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2. Electric Potential and Capacitance
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$64$ 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
A
$1:64$
B
$64:1$
C
$4:1$
D
$1:4$
Solution
(d) $\frac{{{\sigma _{small}}}}{{{\sigma _{Big}}}} = \frac{q}{Q} \times \frac{{{R^2}}}{{{r^2}}} = \frac{q}{{(nq)}} \times \frac{{{{({n^{1/3}}r)}^2}}}{{{r^2}}}$$ = {n^{ – 1/3}} = {(64)^{ – 1/3}} = \frac{1}{4}$
Standard 12
Physics
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