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. \( \sqrt{2} : 1 \)
• B. \( 2^{1/3} : 1 \)
• C. \( 2 : 1 \)
• D. \( 2^{2/3} : 1 \)

Hint:

When two drops coalesce, their volumes add up, but the total surface area doesn't double. The ratio of surface energies is based on the ratio of surface areas before and after the change.

Answer 1

The Correct Option is B

Step 1: Surface energy of a drop.
The surface energy \( E \) of a drop is given by: \[ E = \sigma \cdot A \] where \( \sigma \) is the surface tension and \( A \) is the surface area. For a spherical drop, the surface area is given by \( A = 4 \pi R^2 \).
Step 2: Initial and final surface energy.
- For two small drops, the total surface area before coalescing is: \[ A_{\text{initial}} = 2 \times 4 \pi R^2 = 8 \pi R^2 \] - After coalescence, the new radius of the large drop is \( R_{\text{new}} = 2^{1/3} R \), so the surface area of the large drop is: \[ A_{\text{final}} = 4 \pi (2^{1/3} R)^2 = 4 \pi 2^{2/3} R^2 \]
Step 3: Energy ratio.
The ratio of the surface energies before and after the change is: \[ \frac{E_{\text{final}}}{E_{\text{initial}}} = \frac{4 \pi 2^{2/3} R^2 \sigma}{8 \pi R^2 \sigma} = 2^{1/3} \]
Step 4: Conclusion.
Thus, the ratio of the surface energies is \( 2^{1/3} : 1 \), which is option (B).