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 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 [2001]

Question

Smart Achievers · Accepted Answer

(4)

$From Archimedes principle$

$Upthrust = wt. of fluid displaced$

$Now, F_{bottom} - F_{top} = V ρ g$

$\Rightarrow F_{bottom} = F_{top} + V ρ g$

$= P_{1} \times A + V ρ g$

$= (h ρ g) \times (π R^{2}) + V ρ g$

$= ρ g [π R^{2} h + V]$

Solution

1 $M g$

2 $M g - V ρ g$

3 $M g + π R^{2} h ρ g$

4 $ρ g (V + π R^{2} h)$

Ans.
(4)

$From Archimedes principle$

$Upthrust = wt. of fluid displaced$

$Now, F_{bottom} - F_{top} = V ρ g$

$\Rightarrow F_{bottom} = F_{top} + V ρ g$

$= P_{1} \times A + V ρ g$

$= (h ρ g) \times (π R^{2}) + V ρ g$

$= ρ g [π R^{2} h + V]$

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Solution

1 Mg

2 Mg-Vρg

3 Mg+πR2hρg

4 ρg(V+πR2h)