Fluid Mechanics

JEE Main level, numerical answer, medium

183 of 192

Question 183 of 192

The velocity of upper layer of water in a river is 36kmh−136km{h}^{-1}. Shearing stress between horizontal layers of water is 10−3Nm−2{10}^{-3}N{m}^{-2}. Depth of the river is _____ mm. (Co-efficient of viscosity of water is 10−2Pas{10}^{-2}Pas)

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Answer: 100

The viscous force formula is given by,
F=−ηAdvdyF=-\eta A\dfrac{dv}{dy}
Therefore, shear stress FA=−ηdvdy\dfrac{F}{A}=-\eta \dfrac{dv}{dy}. Here negative sign tells us that the direction of force is opposite to the velocity gradient.
⇒∣FA∣=ηdvdy\Rightarrow \left|\dfrac{F}{A}\right|=\eta \dfrac{dv}{dy}
Now v=36×518=10ms−1v=36\times \dfrac{5}{18}=10m{s}^{-1} and if the depth is dd
dvdy=10d\dfrac{dv}{dy}=\dfrac{10}{d}
⇒10−3=10−2×10d⇒d=100m\Rightarrow {10}^{-3}={10}^{-2}\times \dfrac{10}{d} \Rightarrow d=100m

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