Eight mercury droplets, each of radius 1 mm and charge 0.066 pC, merge to form one droplet. Its potential is (Take \( \frac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.4 V
When the droplets merge, the total volume and the total charge stay the same. The charge on the big drop resides on its surface, so its potential is that of a charged sphere, \( V=Q/4\pi\varepsilon_0R \).
For eight drops of radius r forming one drop of radius R, equating volumes:
\( \displaystyle 8\times\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3} \)
\( \begin{aligned} R &= 2r \\ &= 2\,\mathrm{mm} \\ &= 2\times10^{-3}\,\mathrm{m} \end{aligned} \)
The total charge is:
\( \begin{aligned} Q &= 8q \\ &= 8\times0.066\,\mathrm{pC} \\ &= 0.528\times10^{-12}\,\mathrm{C} \end{aligned} \)
The potential of the big drop is:
\( \begin{aligned} V &= \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{R} \\ &= \frac{9\times10^{9}\times0.528\times10^{-12}}{2\times10^{-3}}\,\mathrm{V} \\ &\approx 2.38\,\mathrm{V} \\ &\approx 2.4\,\mathrm{V} \end{aligned} \)