The velocity profile for laminar flow of water between two fixed parallel plates is given by $u=0.01[1-1000y^{2}]$ where u is in m/s and y is in N-s/m$^{2}$ and the gap between the two plates is 2 cm. Calculate the shear stress on each plate.

Data: $u=0.01[1-1000y^{2}]$, where u$\longrightarrow$m/s, y$\longrightarrow$m

$\mu=10^{-3} N-s/m^{2}$, Gap t=2 cm=2$\times 10^{-2}$ m

To find: $T_{1}=?$, $T_{2}=?$

Plate (1), Let us consider two layers of water, onne is fixed with plate (1) and the other moves with flow,

∴dy=$\cfrac{t}{2} = \cfrac{2 \times 10^{-2}}{2}=10^{-2}m$

We have, $u=0.01[1-1000y^{2}] = 0.01[1-1000(10^{-2})^{2}]$ [$\because$ y=$10^{-2}$ m]

=-0.09 m/s

Now, by using Newton’s law of viscosity,

$T_{1}=\mu \cfrac{du}{dy}=10^{-3} \times \cfrac{(-0.09)}{10^{-2}}=-9 \times 10^{-3} N/m^{2}$

But, du=u-0=u=-0.09 m/s

Plate (2),

Shear stress will be same for plate (2) as distance and viscosity is same. (Refer- Shear stress distribution)

∴$T_{2}=-9 \times 10^{-3} N/m^{2}$