Question

A plate of dimensions (4m * 4m) and 0.01 m away from a fixed plate moves at 0-4 m s ¹. A force of 1 N is required to maintain this speed. Determine the coefficient of viscosity of the fluid between the plates.​

Answer 1

Answer:

Given data,

Area=a2=1×0.1=0.1m2

    

η=0.01poise

η=100.01=0.001deca poise

dv=0.1m/s

F=0.002N

We know that

As F=ηAdxdv

or dx=ηFAdv

=0.0020.001×(0.1)2×0.1=0.0005m

Explanation:

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Answer 2

Answer:

dy=0.025mm

=0.025×10

−3

m

Velocity of upper plate .

=60cm/s=0.6m/s

Force on upper plate

=2

m

2

N

= Shear stress , τ

du = change of velocity =u–0=0.6m/s

dy = change of distance =0.025×10−3m

Viscosity,

μ=

du/dy

τ

=

(0.6/0.25×10

−3

)

2

8.33×10−5

m

2

Ns

=8.33×10

−5

Pas

Explanation:

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