1. Surface energy of a liquid drop

For a liquid drop, $E = T \times A$ where $T$ is surface tension and $A$ is surface area.

Since $T$ is the same for water, surface energy is proportional to surface area: $E \propto A$

2. Let radius of each small droplet be $r$

If there are $1000$ small droplets, total volume before coalescing is $1000 \cdot \frac{4}{3}\pi r^3$

Let radius of the big drop be $R$. Then $\frac{4}{3}\pi R^3 = 1000 \cdot \frac{4}{3}\pi r^3$

So, $R^3 = 1000r^3$ $R = 10r$

3. Initial total surface area

Surface area of one small droplet: $4\pi r^2$

So for $1000$ droplets, $A_i = 1000 \cdot 4\pi r^2 = 4000\pi r^2$

4. Final surface area of big drop

$A_f = 4\pi R^2 = 4\pi (10r)^2 = 400\pi r^2$

5. Compare final and initial surface energy

Since $E \propto A$, $\frac{E_f}{E_i} = \frac{A_f}{A_i} = \frac{400\pi r^2}{4000\pi r^2} = \frac{1}{10}$

Thus, the surface energy becomes $\boxed{\frac{1}{10}\text{th}}$

6. Option check

• A: $\frac{1}{100}$ th ❌
• B: $\frac{1}{10}$ th ✅
• C: 100 times ❌
• D: 10 times ❌

So the correct option is B.