According to Newton’s law of viscosity.
\(\tau = \mu \frac{{du}}{{dy}}\)
& as the profile is assumed to be linear, then
\(\tau = \mu \frac{v}{y}\)
Here, y = 50 mm = 0.05 m, v = 3 m/s, μ = 0.44 kg/ms
\(\tau = \frac{{0.44 \times 3}}{{0.05}} = 26.4\;N/{m^2}\)
Key point
Force per unit area is the stress applied here.