JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
The velocity of water in a river is \(18\, km/h\) 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 \(= 10^{-2}\,,poise\)
- A \(10^{-1}\,N/m^2\)
- B \(10^{-2}\,N/m^2\)
- C \(10^{-3}\,N/m^2\)
- D \(10^{-4}\,N/m^2\)
Answer & Solution
Correct Answer
(B) \(10^{-2}\,N/m^2\)
Step-by-step Solution
Detailed explanation
\(\eta = {10^{ - 2}}poise\) \(v = 18\,km/h = \frac{{18000}}{{3600}} = 5m/s\) \(l = 5\,m\) \(Strain\,rate = \frac{v}{l}\) Coefficient of viscosity, \(\eta = \frac{{shearing\,stress}}{{strain\,rate}}\) \(\therefore \,Shearing\,stress = \eta \times strain\,rate\)…
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