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Q.

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

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a

1:8

b

8:1

c

4:1

d

1:4

answer is B.

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Detailed Solution

Radius of the big drop will be 

R=n1/3r

The ratio of surface charge densities is

σbigσsmall=nq/4πR2q4πr2

=nR2×r2

=nn1/3×r2×r2

=n1/3=431/3

= 4:1

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