A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become:

Find how surface energy changes when 1000 small droplets coalesce into one big drop.

We relate the radii by conserving the total volume: $$1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$ which gives $$R^3 = 1000r^3$$ and hence $$R = 10r$$.

The total surface area of the 1000 small drops is $$A_{\text{small}} = 1000 \times 4\pi r^2$$, while the surface area of the single large drop is $$A_{\text{big}} = 4\pi R^2 = 4\pi(10r)^2 = 400\pi r^2$$.

Since the surface energy is given by $$E = T \times A$$ with constant surface tension $$T$$, the ratio of the energies is $$\frac{E_{\text{big}}}{E_{\text{small}}} = \frac{A_{\text{big}}}{A_{\text{small}}} = \frac{400\pi r^2}{4000\pi r^2} = \frac{1}{10}$$.

Therefore, the surface energy becomes $$\frac{1}{10}$$th of its original value.

The correct answer is Option D: $$\frac{1}{10}$$th.