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In Milikan's oil drop experiment,what is the terminal speed of an uncharged drop of radius $2\times 10^{-5}m$ and density $1.2\times 10^3Kgm^{-3}$?Take the viscocity of air at the temperature of the experiment is $1.8\times 10^{-5}pas$.How much is the viscous force on the drop at the speed?Neglect byoyancy of the drop due to air.

$\begin{array}{1 1}(A)\;4cm/s,4.2\times 10^{-10}N\\(B)\;4.8cm/s,8\times 10^{-9}N\\(C)\;5.8cm/s,3.9\times 10^{-10}N\\(D)\;5cm/s,3.8\times 10^{-9}N\end{array} $

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Terminal speed =$5.8cm/s$
Viscous force =$3.9\times 10^{-10}N$
Radius of drop $r=2\times 10^{-5}m$
$\rho=1.2\times 10^3Kgm^{-3}$
Viscosity of air $\eta=1.8\times 10^{-5}pas$
Terminal velocity $V=\large\frac{2r^2-(\rho-\rho_0)g}{g\eta}$
$\Rightarrow \large\frac{2\times (2\times 10^{-5})^2(1.2\times 10^3-0)\times 9.8}{9\times 1.8\times 10^{-5}}$
$\Rightarrow 5.8cms^{-1}$
Here $\rho_0$(density of air)=0 to neglect byoyancy
$\therefore$ Terminal velocity =$5.8cms^{-1}$
$F=6\pi \eta r V$
$\;\;\;=3.9\times 10^{-10}N$
Hence (C) is the correct answer.
answered May 9, 2014 by sreemathi.v

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