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Q.

There are n electrons on a drop of oil. It is in equilibrium in an electric field of intensity E. If the density of the oil is ρ, then radius of the drop will be

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a

4πeE3nρg1/3

b

3nρg4πeE1/3

c

4πρg3neE1/3

d

3neE4πρg1/3

answer is D.

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Detailed Solution

In equilibrium, q E = m gThere are n electrons on the drop hence, q = n e∴ neE=43πr3ρgor    r=3neE4πρg1/3
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