1. Forces acting on the ball in glycerin

When the ball falls through glycerin, the forces on it are:

• Weight downward: $mg$
• Buoyant force upward: $F_b$
• Viscous force upward: $F_v$

After some time, the ball moves with constant velocity (terminal velocity), so acceleration becomes zero.

Hence, $mg = F_b + F_v$

So, $F_v = mg - F_b$

2. Find the buoyant force

Buoyant force equals the weight of displaced glycerin: $F_b = V d_2 g$ where $V$ is the volume of the ball.

Also, mass of the ball is $m = V d_1$ So, $V = \frac{m}{d_1}$

Substitute into buoyant force: $F_b = \frac{m}{d_1} d_2 g = mg\frac{d_2}{d_1}$

3. Compute viscous force

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

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

But for a ball to fall downward in the liquid, typically $d_1 > d_2$, so this is positive.

Thus the viscous force is $\boxed{mg\left(1 - \frac{d_2}{d_1}\right)}$

4. Match with the options

This corresponds to Option B.

Note: Option C, $mg\left(\frac{d_1}{d_2}-1\right)$, is not dimensionally wrong but is algebraically different and does not follow from force balance.

5. Comparison with stored answer

Stored correct answer: B

Derived answer: B

So, the derived answer agrees with the stored answer.