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.
$\therefore \frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3$
$\Rightarrow R=2r\dots$(i)
The capacitance of a smaller spherical drop is
$C=4\pi {\epsilon }_{0}r\dots$(ii)
The capacitance of bigger drop is
$C^{\prime }=4\pi {\epsilon }_{0}R$
$=2\times 4\pi {\epsilon }_{0}r(\because R=2r)$
$=2C$ [from Eq. (ii)]
$\therefore C=\frac{C^{\prime }}{2}$
$=\frac{1}{2}\mu F(\because C^{\prime }=1\mu F)$