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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
- (2/3)*pi*R²*h*rho*g
- (1/3)*pi*R²*h*rho*g
- pi*R²*h*rho*g
- 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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