Question

An iron ball of radius $0.3cm$ falls through a column of oil of density $0.94gc{m}^{-3}$. If it attains a terminal velocity of $0.54m{s}^{-1}$, what is the viscosity of oil? Density of iron is $7.8gc{m}^{-3}$.
A viscous force of $1.018\times {10}^{-7}N$ acting on a raindrop makes it fall through air with a terminal velocity of $1m{s}^{-1}$. If the viscosity of air is $0.018\times {10}^{-3}PI$, what is the radius of the raindrop ?

Answer 1

Given:

$e=7.8gc{m}^{-2}$

$\sigma =0.94gc{m}^{-3}$

$v_e=\frac{2(e-\sigma )r^{2}g}{9\nu }$

$\nu =\frac{2(e-\sigma )r^2}{9vt}g=\frac{2\times (7.8-0.94)\times (0.3{)}^2\times 980}{9\times 0.5}$

$=268.9$ Paise

radius of raindrop

$r=\sqrt{\frac{9\nu vt}{2g(e-\sigma )}}=\sqrt{\frac{9\times 0.018\times 1}{2\times 980\times (7.8-0.94)}}$

$=\sqrt{\frac{0.162}{3978}}=\sqrt{0.40\times {10}^{-6}}$

$=0.2\times {10}^{-3}cm$