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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2021
  • Hard
  • MCQ

An oil drop of radius \(2\, mm\) with a density \(3\, g\) \(cm ^{-3}\) is held stationary under a constant electric field \(3.55 \times 10^{5}\, V\, m ^{-1}\) in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? (consider \(\left. g =9.81\, m / s ^{2}\right)\)

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Answer & Solution

Correct Answer

(B) \(1.73 \times 10^{10}\)

Step-by-step Solution

Detailed explanation

\(q E = Mg\)

\(neE =\rho\left(\frac{4}{3} \pi r ^{3}\right) \times g\)

\(n \times 1.6 \times 10^{-19} \times 3.55 \times 10^{5}\)

\(=3 \times 10^{3} \times \frac{4}{3} \times \pi \times\left(2 \times 10^{-3}\right)^{3} \times 9.81\)

\(n =173 \times 10^{(3-9-5+19)}\)

\(n =1.73 \times 10^{10}\)

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