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 **