N identical spherical drops charged to the same potential V are combined to form a big drop. The potential of the new drop will be

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V / N

V×N

V×N2/3

answer is D.

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Detailed Solution

If the drops are conducting, then 43πR3=N43πr3⇒R=N1/3r. Final charge Q = NqSo final potential V=QR=NqN1/3r=V×N2/3

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