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