Q. The viscous force acting on a rain drop of radius 0.35 mm falling through air with a velocity of $1\ {ms}^{- 1}$, is $(ɳ = 2 \times 10^{- 4}\ N\ s\ m^{- 2})$
A) $6.6 \times 10^{- 6}\ N$
B) $6.6 \times 10^{- 5}\ N$
C) $1.32 \times 10^{- 7}\ N$
D) $13.2 \times 10^{- 7}\ N$
Explanation
(d) From the formula the viscous force is given by
$F = 6\pi ɳrv$
$\ \ \ = 6 \times \frac{22}{7} \times 2 \times 10^{- 4} \times 0.35 \times 10^{- 3} \times 1$
$\ \ = 13.2 \times 10^{- 7}N\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.