Q129 / 180 · COMEDK 2019 · Physics
A spherical drop of capacitance $1 \mu \mathrm{F}$ is broken into eight drops of equal radius. Then, the capacitance of each small drop is
- A$\frac{1}{8} \mu \mathrm{F}$
- B$8 \mu \mathrm{F}$
- Correct answerC$\frac{1}{2} \mu \mathrm{F}$
- D$\frac{1}{4} \mu \mathrm{F}$
Solution
Given, capacitance of bigger drop $=1 \mu \mathrm{F}$
Let $R$ be the radius of bigger drop and $r$ be the radius of smaller drops.
According to the question, $\frac{4}{3} \pi R^{3}=8 \times \frac{4}{3} \pi r^{3}$
$$
\begin{aligned}
\Rightarrow & R^{3}=8 r^{3} \\
\Rightarrow & R=2 r^{2}
\end{aligned}
$$
The capacitance of bigger drop ,
$$
\begin{aligned}
C &=4 \pi \varepsilon_{0} R \\
&=2 \times 4 \pi \varepsilon_{0} r \quad(\because R=2 r)
\end{aligned}
$$
Capacitance of smaller drop, $C^{\prime}=4 \pi \varepsilon_{0} r$
Thus,
$$
\begin{aligned}
C^{\prime} &=\frac{C}{2} \\
C^{\prime} &=\frac{1}{2} \mu \mathrm{F}
\end{aligned}
$$
Hence,
$C^{\prime}=\frac{1}{2} \mu \mathrm{F}$
Hence, the correct answer is option: 'C'.