Eight small drops, each of radius $r$ and having same charge $q$ are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is
Medium
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(b) By using ${V_{big}} = {n^{2/3}}{v_{small}}$ $==>$ $\frac{{{V_{Big}}}}{{{v_{small}}}} = {(8)^{2/3}} = \frac{4}{1}$
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