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 (a) Mg (b) Mg - Vρ g (c) Mg + π R² hρ g (d) ρ g (V + π R² h)

Answer: (d). Upthrust = (force on bottom, upward) − (force on top, downward) = Vρ g. Force on top surface (gauge): Ftop = ρ g h · π R². Fbottom = Vρ g + π R² hρ g = ρ g(V + π R² h) So the answer is (d).

A hemispherical portion of radius ' R ' is removed from the bottom of… | MHT CET 2024 (4 May, Shift 2)

Q10MHT CET 2024 · 4 MAY · SHIFT 2Difficult

Archimedes' Principle & Floatation

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 ' $\rho$ ', 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

Show answer and solution

Answer: (d)

Upthrust = (force on bottom, upward) − (force on top, downward) $= V\rho g$ .
Force on top surface (gauge): $F_{top} = \rho g h \cdot \pi R^2$ .
$F_{bottom} = V\rho g + \pi R^2 h\rho g = \rho g(V + \pi R^2 h)$
So the answer is (d).

Report a mistake

Archimedes' Principle & Floatation question from MHT CET 2024 (4 May, Shift 2)

More questions on Archimedes' Principle & Floatation