Eight drops of mercury, each of same radius and same charge combine to form a bigger drop. The ratio of…
Eight drops of mercury, each of same radius and same charge combine to form a bigger drop. The ratio of capacitance of the bigger drop to that of each smaller drop is
$8: 1$
$4: 1$
$2: 1$
$1: 1$
Solution
Number of drops, $\mathrm{n}=8$
Let, $\mathrm{r}$ is radius of smaller drop and $\mathrm{R}$ is the radius of Bigger drop.
Volume of smaller drops \& bigger drop remains same
$
8 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi r^3 \Rightarrow R=2 r
$
Capacitance of spherical conductor, $C=4 \pi \epsilon_0 R$ Ratio of capacitance of bigger drop to smaller drop is given as
$
\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=\frac{4 \pi \epsilon_0 \mathrm{R}}{4 \pi \epsilon_0 \mathrm{r}}=\frac{\mathrm{R}}{\mathrm{r}}=\frac{2 \mathrm{r}}{4}=2: 1
$
$\frac{C}{C^{\prime}}=\frac{2}{1}$