Given:

$\begin{aligned} μ &=14 poise=1.4 Ns/m^2 \text{(Dynamic viscosity of oil)} \quad \text{(Since 10 poise}=1 Ns/m^2) \\ v&=2.5 m/s \text{(velocity of the top plate)} \\ h&=1.25 cm \text{(distance between the two plates)} \end{aligned}$

To find:

$\tau \text{(Shear stress in the oil)}$

Sol:

Now, we know by Newton’s law of viscosity for Newtonian Fluids,

$τ∝\dfrac{du}{dy} \text{(rate of strain)}$

$∴τ=μ \dfrac{du}{dy}$

where, μ→Proportionality constant(Coefficient of Dynamic Viscosity)

Now, Since h is small, we can assume the velocity profile to be linear and laminar between the two plates,

$∴ \dfrac{du}{dy}=\dfrac vh=constant=\dfrac{2.5}{1.25 \times 10^{-2}}/s=200/s $

$∴τ=1.4 \times 200=280 N/m^2$

Thus the Shear stress between the oil layers is $280N/m^2.$