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A spherical drop of mercury of diameter 8 cm is charged to a potential of 12 volt. The drop is broken into 8 equal drops, then potential of each drop is

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detailed solution

Correct option is D

8×43πr3=43π×823⇒r=2 cmIf Q be the charge on the larger drop, then charge on each smaller drop is Q/8.Now, 14π∈o.Q4=12⇒Q=48×4π∈o coul.∴V'=14π∈o×Q/82=116×48×4π∈o4π∈o=3 volt

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