A Plate at a distance of 1 mm from a  moves at 60 cm/sec and requires a force of 2 N per unit area i.e. 2 N/m² to maintain speed. Determine fluid viscosity between the plates.

Concept:

According to newton's law of viscosity-

This law states that the shear stress (\(\tau\)) on a fluid element layer is directly proportional to the rate of shear strain. The constant of proportionality is called the coefficient of viscosity.

\(\tau=\mu\, .({du\over dy})\)

Where \(\tau=\) shear stress \(={Force\over Area}\)

\(\mu=\) dynamic viscosity

\({du\over dy}=\) velocity gradient

Calculation:

Given data:

Distance between plates (y) = 1 mm or 1 × 10-3 m

Velocity of moving plate (\(v\)) = 60 cm/s or 0.6 m/s

Force per unit area on plate (F/A) = 2 N/m2

Dynamic viscosity (μ) =?

\(Shear\, stress(\tau)=\mu× ({v\over y})\)

\(\Rightarrow2=\mu×\left ({0.6\over 1\times 10^{-3}}\right)\)

\(\mu={2\times 1\times 10^{-3}\over 0.6}\)

\(\therefore\mu=3.33\times 10^{-3}\, N-s/m^2\)