Correct Answer - Option 1 :

\({\tau _{wall}} = \;\frac{{\pi \mu {U_\infty }}}{{2\delta }}\)
__Concept:__

According to** Newton’s law of viscosity, shear stress **is given by:

**\(\tau = \mu \times \frac{{du}}{{dy}} = \;\mu \times \frac{{d\theta }}{{dy}}\;\)**

τ = shear stress, μ = coefficient of viscosity or absolute viscosity (or dynamic viscosity)

\(\frac{{du}}{{dy}} = Velocity\;gradient\)

\(\frac{{d\theta }}{{dt}}\; = Rate\;of\;angular\;deformation\;or\;Rate\;of\;shear\;strain\)

__Calculation:__

__Given:__

**Velocity profile **

**\(u\left( y \right) = \;{U_\infty } \times \sin \left( {\frac{{\pi y}}{{2\delta }}} \right),\;0 \le y \le \delta \)**

\(\tau = \mu \times \frac{{du}}{{dy}}\)

\(\tau = \mu \times \frac{{du}}{{dy}} = \mu \times \;{U_\infty } \times\frac{\partial }{{\partial y}}\left( {\sin \left( {\frac{{\pi y}}{{2\delta }}} \right)} \right)\)

\( = \mu \times {U_\infty } \times \cos \left( {\frac{{\pi y}}{{2\delta }}} \right) \times \frac{\pi }{{2\delta }}\)

As we have to find the shear stress at the wall,

y = 0

\({\tau _{wall}} = \mu \times {U_\infty } \times \frac{\pi }{{2\delta }} = \frac{{\pi \mu {U_\infty }}}{{2\delta }}\)