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 $1.8 \times 10^{-5} N\,sm ^{-2}$ and its density is negligible as compared to that of water $10^{6} gm ^{-3}$. Terminal velocity of the water drop is:
(Take acceleration due to gravity $=10 \,ms ^{-2}$ )

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

Answer 1

Answer: (D)

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