27 similar drops of mercury are maintained at 10 V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ______ times that of a smaller drop.

We have 27 identical mercury drops, each maintained at a potential of 10 V. Let the radius of each small drop be $$r$$ and charge be $$q$$.

Since the volume is conserved when the drops merge, $$27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$, which gives $$R = 3r$$.

The total charge on the big drop is $$Q = 27q$$.

The potential energy (self-energy) of a charged sphere is $$U = \frac{kQ^2}{2R}$$.

For a small drop: $$U_s = \frac{kq^2}{2r}$$.

For the big drop: $$U_B = \frac{k(27q)^2}{2(3r)} = \frac{729 \, kq^2}{6r} = \frac{243 \, kq^2}{2r}$$.

Therefore, $$\frac{U_B}{U_s} = \frac{243 \, kq^2 / (2r)}{kq^2 / (2r)} = 243$$.

The potential energy of the bigger drop is $$243$$ times that of a smaller drop.