Question

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

• A. $1.018\times {10}^{-2}$ dyne
• B. $2.018\times {10}^{-2}$ dyne
• C. $3.018\times {10}^{-2}$ dyne
• D. $4.018\times {10}^{-2}$ dyne.

Answer 1

The correct option is A $1.018\times {10}^{-2}$ dyne

Here, $r=0.3mm=0.03cm,$
$v=1m{s}^{-1}=100cm{s}^{-1},$

$\eta =18\times {10}^{-5}$ poise.
According to stokes law, force of viscosity on rain drop is
$F=6\pi \eta rv$
$=6\times 3.142\times 18\times {10}^{-5}\times 0.03\times 100$ dyne
$=1.018\times {10}^{-3}$ dyne.