Mechanical Properties of Solids

JEE Main level, single correct, medium

107 of 118

Question 107 of 118

The velocity of water in a river is 18kmh−118km{h}^{-1} near the surface. If the river is 5m5m deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water=10−2poise={10}^{-2}poise.

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Answer: (B) 10−3Nm−2{10}^{-3}N{m}^{-2}

Viscous force =Fv=−ηA⋅dvdh={F}_{v}=-\eta A\cdot \dfrac{dv}{dh}
=10−3×A×18×10003600×5,[η=10−2poiseη=10−3decapoise(mks)]={10}^{-3}\times A\times \dfrac{18\times 1000}{3600\times 5},\left[\begin{matrix}\eta ={10}^{-2}poise \\ \eta ={10}^{-3}decapoise\left(mks\right)\end{matrix}\right]
∴FvA=185×3600=1200×5∴\dfrac{{F}_{v}}{A}=\dfrac{18}{5\times 3600}=\dfrac{1}{200\times 5}
⇒FvA=10−3Nm−2\Rightarrow \dfrac{{F}_{v}}{A}=1{0}^{-3}N{m}^{-2}

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