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The potential of a large liquid drop when eight liquid drops are combined is $20 \mathrm{~V}$. Then the potential of each single drop was
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Verified Answer
The correct answer is:
$5 \mathrm{~V}$
Volume of 8 drops $=$ volume of big drop
$$
8 q=Q
$$
Potential of one small drop $\left(V^{\prime}\right)=\frac{q}{4 \pi \varepsilon_{0} r}$
Similarly, potential of big drop $(V)=\frac{Q}{4 \pi \varepsilon_{0} R}$
Now, $\quad \frac{V^{\prime}}{V}=\frac{q}{Q}+\frac{R}{r}$
$$
\begin{array}{ll}
\Rightarrow \quad \frac{V^{\prime}}{20} & =\frac{q}{8 q} \times \frac{2 r}{r} \text { [from Eqs. (i) and (ii)] } \\
\therefore \quad V^{\prime} & =5 \text { volt }
\end{array}
$$
$$
8 q=Q
$$
Potential of one small drop $\left(V^{\prime}\right)=\frac{q}{4 \pi \varepsilon_{0} r}$
Similarly, potential of big drop $(V)=\frac{Q}{4 \pi \varepsilon_{0} R}$
Now, $\quad \frac{V^{\prime}}{V}=\frac{q}{Q}+\frac{R}{r}$
$$
\begin{array}{ll}
\Rightarrow \quad \frac{V^{\prime}}{20} & =\frac{q}{8 q} \times \frac{2 r}{r} \text { [from Eqs. (i) and (ii)] } \\
\therefore \quad V^{\prime} & =5 \text { volt }
\end{array}
$$
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