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Question: The velocity of water in river is \(180kmh^{- 1}\) near the surface. If the river is 5 m deep, then ...

The velocity of water in river is 180kmh−1180kmh^{- 1} near the surface. If the river is 5 m deep, then the coefficient of viscosity of water is

A

10−2 N m−210^{- 2}\text{ N }\text{m}^{- 2}

B

10−3Nm−210^{- 3}Nm^{- 2}

C

10−4 N m−210^{- 4}\text{ N }\text{m}^{- 2}

D

10−5 N m−210^{- 5}\text{ N }\text{m}^{- 2}

Answer

10−3Nm−210^{- 3}Nm^{- 2}

Explanation

Solution

AS the velocity of water at the bottom of the river is zero.

dv=18kmh−1=18×518=5ms−1dv = 18kmh^{- 1} = 18 \times \frac{5}{18} = 5ms^{- 1}

Also, dx=5m,η=10−2poise=10−3Pasdx = 5m,\eta = 10^{- 2}poise = 10^{- 3}Pas

Force of viscosity, F=ηAdvdxF = \eta A\frac{dv}{dx}

∴\therefore Shearing stress, =FA=ηdvdx= \frac{F}{A} = \eta\frac{dv}{dx}

=10−3×55=10−3Nm−2= \frac{10^{- 3} \times 5}{5} = 10^{- 3}Nm^{- 2}