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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
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Verified Answer
The correct answer is:
$\frac{1}{2} \mu \mathrm{F}$
As, the volume remains constant, so
$\begin{array}{ll} & \frac{4}{3} \pi R^{3}=8\left(\frac{4}{3} \pi r^{3}\right) \\ \Rightarrow \quad & R=2 r \text { or } r=\frac{R}{2} \\ \text { Capacitance of spherical capacitor, } \\ \Rightarrow \quad C & =4 \pi \varepsilon_{0} R \\ \Rightarrow \quad & C R \\ \therefore \quad & \frac{C_{1}}{C_{2}}=\frac{R}{r}=\frac{2 r}{r}=2 \quad \Rightarrow \quad C_{2}=\frac{C_{1}}{2}=\frac{1}{2} \mu \mathrm{F}\end{array}$
$\begin{array}{ll} & \frac{4}{3} \pi R^{3}=8\left(\frac{4}{3} \pi r^{3}\right) \\ \Rightarrow \quad & R=2 r \text { or } r=\frac{R}{2} \\ \text { Capacitance of spherical capacitor, } \\ \Rightarrow \quad C & =4 \pi \varepsilon_{0} R \\ \Rightarrow \quad & C R \\ \therefore \quad & \frac{C_{1}}{C_{2}}=\frac{R}{r}=\frac{2 r}{r}=2 \quad \Rightarrow \quad C_{2}=\frac{C_{1}}{2}=\frac{1}{2} \mu \mathrm{F}\end{array}$
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