Question
Download Solution PDFThe shear stress at a point in a liquid is found to be 0.02 N/m2. The velocity gradient at this point is 0.20 s-1. What will be the viscosity of the liquid (in Poise)?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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 = 0.02 N/m2, du/dy= 0.20 s-1, Viscosity, μ = ?
\(τ = μ \frac{{du}}{{dy}}\)
\(0.02 = \mu \times 0.2= 0.1 \frac{Ns}{m^2}=1 ~poise\)
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