Two identical containers X and Y are connected at the bottom by a thin tube of negligible volume. The tube has a valve in it, as shown in the figure. Initially container X has a liquid filled up to height h in it and container Y is empty. When the valve is opened, both containers have equal amount of liquid in equilibrium. If the initial (before the valve is opened) potential energy of the liquid is P1 and the final potential energy is P2 then :

The potential energy (PE) of a liquid column is determined by the position of its center of mass. For a uniform liquid column of height h, the center of mass lies at h/2.

• Initial State (P1): All liquid is in container X. If the total mass is M, then P1 = M × g × (h/2).
• Final State (P2): When the valve is opened, the liquid redistributes until the heights are equal. Since the containers are identical, the liquid height in each becomes h/2. The total mass M remains the same, but the new center of mass for the liquid in both containers is now at (h/2) / 2 = h/4.
• Comparison: Thus, P2 = M × g × (h/4).

By comparing the two expressions, we find that P1 = 2P2. The reduction in potential energy occurs because the liquid's average height decreases, with the energy difference typically dissipated as heat due to fluid friction.