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In Millikan's oll drop experiment, what is the terminal speed of an uncharged drop of radius $2.0 \times 10^{-5} \;m$ and density $1.2 \times 10^{3} \;kg m ^{-3} .$ Take the viscosity of air at the temperature of the experiment to be $1.8 \times 10^{-5}\; Pa\; s$. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to atr.
Solution
Terminal speed $=5.8 cm / s ;$ Viscous force $=3.9 \times 10^{-10} N$
Radius of the given uncharged drop, $r=2.0 \times 10^{-5} m$
Density of the uncharged drop, $\rho=1.2 \times 10^{3} kg m ^{-3}$
Viscosity of air, $\eta=1.8 \times 10^{-5} Pa s$
Density of air $\left(\rho_{o}\right)$ can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, $g=9.8 m / s ^{2}$
Terminal velocity ( $v$ ) is given by the relation
$v=\frac{2 r^{2} \times\left(\rho-\rho_{0}\right) g }{9 \eta}$
$=\frac{2 \times\left(2.0 \times 10^{-5}\right)^{2}\left(1.2 \times 10^{3}-0\right) \times 9.8}{9 \times 1.8 \times 10^{-3}}$
$=5.807 \times 10^{-2} m s ^{-1}$
$=5.8 cms ^{-1}$
Hence, the terminal speed of the drop is $5.8 cm s ^{-1}$ The viscous force on the drop is given by:
$F=6 \pi \eta r v$
$\therefore F=6 \times 3.14 \times 1.8 \times 10^{-5} \times 2.0 \times 10^{-5} \times 5.8 \times 10^{-2}$
$=3.9 \times 10^{-10} N$
Hence, the viscous force on the drop is $3.9 \times 10^{-10} \,N$
