In Millikan's oil drop experiment, an oil drop carrying a charge Q is held stationary by a potential difference 2400 V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600 V. What is the charge on the second drop?

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3Q2

answer is B.

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Detailed Solution

In balance condition,QE=mg⇒QVd=43πr3ρg ⇒Q∝r3V⇒Q1Q2=r1r23×V2V1 ⇒QQ2=rr/23×6002400=2⇒Q2=Q/2

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