A small ball of mass M and density $\rho$ is dropped in a viscous liquid of density $\rho_{0}$. After some time, the ball falls with a constant velocity. What is the viscous force on the ball?

(A) $\mathrm{F}=\operatorname{Mg}\left(1-\frac{\rho_{0}}{\rho}\right)$

(B) $\mathrm{F}=\mathrm{Mg}\left(1+\frac{\rho}{\rho_{0}}\right)$

(C) $\mathrm{F}=\mathrm{Mg}\left(1+\frac{\rho_{0}}{\rho}\right)$

(D) $\mathrm{F}=\mathrm{Mg}\left(1 \pm \rho \rho_{0}\right)$

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Fluid Mechanics

A small ball of mass M and density

ρ

is dropped in a viscous liquid of density

ρ_{0}

. After some time, the ball falls with a constant velocity. What is the viscous force on the ball?

(A) $F = Mg (1 - \frac{ρ _{0}}{ρ})$ ✓ Correct
(B) $F = Mg (1 + \frac{ρ}{ρ _{0}})$
(C) $F = Mg (1 + \frac{ρ _{0}}{ρ})$
(D) $F = Mg (1 \pm ρ ρ_{0})$

Answer: (A)

Solution

When the ball falls with constant velocity, the net force is zero. Three forces act on the ball: weight

M g

(downward), buoyant force

ρ_{0} V g

(upward), and viscous force

F_{vis}

(upward).

For equilibrium:

F_{vis} + ρ_{0} V g = ρ V g

Since

M = ρ V

, we have

V = M / ρ

. Substituting:

F_{vis} = (ρ - ρ_{0}) V g = (ρ - ρ_{0}) \frac{M}{ρ} g = M g (1 - \frac{ρ _{0}}{ρ})

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option

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