Question

A spherical drop of capacitance $1\mu \text{F}$ having an energy of $64\text{mJ}$ is at $4\text{V}$. It is broken into eight drops of equal radii. The capacitance, potential and energy of each small drop is

• A. $\frac{1}{4}\mu \text{F},4\text{V}$ and $2\text{mJ}$
• B. $\frac{1}{8}\mu \text{F},0.2\text{V}$ and $20\text{mJ}$
• C. $\frac{1}{2}\mu \text{F},1\text{V}$ and $2\text{mJ}$
• D. $\frac{1}{2}\mu \text{F},2\text{V}$ and $2\text{mJ}$

Answer 1

The correct option is C $\frac{1}{2}\mu \text{F},1\text{V}$ and $2\text{mJ}$

(i) For spherical bodies capacitance is directly proportional to its radius i.e., $C\propto R$

$C_b=n^{\frac{1}{3}}{C}_s$

$1\times {10}^{-6}=(8{)}^{\frac{1}{3}}{C}_s=(2^3{)}^{\frac{1}{3}}{C}_s$

$C_s=\frac{1}{2}\times {10}^{-6}\text{F}=\frac{1}{2}\mu \text{F}$

(ii) $V_b=n^{\frac{2}{3}}{V}_s$

$4=n^{\frac{2}{3}}{V}_s=4V_s$

$V_s=\frac{4}{4}=1\text{V}$

(iii) $E_b=n^{\frac{5}{3}}{E}_s$

$64\times {10}^{-3}=(8{)}^{\frac{5}{3}}{E}_s=(2^3{)}^{\frac{5}{3}}{E}_s$

$64\times {10}^{-3}=2^5{E}_s=32E_s$

$E_s=\frac{64}{32}\times {10}^{-3}$

$=2\times {10}^{-3}\text{J}$

$E_s=2\text{mJ}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (c) is the correct answer.