_0 (2L) = \left( P_0 - {{Mg} \over A} \right) x$ Now, solve for $x$: $x = {{P_0 (2L)} \over {P_0 - {{Mg} \over A}}}$ Substitute $A = \pi R^2$ back into the expression: $x = {{P_0 (2L)} \over {P_0 - {{Mg} \over {\pi R^2}}}}$ To simplify and match the options, multiply the numerator and denominator by $\pi R^2$: $x = {{P_0 (2L) (\pi R^2)} \over {\left( P_0 - {{Mg} \over {\pi R^2}} \right) (\pi R^2)}}$ $x = {{P_0 \pi R^2 (2L)} \over {P_0 \pi R^2 - Mg}}$ Rearranging this expression to match the format of the options: $x = \left( {{{P_0 \pi R^2} \over {P_0 \pi R^2 - Mg}}} \right) (2L)$

5. Compare with the given options: The derived expression for $x$ matches option D.

   A: $\left( {{{2{P_0}\pi {R^2}} \over {\pi {R^2}{P_0} + Mg}}} \right)(2L)$ - Incorrect. B: $\left( {{{{P_0}\pi {R^2} - Mg} \over {\pi {R^2}{P_0}}}} \right)(2L)$ - Incorrect. C: $\left( {{{{P_0}\pi {R^2} + Mg} \over {\pi {R^2}{P_0}}}} \right)(2L)$ - Incorrect. D: $\left( {{{{P_0}\pi {R^2}} \over {\pi {R^2}{P_0} - Mg}}} \right)(2L)$ - Correct.