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Question

# Two identical mercury drops, each of radius r are charged to the same potential V.If the mercury drops coalesce to form a big drop, each of radius R then the potential of the combined drop will be

A
(3)1/2V
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B
(3)2/3V
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C
(2)2/3V
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D
(2)3/2V
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Solution

## The correct option is C (2)2/3VGiven number drops =2 with radius of small drop=r and potential =V.Radius of bigger drop=RNow potential each small drop V=kqrVolume of 2 Hg drop=Volume of one large drop2×43πr3=43πR3R=(2)13r -------------(A) {here charge remains conserved}Now charge on small drop =q and charge on large drop containing 2 drop=2qThus potential of 1 large drop Vl=kqR=k2q21/3r=22/3kqrTherefore Vl=22/3V

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