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JEE Main 201419 AprilOnline
Exam
JEE Main
Year
2014
Date
19 April
Mode
Online
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2014 · 19 April · Online · Physics · Mechanical Properties of Solids

The velocity of water in a river is 18km h ^ - 1 near the surface....

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.

Options

  1. A10−4Nm−2{10}^{- 4}N{m}^{- 2}
  2. B10−3Nm−2{10}^{- 3}N{m}^{- 2}
  3. C10−2Nm−2{10}^{- 2}N{m}^{- 2}
  4. D10−1Nm−2{10}^{- 1}N{m}^{- 2}

Correct answer

B. 10−3Nm−2{10}^{- 3}N{m}^{- 2}

Step-by-step solution

Viscous force =Fv=−ηA⋅dvdh= {F}_{v} = - ηA \cdot \dfrac{dv}{dh}

= {10}^{- 3} \times A \times \dfrac{18 \times 1000}{3600 \times 5}, \left[η=10−2poiseη=10−3decapoise(mks)\begin{array}{ll}η = {10}^{- 2}poise \\ η = {10}^{- 3}decapoise\left(mks\right)\end{array}\right]

∴FvA=185×3600=1200×5∴ \dfrac{{F}_{v}}{A} = \dfrac{18}{5 \times 3600} = \dfrac{1}{200 \times 5}

⇒FvA=10−3Nm−2⇒ \dfrac{{F}_{v}}{A} = 1{0}^{- 3} N {m}^{- 2}

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