512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ______ V in Volt.

We have 512 identical mercury drops, each charged to a potential of 2 V. Let each small drop have radius $$r$$ and charge $$q$$. The potential of each small drop is $$V = \frac{kq}{r} = 2\,\text{V}$$.

When 512 drops merge into one large drop, charge is conserved: $$Q = 512q$$. Volume is also conserved: $$\frac{4}{3}\pi R^3 = 512 \times \frac{4}{3}\pi r^3$$, which gives $$R = 8r$$ (since $$512^{1/3} = 8$$).

The potential of the large drop is $$V' = \frac{kQ}{R} = \frac{k(512q)}{8r} = 64 \times \frac{kq}{r} = 64 \times 2 = 128\,\text{V}$$.

Therefore, the potential of the combined drop is $$128$$ V.