Question

A plate of area $4{\text{m}}^2$ is made to move horizontally with a speed of $4\text{m/s}$ by applying a horizontal tangential force over the free surface of a liquid. If the depth of the liquid is $2\text{m}$ and the liquid is in contact with the bed which is stationary. Co-efficient of viscosity of liquid is $0.01\text{poise}$. The tangential force needed to move the plate is (in ${10}^{-3}\text{N})$

Answer 1

Given, velocity of free surface of liquid, $v_f=4\text{m/s}$,
velocity of the liquid in conatct with bed, $v_b=0\text{m/s}$

Velocity gradient between the bed and free surafce,
$\frac{\Delta v}{\Delta y}=\frac{v_f-v_b}{\text{depth}}$

$\Rightarrow \frac{\Delta v}{\Delta y}=\frac{4-0}{2-0}$

$\Rightarrow \frac{\Delta v}{\Delta y}=2{\text{sec}}^{-1}$

From newton's law of viscosity,

$|F|=\eta A\frac{\Delta v}{\Delta y}....\left(1\right)$

Where,
$|F|=$ The tangential force needed to move the liquid surface,
$\eta =0.01\text{poise}=$ Co-efficient of viscosity,
$A=4{\text{m}}^2$, surface area on which fluid is sliding.

Substituting the values in equation $\left(1\right)$,
$|F|=(0.01\times {10}^{-1})\times 4\times 2$

$=8\times {10}^{-3}\text{N}$

Accepted Answer: 8, 8.0, 8.00