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An experiment was set-up to measure an unknown fluid's viscosity. Two flat plates are separated by a gap of 4.5 mm and move relative to each other at a velocity of 5 m/s. The space between them is occupied by the unknown viscosity. The motion of the plates is resisted by a shear stress of 10 Pa due to the viscosity of the fluid. Assuming that the velocity gradient of the fluid is constant, determine the coefficient of viscosity of the fluid.

Respuesta :

Answer:[tex]\mu =9\times 10^{-3}Pa-s[/tex]

Explanation:

Distance between Plates(dy)=4.5 mm

Relative Velocity(du)=5 m/s

We know shear stress is given by [tex]\tau =10 Pa[/tex]

[tex]\tau =\frac{\mu du}{dy}[/tex]

where du=relative Velocity

dy=Distance between Plates

[tex]10=\frac{\mu \times 5}{4.5\times 10^{-3}}[/tex]

[tex]\mu =9\times 10^{-3}Pa-s[/tex]

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