64 small drops of mercury, each of radius and charge
, come together to form a big drop. Therefore, the ratio of surface density of charge of each small drop with that of the big drop is,
One of the 64 small drops has a surface area of
Where is its radius. Now these small drops coalesce into a large drop, the total volume hence must be the same as in
where must be the radius of the larger sphere.
is the volume of one small drop. Hence,
Also, the total charge on the larger drop is the sum of the charge on the individual drop, hence
where is the charge on the larger drop and
is the charge on the smaller drop. Hence, the surface densities of the two surfaces are:
Hence, the ratio of small drop to large drop is
By elimination and reduction, we have
Therefore, option (B) is the correct answer.
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