Question

A cylindrical glass tumbler is lowered upside-down into water and floats vertically, with the internal surface of its bottom level with the surface of the water in the vessel. The weight of the glass is 408 g; the area of its bottom is 10cm². The air pressure in the glass before submersion was 76cm of mercury. What portion of the tumbler will the air occupy after submersion?

Answer 1

Answer: 0.96 of the initial volume.

Explanation:

The air in the tumbler after submersion will be under a pressure of P + Q/S d₀, where P is the air pressure before submersion, Q the volume of the tumbler, S the area of its bottom, and d₀ the density of water.