# The shear stress at a point in a liquid is found to be 0.05 N/m2. The velocity gradient at the point is 0.2 s-1. What will be its viscosity?

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The shear stress at a point in a liquid is found to be 0.05 N/m2. The velocity gradient at the point is 0.2 s-1. What will be its viscosity?
1. 0.01 Ns/m2
2. 1.0 poise
3. 2.5 poise
4. 2.5 Ns/m2

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Correct Answer - Option 3 : 2.5 poise

Concept:

According to Newton’s law of viscosity, $τ = μ \frac{{du}}{{dy}}$  the shear stress is directly proportional to the rate of shear strain or the rate of angular deformation of a fluid particle. The fluid-particle tends to deform continuously when it is in motion.

$τ = μ \frac{{du}}{{dy}}$

Newton’s law of viscosity is a relationship between shear stress and the rate of shear strain.

Calculation:

Given:

τ = 0.05 N/m2, du/dy= 0.2 s-1, Viscosity, μ = ?

$τ = μ \frac{{du}}{{dy}}$

$0.05 = \mu \times 0.2= 0.25 \frac{Ns}{m^2}=2.5 ~poise$