Solution:
By conserving volume, \(R = n^{1/3}r = 64^{1/3}r = 4r\). The total charge is \(Q = 64q\). Thus, potential of the big drop is \(V' = \frac{kQ}{R} = \frac{64kq}{4r} = 16V\).
Sixty four identical drops of water having equal charge combine to form a bigger drop. The factor by which potential of bigger drop change in comparison to a small drop is
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