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A uniformly tapering vessel placed in vacuum holds a liquid of density 900 kg/m³. Taking g = 10 m/s², the force exerted on the base of the vessel by the liquid is best given by which value? (Standard data: base area 0.0004 m² and liquid depth 0.2 m as in the original figure.)
- 0.72 N
- 7.2 N
- 9.0 N
- 14.4 N
Correct answer: 0.72 N
Solution
The force on the base equals the hydrostatic pressure at the base times the base area. Pressure P = rho*g*h. With rho = 900 kg/m³, g = 10 m/s², h = 0.2 m, P = 900*10*0.2 = 1800 Pa. With base area A = 0.0004 m², F = P*A = 1800*0.0004 = 0.72 N. (Note: this is the force on the base, generally different from the weight of liquid in a tapering vessel due to the hydrostatic paradox.)
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