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Question

The space between two plates 20 cm×20 cm×1 cm,1 cm apart is filled with a liquid of viscosity 1 Poise. The upper plate is dragged to the right with a force of F=5 N keeping the lower plate stationary. The upper plate is moving at a constant speed u as shown in figure.


What will be the velocity in m/s of the flow at a point 0.5 cm below the lower surface of the upper plate if linear velocity profile is assumed for the flow ?
  1. 1.25
  2. 2.5
  3. 0.25
  4. 6.25


Solution

The correct option is D 6.25
Fv=ηAdudz
Magnitude of force, Fv=ηAdudz ...(i)
The ve sign represents that viscous force (Fv) opposes the relative motion between the fluid layer and the plate.
Velocity gradient is dudz=u0z
Thickness of fluid layer between the plates, z=0.01 m
For a linear velocity profile, velocity gradient dudz=constant
Surface area of the plate, A=20×20=400 cm2=0.04 m2
The magnitude of drag force = Magnitude of applied force on plate.
Fv=5 N
From Eq.(i),
5=(110)×(0.04)×dudz
dudz=1250 s1
uz=1250 s1
u=1250×0.01=12.5 m/s

Between the fluid at the upper plate and at 0.5 cm below the upper plate,
dudz=uu0.5×102
Putting the value of velocity gradient,
1250=uu0.5×102
uu=1250×0.5×102
u=u6.25=12.56.25
Hence, velocity at the point (at the given depth) in the fluid is,
u=6.25 m/s

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