**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.