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If eight identical drops are joined to form a bigger drop, the potential of the bigger drop compared to that of a smaller drop will be

1. Double
2. Four times
3. Eight times
4. One time

Correct answer: Four times

Solution

When 8 identical drops merge, the volume becomes 8 times, so the radius becomes $8^{1/3}=2$ times. For a charged conducting sphere, potential is proportional to radius for the same charge density after coalescence, giving a factor of 4 in this standard result for identical charged drops. Hence the bigger drop’s potential is four times that of a smaller drop.

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