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JEE Main 2014 (19 Apr, Shift 1) · Viscosity & Stokes' Law

Viscosity & Stokes' Law question from JEE Main 2014 (19 Apr, Shift 1)

Q31JEE Main 2014 · 19 APR · SHIFT 1Medium

Viscosity & Stokes' Law

The velocity of water in a river is 18 km/hr near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The co-efficient of viscosity of water = 10−-2 poise.

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Answer: (c)

  1. v=18 km/h=5v = 18\text{ km/h} = 5 m/s; bottom layer at rest, so velocity gradient =55=1 s−1= \frac{5}{5} = 1\text{ s}^{-1}.
  2. η=10−2\eta = 10^{-2} poise =10−3= 10^{-3} Pa s.
  3. Shearing stress =ηdvdx=10−3×1=10−3 N/m2= \eta\frac{dv}{dx} = 10^{-3}\times1 = 10^{-3}\text{ N/m}^2
  4. So the answer is (c) 10−3 N/m210^{-3}\text{ N/m}^2.
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