Question

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

• A. 1 : 4
• B. 4 : 1
• C. 1: 8
• D. 8 : 1

Answer 1

Answer: (B)

4 : 1

Answer 2

q ′ = 64 q … (i)
A ′ = 16 A … (ii)

Dividing (i) & (ii),
σ′ = 4σ
⇒$\frac{ σ´}{σ}$ =$\frac{4}{1}$

$\therefore ,$ the correct option is (B): 4:1