In Millikan's oil drop

Question:
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).

$3.8 \times 10^{-11} \mathrm{~N}$
$3.9 \times 10^{-10} \mathrm{~N}$
$1.8 \times 10^{-10} \mathrm{~N}$
$5.8 \times 10^{-10} \mathrm{~N}$

Correct Option: , 2

Solution:

Viscous force $=$ Weight

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

$=3.9 \times 10^{-10}$

In Millikan's oil drop

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