The velocity of water in a river is $18\, km/h$ 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 $= 10^{-2}\,,poise$
JEE MAIN 2014, Medium
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$\eta = {10^{ - 2}}poise$
$v = 18\,km/h = \frac{{18000}}{{3600}} = 5m/s$
$l = 5\,m$
$Strain\,rate = \frac{v}{l}$
Coefficient of viscosity,
$\eta = \frac{{shearing\,stress}}{{strain\,rate}}$
$\therefore \,Shearing\,stress = \eta \times strain\,rate$
$ = {10^{ - 2}} \times \frac{5}{5} = {10^{ - 2}}N{m^{ - 2}}$
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