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)

$\frac{1}{2} \mu F$

Answer 2

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 E (ii)] $\therefore C = \frac{C'}{2}$ $= \frac{1}{2} \mu F \,\,(\because C' = 1\mu F)$