Sixty four conducting drops each of radius $$0.02$$ m and each carrying a charge of $$5$$ $$\mu$$C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be

When 64 small conducting drops combine to form a bigger drop, the total charge and total volume are conserved.

Volume is conserved:

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

$$R^3 = 64r^3$$

$$R = 4r$$

where $$r = 0.02$$ m is the radius of each small drop and $$R$$ is the radius of the bigger drop.

Total charge is conserved:

$$Q = 64 \times q = 64 \times 5 = 320 \text{ } \mu\text{C}$$

$$\sigma_s = \frac{q}{4\pi r^2}$$

$$\sigma_B = \frac{Q}{4\pi R^2} = \frac{64q}{4\pi (4r)^2} = \frac{64q}{4\pi \times 16r^2} = \frac{4q}{4\pi r^2}$$

$$\frac{\sigma_B}{\sigma_s} = \frac{\frac{4q}{4\pi r^2}}{\frac{q}{4\pi r^2}} = 4$$

Therefore, $$\sigma_B : \sigma_s = 4 : 1$$

Hence, the correct answer is Option B.