The velocity of water in a river is 18kmh−1 near the surface. If the river is 5m deep, then the shearing stress between the surface layer and the bottom layer is: (Given - coefficient of viscosity of water, η=10−3Pa.s)
A
10−2Nm−2
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B
10−3Nm−2
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C
10−4Nm−2
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D
10−5Nm−2
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Solution
The correct option is B10−3Nm−2 Since the velocity of water in contact with the bottom of the river is zero, Δv=18kmh−1=18×518=5ms−1 Also, Δz=5m,η=10−3Pa.s Velocity gradient is given as, ΔvΔz=55=1s−1 Viscous force is, Fv=ηAdvdz Shearing stress will be, σs=FvA=ηdvdz=ηΔvΔz ⇒σs=10−3×1=10−3Nm−2