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

We have a mercury drop of radius $$R = 10^{-3}$$ m broken into $$n = 125$$ equal droplets, with surface tension $$T = 0.45$$ N/m.

By conservation of volume:

$$\frac{4}{3}\pi R^3 = 125 \times \frac{4}{3}\pi r^3$$

so $$R^3 = 125r^3$$, giving $$r = \frac{R}{5} = \frac{10^{-3}}{5} = 2 \times 10^{-4}$$ m.

The initial surface area is $$A_i = 4\pi R^2 = 4\pi \times 10^{-6}$$ m$$^2$$. The final total surface area (for all 125 droplets) is

$$A_f = 125 \times 4\pi r^2 = 125 \times 4\pi \times (2 \times 10^{-4})^2 = 125 \times 4\pi \times 4 \times 10^{-8} = 20\pi \times 10^{-6} \text{ m}^2$$

Now the change in surface area is

$$\Delta A = A_f - A_i = 20\pi \times 10^{-6} - 4\pi \times 10^{-6} = 16\pi \times 10^{-6} \text{ m}^2$$

So the gain in surface energy is (since surface energy equals surface tension times area)

$$\Delta E = T \times \Delta A = 0.45 \times 16\pi \times 10^{-6} = 7.2\pi \times 10^{-6} = 22.6 \times 10^{-6} \text{ J} = 2.26 \times 10^{-5} \text{ J}$$

Hence, the correct answer is $$2.26 \times 10^{-5}$$ J.