Question

A spherical drop of capacitance 1 $\mu$F is broken into eight drops of equal radius. Then, the capacitance of each small drop is

• A. $\frac{1}{2} \mu F$
• B. $\frac{1}{4} \mu F$
• C. $\frac{1}{8} \mu F$
• D. $8 \,\mu F$

Answer 1

Answer: (A)

Let $R$ and $r$ be the radii of bigger and each smaller drop respectively.
$\therefore \frac{4}{3} \pi R^{3} =8 \times \frac{4}{3} \pi r^{3} $
$\Rightarrow R =2 r \dots$(i)
The capacitance of a smaller spherical drop is
$C=4 \pi \varepsilon_{0} r\dots$(ii)
The capacitance of bigger drop is
$C' = 4\pi \varepsilon_0 R$
$ = 2 \times 4\,\pi \varepsilon_0 r \,\,(\because R = 2r)$
$= 2C \,$ [from Eq. (ii)]
$\therefore C = \frac{C'}{2}$
$ = \frac{1}{2} \mu F \,\,(\because C' = 1\mu F)$