Question

A water drop of radius $1\mu m$ falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is $1.8\times 10^{-5}Ns{m}^{-2}$ and its density is negligible as compared to that of water $10^{6}g{m}^{-3}$. Terminal velocity of the water drop is:
(Take acceleration due to gravity $=10m{s}^{-2}$ )

• A. $145.4\times 10^{-6}m{s}^{-1}$
• B. $118.0\times 10^{-6}m{s}^{-1}$
• C. $132.6\times 10^{-6}m{s}^{-1}$
• D. $123.4\times 10^{-6}m{s}^{-1}$

Answer 1

Answer: (D)

$6\pi \eta r{v}_t=\frac{4}{3}\pi r^3\rho g$
$v_t=\frac{4}{3}\times \frac{\pi r^3\rho g}{6\pi \eta r}$
$v_t=\frac{4}{3}\times \frac{{\pi }^3\rho g}{6\pi \eta r}=\frac{2\times 10^{-12}\times 10^3\times 10}{9\times 1.8\times 10^{-5}}$
$=123.4\times 10^{-6}m/s$