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Question

512 identical drops of mercury are charged to a potential of 2 V each. The drops are then joined to form a bigger drop. The potential of this big drop is V

A
128.0
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B
128
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C
128.00
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Solution

Total number of drops , N=512

Potential of the each small drop , Vs=14πϵ0 qr ......(1)

Potential of the big drop , Vb=14πϵ0 QR=14πϵ0 NqR .....(2)

We know that, Volume of bigger drop is N times the volume of each small drop.

43πR3=N(43 πr3)

R=N13 r

Equation (2) becomes

Vb=14πϵ0 NqN13r=N23[14πϵ0 qr]

=N23 Vs

Potential of the big drop Vb=(512)23 (2)

=(83)23 2=64×2=128 V

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