Equilibrium Distance of Piston in Sealed Cylinder

To find the distance of the piston from the top at equilibrium, we analyze the states of the trapped gas inside the cylinder and the forces acting on the piston.

1. Initial State of the Trapped Gas:

• The cylinder has a circular cross-section of radius $R$, so its cross-sectional area is:
  $A = \pi R^2$
• When the hole at the top is sealed, the piston is at a distance of $2L$ from the top.
• Since the hole was open to the atmosphere before being sealed, the initial pressure of the trapped air inside the cylinder above the piston is the atmospheric pressure $P_0$.
• Thus, the initial volume of the trapped air is:
  $V_1 = A(2L) = \pi R^2 (2L)$
  and its initial pressure is:
  $P_1 = P_0$

2. Final Equilibrium State:

• Let the piston move to a new equilibrium position at a distance $y$ from the top.
• The process occurs isothermally at constant temperature $T$. By Boyle's Law ($P_1 V_1 = P V$):
  $P_0 \cdot [A(2L)] = P \cdot [Ay]$
  $y = \left(\frac{P_0}{P}\right)(2L)$
  where $P$ is the final pressure of the gas trapped above the piston.

3. Force Balance at Equilibrium:

• The forces acting on the piston of mass $M$ are:

  • Downward force exerted by the trapped gas: $F_{\text{gas}} = P A$
  • Downward gravitational force: $W = Mg$
  • Upward force exerted by the atmospheric pressure from the open bottom: $F_{\text{atm}} = P_0 A$

• Setting the net vertical force to zero for equilibrium:
  $\sum F_y = 0 \implies P A + Mg = P_0 A$
  $P A = P_0 A - Mg$
  $P = \frac{P_0 A - Mg}{A} = \frac{P_0 \pi R^2 - Mg}{\pi R^2}$

4. Calculating the Distance $y$:
Substituting the value of $P$ into the relation for $y$:
$y = \frac{P_0}{\left(\frac{P_0 \pi R^2 - Mg}{\pi R^2}\right)} (2L) = \left(\frac{P_0 \pi R^2}{\pi R^2 P_0 - Mg}\right)(2L)$

Therefore, the correct option is (D).