Question

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

• A. $1.8\times {10}^{-10}\text{N}$
• B. $3.8\times {10}^{-11}\text{N}$
• C. $5.8\times {10}^{-10}\text{N}$
• D. $4\times {10}^{-10}\text{N}$

Answer 1

The correct option is D $4\times {10}^{-10}\text{N}$

When the drop approaches terminal velocity, we can say that :
$\text{Viscous force= Weight of the drop}$
$\text{Viscous force}=\frac{4\pi r^3}{3}\times \rho \times g$
$\text{Viscous force}=\frac{4\pi (2\times {10}^{-5}{)}^3}{3}\times 1.2\times {10}^3\times 10$
$\text{Viscous force}=4\times {10}^{-10}\text{N}$