The velocity of a small ball of mass $$m$$ and density $$d_1$$, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is $$d_2$$, then the viscous force acting on the ball will be

When the ball reaches terminal (constant) velocity, the net force on it is zero. The three forces acting on the ball are:

1. Weight (downward): $$W = mg$$

2. Buoyant force (upward): $$F_b = \frac{m}{\rho_1} \times d_2 \times g = mg\frac{d_2}{d_1}$$

Here, the volume of ball $$= \frac{m}{d_1}$$, so buoyant force $$= \frac{m}{d_1} \times d_2 \times g$$.

3. Viscous force (upward): $$F_v$$

At terminal velocity, the net force is zero:

$$mg = F_b + F_v$$

$$mg = mg\frac{d_2}{d_1} + F_v$$

$$F_v = mg - mg\frac{d_2}{d_1}$$

$$F_v = mg\left(1 - \frac{d_2}{d_1}\right)$$

The correct answer is Option B.