Question

An oil drop falls through air with a terminal velocity of $5\times {10}^{-4}m/s$. The radius of the drop will be? Neglect density of air as compared to that of oil.

(Viscosity of air = $\frac{18\times {10}^{-5}}{5}N-s/m^2$, g = 10 $m/s^2$, density of oil = 900 kg/$m^3$)

• A. $2.5\times {10}^{-6}m$
• B. $2\times {10}^{-6}m$
• C. $3\times {10}^{-6}m$
• D. $4\times {10}^{-6}m$

Answer 1

The correct option is C $3\times {10}^{-6}m$

$v=\frac{2}{9}{r}^{2}g\frac{d_1-d_2}{\eta }$

where $v=$terminal velocity$=5\times {10}^{-4}m/s$

$r=$radius of oil drop

$g=$acceleration due to gravity$=10m/s^2$

$d_1=$density of oil$=900kg/m^3$

$d_2=$density of air(neglected)

As $v=\frac{2}{9}{r}^{2}g\frac{d_1-d_2}{\eta }$

$5\times {10}^{-4}=\frac{2}{9}{r}^2\times 10\times 5\times \frac{900}{18\times {10}^{-5}}$

$r^2=9\times {10}^{-12}$

$r=3\times {10}^{-6}m$

Hence the correct option is (C).