1. Forces acting on the ball in the liquid

When the ball falls through a viscous liquid, the following forces act on it:

• Weight downward: $W = mg$
• Buoyant force upward: $F_b = \rho_0 V g$
• Viscous force upward (opposes motion): let it be $F_v$

Here, $V$ is the volume of the ball.

2. Condition at constant velocity

After some time, the ball attains terminal velocity, so its acceleration becomes zero. Hence, net force on the ball is zero:

$mg = F_b + F_v$

Therefore,

$F_v = mg - F_b$

3. Express buoyant force in terms of $m$ and densities

Since density of the ball is $\rho$,

$\rho = \frac{m}{V} \implies V = \frac{m}{\rho}$

So the buoyant force becomes

$F_b = \rho_0 V g = \rho_0 \left(\frac{m}{\rho}\right) g$

$F_b = mg\frac{\rho_0}{\rho}$

4. Find viscous force

Substitute into $F_v = mg - F_b$

$F_v = mg - mg\frac{\rho_0}{\rho}$

$F_v = mg\left(1 - \frac{\rho_0}{\rho}\right)$

5. Match with options

This corresponds to:

Option A: $mg\left(1-\frac{\rho_0}{\rho}\right)$

So the correct answer is A.