The velocity of water in a river is $$18\,km/hr$$   near the surface. If the river is $$5\,m$$  deep, find the shearing stress between the horizontal layers of water. The co-efficient of viscosity of water $$ = {10^{ - 2}}{\text{poise}}{\text{.}}$$

Solution :
$$\eqalign{ & \eta = {10^{ - 2}}{\text{poise}} \cr & v = 18\,km/h = \frac{{1800}}{{3600}} = 5\,m/s\,\,l = 5\,cm \cr & {\text{Strain rate}} = \frac{v}{l} \cr & {\text{Coefficient of viscosity,}}\,\eta {\text{ = }}\frac{{{\text{shearing}}\,{\text{stress}}}}{{{\text{strain}}\,{\text{rate}}}} \cr & \therefore {\text{Shearing}}\,{\text{stress}} = \eta \times {\text{strain}}\,{\text{rate}} \cr & = {10^{ - 2}} \times \frac{5}{5} = {10^{ - 2}}N{m^{ - 2}} \cr} $$