Question

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is $2.0\times {10}^{-5}kg{m}^{-1}{s}^{-1}$, the terminal velocity of the drop will be:
(The density of water = ${10}^{3}kg{m}^{-3}$ and $g=10m{s}^{-2}$)

• A. $1.0\times {10}^{-4}m{s}^{-1}$
• B. $2.0\times {10}^{-4}m{s}^{-1}$
• C. $2.5\times {10}^{-4}m{s}^{-1}$
• D. $5.0\times {10}^{-4}m{s}^{-1}$

Answer 1

Answer: (C)

$2.5\times {10}^{-4}m{s}^{-1}$

Here, $r=0.0015mm=0.0015\times {10}^{-3}m$
$\eta =2.0\times {10}^{-5}kg{m}^{-1}{s}^{-1}$
$\rho =1.0\times {10}^{3}kg{m}^{-3}$
$g=10m{s}^{-2}$
Neglecting the density of air, the terminal velocity of the water drop is
$v_T=\frac{2}{9}\frac{r^2\rho g}{\eta }=\frac{2\times (0.0015\times {10}^{-3}{)}^2\times 1.0\times {10}^3\times 10}{9\times 2.0\times {10}^{-5}}$
$v_T=2.5\times {10}^{-4}m{s}^{-1}$