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Q.

64 small drops of mercury, each of radius r and charge q coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is

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a

1 : 64

b

64 : 1

c

4 : 1

d

1 : 4

answer is D.

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Detailed Solution

σsmallσBig=qQ×R2r2=q(nq)×(n1/3r)2r2=n−1/3=(64)−1/3=14
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