Question

$64$ spherical drops each of radius $2cm$ and carrying $5C$ charge combine to form a bigger drop. Its capacity is:

• A. $\frac{8}{9}\times {10}^{-11}F$
• B. $90\times {10}^{-11}F$
• C. $1.1\times {10}^{-11}F$
• D. $9\times {10}^{11}F$

Answer 1

Answer: (A)

$\frac{8}{9}\times {10}^{-11}F$

Let $Q$ be the charge on each drop
$\Rightarrow$ Voltage on each drop: $V=\frac{KQ}{r}$
We know: $C=\frac{Q}{V}$
or, $C=\frac{Qr}{KQ}$
or, $C=\frac{r}{k}$

Equating the sum of volume of drops to the volume of bigger drop:
$64\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
So, radius of final drop: $R=(64{)}^{\frac{1}{3}}r=4r$
$\therefore C^{\prime }=\frac{R}{K}$$=\frac{4\times 2\times {10}^{-2}}{9\times {10}^9}$$=\frac{8}{9}\times {10}^{-11}F$