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A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density ρ. It stays vertical inside the liquid. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is(a) Mg (b) Mg - Vρg(c) Mg + πR2 hρg(d) ρg(V + πR2 h) |
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Answer» Correct Answer is:(d) ρg(V + πR2 h) Net upward buoyancy force on the cylinder = weight of liquid displaced by it = ρgV = (upward force on the bottom) – (downward force on the top) ∴ the force on the bottom is ρgV + (hρg)πR2 = ρg(V + πR2 h). |
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