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Q.

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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a

Q4

b

Q2

c

Q

d

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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