QJEEM21Ph5cfa: JEE Main 2021 Physics - Class 12 JEEM21Ph5cfa Solution: An oil drop of radius 2 mm with a density 3g cm − 3 is held…

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QJEEM21Ph5cfa: JEE Main 2021 Physics - Class 12 JEEM21Ph5cfa Solution: An oil drop of radius 2 mm with a density 3g cm − 3 is held…

An oil drop of radius 2 mm with a density 3g cm^{$-$ 3} is held stationary under a constant electric field 3.55

\times

10⁵ V m^{$-$ 1} in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (consider g = 9.81 m/s²)

Multiple Choice Options

A 48.8 × 1011
B 1.73 × 1010
C 17.3 × 1010
D 1.73 × 1012

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Accepted Answer

JEE Main 2021 (Online) 18th March Morning Shift Physics - Electrostatics Question 129 English Explanation

F_e = qE = (ne)E

F_e = mg

(ne)E = mg

n = \frac{m g}{e E} = \frac{ρ \frac{4}{3} π R^{3} \times g}{e E}

n = \frac{3000 \times \frac{4}{3} \times 3.14 \times 8 \times 10^{- 9} \times 9.8}{1.6 \times 10^{- 19} \times 3.55 \times 10^{5}}

n = \frac{984704 \times 10^{5}}{5.68} = 1.73 \times 10^{10}

n = 1.73 \times 10^{10}

QJEEM21Ph5cfa: JEE Main 2021 Physics - Class 12 JEEM21Ph5cfa Solution: An oil drop of radius 2 mm with a density 3g cm − 3 is held…

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