In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 × 10 - 5 m and density 1.2 × 103 kgm - 3? Take viscosity of liquid = 1.8 × 10 - 5 Nsm ⁻². (Neglect buoyancy due to air). (a) 3.8 × 10 - 11 N (b) 3.9 × 10 - 10 N (c) 1.8 × 10 - 10 N (d) 5.8 × 10 - 10 N

Answer: (b). The drop falls at terminal velocity and buoyancy is neglected, so Fv = mg = 4/3π r³ρ g. r³ = (2×10⁻⁵)³ = 8×10⁻¹⁵ m³ Fv = 4/3π(8×10⁻¹⁵)(1.2×10³)(9.8) ≈ 3.9×10⁻¹⁰ N So the answer is (b) 3.9×10⁻¹⁰ N.

In Millikan's oil drop experiment, what is viscous force acting on an… | JEE Main 2021 (27 Aug, Shift 1)

Q26JEE Main 2021 · 27 AUG · SHIFT 1Medium

Viscosity & Stokes' Law

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 $\times$ 10 $-$ 5 m and density 1.2 $\times$ 103 kgm $-$ 3? Take viscosity of liquid = 1.8 $\times$ 10 $-$ 5 Nsm ${}^{-2}$ . (Neglect buoyancy due to air).

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Answer: (b)

The drop falls at terminal velocity and buoyancy is neglected, so $F_v = mg = \frac{4}{3}\pi r^3\rho g$ .
$r^3 = (2\times10^{-5})^3 = 8\times10^{-15}\text{ m}^3$
$F_v = \frac{4}{3}\pi(8\times10^{-15})(1.2\times10^3)(9.8) \approx 3.9\times10^{-10}$ N
So the answer is (b) $3.9\times10^{-10}$ N.

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