Question

Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

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

Hint:

Volume is conserved when drops merge, which gives the radius of the big drop. Surface energy is surface tension times area.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The surface energy of a drop is \(E = T\times\text{area}\). When two drops merge, the total volume stays the same but the area changes.

Step 2: Find the radius of the big drop.
\[ \frac{4}{3}\pi R^3 = 2\times\frac{4}{3}\pi r^3 \Rightarrow R = 2^{1/3}r \]

Step 3: Compare the areas.
Before: \(2\times 4\pi r^2 = 8\pi r^2\). After: \(4\pi R^2 = 4\pi\,2^{2/3}r^2\).
\[ \frac{E_{after}}{E_{before}} = \frac{4\pi\,2^{2/3}r^2}{8\pi r^2} = \frac{2^{2/3}}{2} = 2^{-1/3} \]

Step 4: Check the options.
The ratio is less than 1 because area decreases on coalescing. So options (A) and (B), which are above 1, cannot be right. \(1 : 2\) would hold only if the radius did not change.

Final Answer:
The ratio is \(2^{-1/3} : 1\), option (C). \[ \boxed{2^{-1/3} : 1} \]