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

(D) $3.9 \times 10^{-10} \mathrm{~N}$

(A) $5.8 \times 10^{-10} \mathrm{~N}$

(B) $1.8 \times 10^{-10} \mathrm{~N}$

(C) $3.8 \times 10^{-10} \mathrm{~N}$

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Fluid Mechanics

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0

\times 1 0^{- 5} m

and density

1.2 \times 1 0^{3} kgm^{- 3}

? Take viscosity of liquid

= 1.8 \times 1 0^{- 5} Nsm^{- 2}

. (Neglect buoyancy due to air).

Answer: (D)

At terminal velocity with buoyancy neglected, the viscous drag balances the weight of the drop:

F = m g = ρ \frac{4}{3} π r^{3} g .

With

ρ = 1.2 \times 1 0^{3} kg m^{- 3}

r = 2.0 \times 1 0^{- 5} m

and

g = 9.8 m s^{- 2}

F = 1.2 \times 1 0^{3} \times \frac{4}{3} π (2.0 \times 1 0^{- 5})^{3} \times 9.8 \approx 3.9 \times 1 0^{- 10} N .

The viscous force on the drop is

3.9 \times 1 0^{- 10}

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option