A mercury drop of radius $$10^{-3}$$ m is broken into 125 equal size droplets. Surface tension of mercury is $$0.45 \text{ N m}^{-1}$$. The gain in surface energy is:

Original mercury drop: radius $$R = 10^{-3}$$ m. Broken into 125 equal droplets.

Volume conservation: $$\frac{4}{3}\pi R^3 = 125 \times \frac{4}{3}\pi r^3$$, so $$r = R/5 = 2 \times 10^{-4}$$ m.

Surface tension $$T = 0.45$$ N/m.

Initial surface area = $$4\pi R^2$$.
Final surface area = $$125 \times 4\pi r^2 = 125 \times 4\pi(R/5)^2 = 125 \times 4\pi R^2/25 = 20\pi R^2$$.

Change in surface area = $$20\pi R^2 - 4\pi R^2 = 16\pi R^2$$.

Gain in surface energy = $$T \times \Delta A = 0.45 \times 16\pi \times (10^{-3})^2$$

$$= 0.45 \times 16 \times 3.14 \times 10^{-6} = 0.45 \times 50.27 \times 10^{-6} = 22.6 \times 10^{-6} = 2.26 \times 10^{-5}$$ J.

The correct answer is Option A: $$2.26 \times 10^{-5}$$ J.