Question

64 water drops having equal charges combine to form one bigger drop. The capacitance of bigger drop, as compared to that of smaller drop will be :

• A. 4 times
• B. 8 times
• C. 16 times
• D. 64 times

Answer 1

Answer: (A)

4 times

Let R be the radius of bigger drop after combining n drops and r be the radius of small drops.
$\frac{4}{3}\pi R^3=n\frac{4}{3}\pi r^3\Rightarrow R=n^{\frac{1}{3}}r$
now, $Q_{big}=n{Q}_{small},V_{big}=k\frac{Q_{big}}{R}=k\frac{n{Q}_{small}}{n^{\frac{1}{3}}r}=n^{\frac{2}{3}}k\frac{Q_{small}}{r}=n^{\frac{2}{3}}{V}_{small}$
$Q_{big}=C_{big}{V}_{big}$
or $C_{big}=\frac{Q_{big}}{V_{big}}=\frac{n{Q}_{small}}{n^{\frac{2}{3}}{V}_{small}}$
$=n^{\frac{1}{3}}{C}_{small}$
here, $n=64,C_{big}=(64{)}^{\frac{1}{3}}{C}_{small}=4C_{small}$