Solution 15
We are asked to find the air pressure when the smaller weightless box touches the ceiling of the larger box.
Analysis: The problem states that the smaller box is “weightless.” In a typical hydrostatic setup involving buoyancy, a submerged object rises if the buoyant force exceeds its weight. Since the weight is zero, any volume of displaced fluid would cause it to rise.
However, note the configuration of the holes. Hole A is in the smaller box, and Hole B is in the larger box. Since the process is slow (quasi-static) and involves open communication between the water reservoirs via the holes, the pressure distribution is governed by hydrostatics.
Since the smaller box is weightless, the only way for it to remain in a stable specific position (like touching the ceiling) without simply being pinned there by maximum buoyancy is if there is a pressure differential. However, because of the hole A, the interior of the small box communicates with the surrounding fluid. The air pressure inside the small box and the air pressure in the space of the large box above the water are interacting with the same water level systems.
Specifically, if the box is weightless, the condition of it floating at a specific level or pressing against the ceiling usually implies a balance of forces. But with the box being massless and open to the fluid, the air pressure inside it tends to equalize with the surroundings relative to the water level.