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Question

A plate of area 4 m2 is made to move horizontally with a speed of 4 m/s by applying a horizontal tangential force over the free surface of a liquid. If the depth of the liquid is 2 m and the liquid is in contact with the bed which is stationary. Co-efficient of viscosity of liquid is 0.01 poise. The tangential force needed to move the plate is (in 103 N)

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Solution

Given, velocity of free surface of liquid, vf=4 m/s,
velocity of the liquid in conatct with bed, vb=0 m/s

Velocity gradient between the bed and free surafce,
ΔvΔy=vfvbdepth

ΔvΔy=4020

ΔvΔy=2 sec1

From newton's law of viscosity,

|F|=ηAΔvΔy....(1)

Where,
|F|= The tangential force needed to move the liquid surface,
η=0.01 poise= Co-efficient of viscosity,
A=4 m2, surface area on which fluid is sliding.

Substituting the values in equation (1),
|F|=(0.01×101)×4×2

=8×103 N

Accepted Answer: 8, 8.0, 8.00

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