The velocity of water in a river is 18 km h^-1 near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water = 10^-2 poise.

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Solution

The correct option is B
The velocity gradient in vertical direction is
$\frac{d v}{d z} = \frac{18 k m h^{1}}{5 m} = 1.0 s^{1}$
The magnitude of the force of viscosity is
$F = η A \frac{d v}{d z}$
The shearing stress is
$\frac{F}{A} = η A \frac{d v}{d z} = (10^{- 2} p o i s e) (1.0 s^{- 1}) = 10^{- 3} \frac{N}{m^{2}}$

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The velocity of water in a river is 18 km h-1 near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water = 10-2 poise.

The correct option is B The velocity gradient in vertical direction is dvdz=18kmh15m=1.0s1 The magnitude of the force of viscosity is F=ηAdvdz The shearing stress is FA=ηAdvdz=(10−2poise)(1.0s−1)=10−3Nm2

The velocity of water in a river is 18 km h^-1 near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water = 10^-2 poise.