The velocity of water in river is 180 km `h^(-1)` near the surface .If the river is 5 m deep,then the shearing stress between the surface layer and the bottom layer is ( cofficient of viscosity of water `eta =10^(-3)` Pa s)

Options

A. `0.6 xx 10^(-3) Nm^(-2)`
B. `0.8 xx 10^(-3) Nm^(-2)`
C. `0.5 xx 10^(-3) Nm^(-2)`
D. `10^(-3) Nm^(-2)`

Answer

C

Explanation

Here, `eta = 10^(-3) N m^(-2)s, v= 5 ms^(-1), l = 10m`,
As coefficient of viscosity `eta= ("Shearing stress")/("Strain rate")`
`:.` Shearing stress `= eta xx` strain rate
But strain rate `= (v)/(l)`
`:.` Shearing stress `= (eta v)/(l) = ((10^(-3) N m^(-2) s) (5 ms^(-1)))/((10m))`
`=0.5 xx 10^(-3) N m^(-2)`