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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 small drops $=$ Volume of big drop
$\therefore\left(\frac{4}{3} \pi r^{3}\right) \times 8=\frac{4}{3} \pi R^{3}$ $\Rightarrow 2 r=R$ $\ldots$ (i)
According to charge conservation
(.) (ii)
$$
8 \mathrm{q}=\mathrm{Q}
$$
Potential of one small drop $\left(\mathrm{V}^{\prime}\right)=\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}}$
Similarly, potential of big drop $(\mathrm{V})=\frac{Q}{4 \pi \varepsilon_{0} R}$
Now, $\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\mathrm{q}}{\mathrm{Q}} \times \frac{\mathrm{R}}{\mathrm{r}} \Rightarrow \frac{\mathrm{V}^{\prime}}{20}=\frac{9}{8 \mathrm{q}} \times \frac{2 \mathrm{r}}{\mathrm{r}}$ $\therefore \mathrm{V}^{\prime}=5 \mathrm{~V}$
$\therefore\left(\frac{4}{3} \pi r^{3}\right) \times 8=\frac{4}{3} \pi R^{3}$ $\Rightarrow 2 r=R$ $\ldots$ (i)
According to charge conservation
(.) (ii)
$$
8 \mathrm{q}=\mathrm{Q}
$$
Potential of one small drop $\left(\mathrm{V}^{\prime}\right)=\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}}$
Similarly, potential of big drop $(\mathrm{V})=\frac{Q}{4 \pi \varepsilon_{0} R}$
Now, $\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\mathrm{q}}{\mathrm{Q}} \times \frac{\mathrm{R}}{\mathrm{r}} \Rightarrow \frac{\mathrm{V}^{\prime}}{20}=\frac{9}{8 \mathrm{q}} \times \frac{2 \mathrm{r}}{\mathrm{r}}$ $\therefore \mathrm{V}^{\prime}=5 \mathrm{~V}$
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