Q. The velocity of the surface layer of water in a river of depth $10\ m$ is 5 $m\ s^{- 1}$. The shearing stress between the surface layer and the bottom layer is (Coefficient of viscosity of water, $\eta = 10^{- 3}$ SI units)
A) $0.6 \times 10^{- 3}N\ m^{- 2}$
B) $0.8 \times 10^{- 3}N\ m^{- 2}$
C) $0.5 \times 10^{- 3}N\ m^{- 2}$
D) $10^{- 3}N\ m^{- 2}$
Explanation
(c) Here, $\eta = 10^{- 3}Nm^{- 2}s,v = 5\ m\ s^{- 1};l = 10\ m$
Strain rate $= \frac{v}{l}$
Coefficient of viscosity, $\eta = \frac{Shearing\ stress}{Strain\ rate}$
$\therefore$ Shearing stress $= \eta \times$ Strain rate
$= \frac{(10^{- 3}N\ m^{- 2}s)(5\ m\ s^{- 1})}{(10\ m)} = 0.5 \times 10^{- 3}\ N\ m^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.