Correct Answer - Option 2 : 7.5 × 10
-5 Ns/m
2
Concept:
Shear stress due to viscosity between the parallel plates is given by
\(τ = μ\frac {du}{dy}\)
Where μ = Viscosity, \(\frac {du}{dy}\) = Rate of shear strain
Calculation:
Given:
τ = 1.5 N/m2, u = 40 cm/s = 0.4 m/s, y = 0.02 mm = 0.00002 m
Using \(τ = μ\frac {du}{dy}\)
\(1.5 = μ\frac {0.4}{0.00002}\)
μ = 7.5 × 10-5 Ns/m2