The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium, the height $H$ of the water column in the cylinder satisfies :

Step 1: Define the initial and final states of the trapped gas.

Initially, the cylinder of length $L_0$ is filled with air at atmospheric pressure $P_0$. When the cylinder is sealed and dipped, water rises to a height $H$, leaving a trapped gas column of length $L_0 - H$.

Step 2: Apply Boyle's Law to the trapped gas.

Assuming the temperature remains constant, the pressure of the trapped gas $P_{\text{gas}}$ satisfies the isothermal relation.

$$P_0 L_0 = P_{\text{gas}}(L_0 - H)$$ $$P_{\text{gas}} = \frac{P_0 L_0}{L_0 - H}$$

Step 3: Determine the pressure at the water-gas interface.

The top of the cylinder is level with the free surface of the water in the tank, meaning the depth of the water-gas interface inside the cylinder below the free surface is $L_0 - H$.

$$P_{\text{gas}} = P_0 + \rho g(L_0 - H)$$

Step 4: Equate the pressure expressions to find the equilibrium relation.

Equate the two expressions for $P_{\text{gas}}$.

$$\frac{P_0 L_0}{L_0 - H} = P_0 + \rho g(L_0 - H)$$

Multiply both sides by $L_0 - H$.

$$P_0 L_0 = P_0(L_0 - H) + \rho g(L_0 - H)^2$$

Rearrange the terms into a quadratic form.

$$\rho g(L_0 - H)^2 + P_0(L_0 - H) - L_0 P_0 = 0$$

Answer: $\rho g(L_0 - H)^2 + P_0(L_0 - H) - L_0 P_0 = 0$