Question

Two small drops of mercury each of radius \( R \) coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

• A. \( \frac{2}{3} : 1 \)
• B. \( \sqrt{2} : 1 \)
• C. \( \frac{1}{2} : 1 \)
• D. \( 2 : 1 \)

Hint:

When drops of the same material coalesce, the total surface energy decreases due to the increase in volume and decrease in total surface area.

Answer 1

The Correct Option is C

Step 1: Surface energy formula.
Surface energy is directly proportional to the surface area of the drop. The surface area of a spherical drop is \( A = 4\pi r^2 \), and the surface energy is proportional to the surface area.
Step 2: Before and after coalescence.
Initially, there are two drops, each with radius \( R \), and after coalescence, the radius of the large drop becomes \( R_{\text{large}} = 2^{1/3} R \). The total surface energy before the change is proportional to \( 2 \times R^2 \), and after the change, it is proportional to \( R_{\text{large}}^2 \).
Step 3: Conclusion.
Thus, the ratio of the total surface energies before and after the change is \( \frac{1}{2} : 1 \), so the correct answer is (C).