2. A Plate 0.02 mm distance from a fixed Plate moves at a velocity Of 0.6mls and requires a force of 1.962 N Per unit area to maitain this Speed. Determine the viscosity of the fluid between the plates?​

6.54 × 10⁻⁵ Pa-s

Explanation:

Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m

Since F = μAu/y

F/A = μu/y where F/A = force per unit area

Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²

So,  μ = F/A ÷ u/y

substituting the values of the variables into the equation, we have

μ = F/A ÷ u/y

μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m

μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s

μ = 6.54 × 10⁻⁵ Ns/m²

μ = 6.54 × 10⁻⁵ Pa-s

1 the 4 is sig