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(D) According to Newton's law of viscosity,the shearing stress $	au$ is given by $	au = \eta \frac{dv}{dy}$.Here,$	au = 10^{-3} \, N/m^2$,$\eta = 1

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$100$

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(D) According to Newton's law of viscosity,the shearing stress $	au$ is given by $	au = \eta \frac{dv}{dy}$.Here,$	au = 10^{-3} \, N/m^2$,$\eta = 10^{-2} \, Pa \cdot s$,and the velocity gradient $\frac{dv}{dy} = \frac{v}{h}$,where $v$ is the velocity of the upper layer and $h$ is the depth of the river.First,convert the velocity $v$ into $SI$ units: $v = 36 \, km/h = 36 	imes \frac{5}{18} \, m/s = 10 \, m/s$.Substituting the values into the formula: $10^{-3} = 10^{-2} 	imes \frac{10}{h}$.Rearranging for $h$: $h = \frac{10^{-2} 	imes 10}{10^{-3}} = \frac{10^{-1}}{10^{-3}} = 10^2 = 100 \, m$.Thus,the depth of the river is $100 \, m$.

The velocity of the upper layer of water in a river is $36 \, km/h$. The shearing stress between horizontal layers of water is $10^{-3} \, N/m^2$. The depth of the river is $h$. (The coefficient of viscosity of water is $10^{-2} \, Pa \cdot s$). Find the value of $h$ in meters.

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