Surface Energy Released When Eight Mercury Drops Coalesce

To find the surface energy released when eight mercury drops coalesce to form a single larger drop, we follow these steps:

1. Conservation of Volume: Let $r$ be the radius of each small mercury drop and $R$ be the radius of the newly formed big drop. Since the total volume remains constant during coalescence: $\frac{4}{3} \pi R^3 = 8 \times \left(\frac{4}{3} \pi r^3\right)$

Taking the cube root on both sides: $R^3 = 8r^3 \implies R = 2r$

2. Surface Area Calculation:

• The initial total surface area of the $8$ small drops is: $A_i = 8 \times (4\pi r^2) = 32\pi r^2$

• The final surface area of the big drop is: $A_f = 4\pi R^2 = 4\pi (2r)^2 = 16\pi r^2$

3. Change in Surface Area: The decrease in surface area ($\Delta A$) is given by: $\Delta A = A_i - A_f = 32\pi r^2 - 16\pi r^2 = 16\pi r^2$

4. Surface Energy Released: The surface energy released in this process is equal to the surface tension $S$ multiplied by the decrease in surface area $\Delta A$: $\text{Energy Released} = S \times \Delta A = 16\pi r^2 S$

Thus, the correct option is B ($16 \pi r^2 S$).