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Q.

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 $ρ$ where it stays vertical. 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 (Figure)

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a

Mg

b

$M g - V ρ g$

c

$M g + {πR}^{2} h ρ g$

d

$ρ g (V + {πR}^{2} h)$

answer is D.

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Detailed Solution

From Archimedes' principle, F_b $-$ F_t = upthrust = weight of volume V of the liquid. Here F_b and F_t denote the force exerted by the liquid on the bottom and the top of the cylinder respectively. Upthrust = $ρ$ Vg and F_t = ( $π$ R² h) pg. Hence

F_b = $(π R^{2} h) ρ g + ρ V g = ρ g ({πR}^{2} h + V)$

Hence the correct choice is ( 4).

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