Question

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is _____ ($S$ is the surface tension of mercury).

• A. $8\pi r^2 S$
• B. $16\pi r^2 S$
• C. $64\pi r^2 S$
• D. $4\pi r^2 S$

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When drops coalesce, the total volume remains constant but the total surface area decreases. This reduction in surface area leads to the release of surface energy ($E = S \cdot \Delta A$).
Step 2: Detailed Explanation:
Let $R$ be the radius of the big drop.
Volume conservation: $8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R^3 = 8r^3 \implies R = 2r$.
Initial surface area $A_i = 8 \times 4\pi r^2 = 32\pi r^2$.
Final surface area $A_f = 4\pi R^2 = 4\pi (2r)^2 = 16\pi r^2$.
Decrease in area $\Delta A = A_i - A_f = 32\pi r^2 - 16\pi r^2 = 16\pi r^2$.
Energy released $E = S \cdot \Delta A = 16\pi r^2 S$.
Step 3: Final Answer:
The energy released is $16\pi r^2 S$.