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 ?1.252.50.256.25

Solution

## The correct option is D 6.25Fv=−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=u−0z 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 s−1 ⇒uz=1250 s−1 ∴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=u−u′0.5×10−2 Putting the value of velocity gradient, 1250=u−u′0.5×10−2 u−u′=1250×0.5×10−2 ⇒u′=u−6.25=12.5−6.25 Hence, velocity at the point (at the given depth) in the fluid is, u′=6.25 m/s

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