Question

A rain drop of radius $0.3\, mm$ has a terminal velocity of $1\, m / s$ in air. The viscosity of air is $18 \times 10^{-5}$ poise. The viscous force on the drop is

• A. $16.95 \times 10^{-7} N$
• B. $1.695 \times 10^{-8} N$
• C. $1.017 \times 10^{-11} N$
• D. $101.73 \times 10^{-9} N$

Answer 1

Answer: (D)

From Stoke's formula
$F=6 \pi \eta r v$
Given, $\eta=18 \times 10^{-5}=18 \times 10^{-6} kg$ / mass,
$r=0.3\, mm$
$=0.3 \times 10^{-3} m$,
$v=1\, m / s$
$\therefore F=6 \times 3.14 \times 18 \times 10^{-6} \times 0.3 \times 10^{-3} \times 1$
$\Rightarrow F=101.73 \times 10^{-9} N$