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A heavy hollow cone of base radius R and height h sits apex-up on a table with its flat circular base down. It is completely filled with water of density rho. The base rim is sealed watertight to the table and there is a tiny hole at the apex. Ignoring atmospheric pressure, the total upward force exerted by the water on the cone is

1. (2/3)*pi*R²*h*rho*g
2. (1/3)*pi*R²*h*rho*g
3. pi*R²*h*rho*g
4. None

Correct answer: (2/3)*pi*R²*h*rho*g

Solution

The water pushes down on the table over the full base area with pressure rho*g*h, giving a downward force pi*R²*rho*g*h. The actual weight of water in the cone is only (1/3)*pi*R²*h*rho*g. The difference is supplied by the upward force the slanted cone walls exert on the water; by Newton's third law the water exerts an equal and opposite (upward) force on the cone of magnitude (2/3)*pi*R²*h*rho*g.

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