Question

The space between two plates $20\text{cm}\times 20\text{cm}\times 1\text{cm},1\text{cm}$ apart is filled with a liquid of viscosity $1\text{Poise}$. The upper plate is dragged to the right with a force of $F=5\text{N}$ 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 $\text{m/s}$ of the flow at a point $0.5\text{cm}$ below the lower surface of the upper plate if linear velocity profile is assumed for the flow ?

• A. $1.25$
• B. $2.5$
• C. $6.25$
• D. $0.25$

Answer 1

The correct option is C $6.25$

$F_v=-\eta A\frac{du}{dz}$
Magnitude of force, $F_v=\eta A\frac{du}{dz}...\left(i\right)$
The $-ve$ sign represents that viscous force $\left(F_v\right)$ opposes the relative motion between the fluid layer and the plate.
Velocity gradient is $\frac{du}{dz}=\frac{u-0}{z}$
Thickness of fluid layer between the plates, $z=0.01\text{m}$
For a linear velocity profile, velocity gradient $\frac{du}{dz}=\text{constant}$
Surface area of the plate, $A=20\times 20=400{\text{cm}}^2=0.04{\text{m}}^2$
The magnitude of drag force $=$ Magnitude of applied force on plate.
$\therefore F_v=5\text{N}$
From Eq.$\left(i\right)$,
$5=\left(\frac{1}{10}\right)\times \left(0.04\right)\times \frac{du}{dz}$
$\Rightarrow \frac{du}{dz}=1250{\text{s}}^{-1}$
$\Rightarrow \frac{u}{z}=1250{\text{s}}^{-1}$
$\therefore u=1250\times 0.01=12.5\text{m/s}$

Between the fluid at the upper plate and at $0.5\text{cm}$ below the upper plate,
$\frac{du}{dz}=\frac{u-u^{\prime }}{0.5\times {10}^{-2}}$
Putting the value of velocity gradient,
$1250=\frac{u-u^{\prime }}{0.5\times {10}^{-2}}$
$u-u^{\prime }=1250\times 0.5\times {10}^{-2}$
$\Rightarrow u^{\prime }=u-6.25=12.5-6.25$
Hence, velocity at the point (at the given depth) in the fluid is,
$u^{\prime }=6.25\text{m/s}$