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Q.

An oil drop having charge 2e is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg/m3, the radius of the drop will be

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a

2.0×10−6m

b

1.7×10−6m

c

1.4×10−6m

d

1.1×10−6m

answer is B.

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Detailed Solution

In equilibrium QE=mg⇒Q⋅Vd=mg=43πr3ρg⇒2×1.6×10−19×120002×10−2=43πr3×900×10⇒r=1.7×10−6m
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