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

$\begin{array}{l}Conserving volume, \\ V_{\text{big }}=512V_{\text{small }}\\ \Rightarrow \frac{4}{3}\pi R^3=8^3\frac{4}{3}\pi r^3\\ \Rightarrow R=8r\\ Now, for potential,\\ V_{\text{small }}=\frac{Kq}{r} and\\ V_{\text{big }}=\frac{K{q}^{\text{'}}}{R}\\ here conserving charge gives ,q^{\text{'}}=512q\\ \Rightarrow V_{\text{big }}=\frac{K\times 512q}{8r}=\frac{512}{8}\frac{\mathrm{Kq}}{r}=64V_{\text{small }}=64\times 2\\ \Rightarrow V_{\text{big }}=128\text{ volt }\end{array}$