SATHEE: Properties of Matter 2 Question 5

Properties of Matter 2 Question 5

5. 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 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

$(2001, 2 M)$

(a) $M g$

(b) $M g - V ρ g$

(d) $ρ g (V + π R^{2} h)$

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Answer:

Correct Answer: 5. (a)

Solution:

$F_{2} - F_{1} =$ upthrust

$∴ F_{2} = F_{1} +$ upthrust

$F_{2} = (p_{0} + ρ g h) π R^{2} + V ρ g$

$= p_{0} π R^{2} + ρ g (π R^{2} h + V)$

$∴$ Most appropriate option is (d).

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Properties of Matter 2 Question 5

5. 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 mass M. It is suspended by a string in a liquid of density ρ, where it stays vertical.The

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5. 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 mass $M$ . It is suspended by a string in a liquid of density $ρ$ , where it stays vertical.The