A hemispherical portion of radius -R is removed from the bottom of a cylinder of radius R. The volume of 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

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$ρg (V + {πR}^{2} h)$

M g

$Mg - Vρg$

$Mg + {πR}^{2} hρg$

answer is D.

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Detailed Solution

The net upward force on the bottom of the cylinder = weight of the liquid displaced by cylinder + thrust on the upper surface of cyliner due to i column of
liquid
$= Vρg + hρg \times {πR}^{2} = ρg [V + {πR}^{2} h]$

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