Question

The velocity of water in river is $9km/hr$ of the upper surface. The river is $10m$ deep. If the coefficient of viscosity of water is ${10}^{-2}poise$, then the shearing stress between horizontal layers of water is

• A. $0.25\times {10}^{-2}N/m^2$
• B. $0.25\times {10}^{-3}N/m^2$
• C. $0.5\times {10}^{-3}N/m^2$
• D. $0.75\times {10}^{-3}N/m^2$

Answer 1

The correct option is B $0.25\times {10}^{-3}N/m^2$

By Newton's law of viscosity, we have:

$F=\eta A\frac{dv}{dx}$

$\Rightarrow \tau =\frac{F}{A}=\eta \frac{dv}{dx}$

The velocity gradient is also known as the strain rate. The law states that shearing stress in a fluid is directly proportional to the strain rate. The constant of proportionality is called the coefficient of viscosity. A fluid has the property to yield under infinitesimal shear stress unlike solid objects, so stress is expressed in terms of strain rate and not strain.

Note that the SI unit of viscosity is decapoise, where 1 decapoise = 10 poise.

Substituting the values, we have:

Shearing stress $=\tau ={10}^{-3}\times \frac{5\times 9}{18\times 10}=0.25\times {10}^{-3}N/m^2$