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The distance between two pistons is 0.02 mm and the viscous fluid flowing through them produces a force of 1.5 N/m2 to keep these two plates moving at a speed of 40 cm/s. Calculate the fluid viscosity in the middle of the plates.
1. 5.25 × 10-5 Ns/m2
2. 7.5 × 10-5 Ns/m2
3. 5.14 × 10-5 Ns/m2
4. 1.2 × 10-5 Ns/m2

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Correct Answer - Option 2 : 7.5 × 10-5 Ns/m2

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 

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