If 'n' identical water drops assumed spherical each charged to a potential energy U coalesce to a single drop, the potential energy of the single drop is: (Assume that drops are uniformly charged).
A
n2/3U
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B
n3/2U
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C
n4/3U
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D
n5/3U
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Solution
The correct option is An2/3U Potential drop in 1 drop be= kqr= U (Let)
According to question, volume of n drops is equal to volume of the bigger drop.
So, n43πr3=43πR3-----(1)
where (r= radius of smaller drop and R= radius of bigger drop)