Question

$64$ small drops of mercury each of radius $r$ and change $q$ coalesce to from a big drop. The ratio of the surface charge density of each small drop with that of big drop is:

• A. $4:1$
• B. $1:4$
• C. $1:64$
• D. $64:1$

Answer 1

The correct option is B $1:4$

Formulae:

$\frac{{\sigma }_{small}}{{\sigma }_{big}}=\frac{q}{Q}\times \frac{R^2}{r^2}$

$=\frac{q}{nq}\times \frac{(n^{\frac{1}{3}}r{)}^2}{r^2}$

$=n^{-\frac{1}{3}}$

$={64}^{-\frac{1}{3}}$

$=\frac{1}{4}$

hence the ratio is $1:4$