A spherical solid ball of volume V is made of a material of density ρ₁. It is falling through a liquid of density ρ₂ (ρ₂ 0). The terminal speed of the ball is (a) √((Vg(ρ₁ - ρ₂))/(k)) (b) (Vgρ₁)/(k) (c) √((Vgρ₁)/(k)) (d) (Vg(ρ₁ - ρ₂))/(k)

Answer: (a). At terminal speed: weight = buoyancy + viscous force. Vρ₁ g = Vρ₂ g + kv² v = √((Vg(ρ₁ - ρ₂))/(k)) So the answer is (a).

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Q32JEE Main 2008 · AIEEEMedium

Viscosity & Stokes' Law

A spherical solid ball of volume $V$ is made of a material of density $\rho_1$ . It is falling through a liquid of density $\rho_2\ (\rho_2 < \rho_1)$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v,$ i.e., $F_{viscous} = -kv^2\ (k > 0).$ The terminal speed of the ball is

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Answer: (a)

At terminal speed: weight $=$ buoyancy $+$ viscous force.
$V\rho_1 g = V\rho_2 g + kv^2$
$v = \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}}$
So the answer is (a).

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