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CGP EDU Academic Team
Published on: September 12, 2026
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
Text Solution
Verified by ExpertsThe correct answer is:
D
$\frac{\sigma_{small}}{\sigma_{Big}} = \frac{q}{Q} \times \frac{R^{2}}{r^{2}} = \frac{q}{(nq)} \times \frac{\left(n^{1/3} r\right)^{2}}{r^{2}}$
$= n^{-1/3} = (64)^{-1/3} = \frac{1}{4}$
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