64 identical drops each charged upto potential of $$10$$ mV are combined to form a bigger drop. The potential of the bigger drop will be _____ mV.

$$V_{\text{small}} = \frac{kq}{r} = 10\text{ mV}$$

Equating total volume to find the radius $$R$$ of the combined large drop:

$$\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3 \implies R = 4r$$

$$Q = 64q$$

$$V_{\text{large}} = \frac{kQ}{R} = \frac{k(64q)}{4r} = 16 \left(\frac{kq}{r}\right)$$

$$V_{\text{large}} = 16 \times 10 = 160\text{ mV}$$