Correct Answer - Option 1 : 16.7 N / m

^{2} and 7.3 N / m

^{2}
__Concept:__

The shear stress distribution over a solid wall assuming the fluid is Newtonian is given as

τ = μ du/dy

where

μ is the dynamic viscosity

du/dy is the velocity gradient

__Calculation:__

Given,

μ = 5 Poise = 5 × 10^{-1 }N/m^{2}

u = 3 sin (5 πy)

\(\frac{{du}}{{dy}} = 3 \times 5\pi \cos \left( {5\pi y} \right)\)

\({\left. {\frac{{du}}{{dy}}} \right|_{y = 0.05}} = 15\pi \cos \left( {5\pi \times .05} \right) = \frac{{15\pi }}{{\sqrt 2 }} = 33.31\)

\({\left. {\frac{{du}}{{dy}}} \right|_{y = 0.12\;m}} = 15\pi \cos \left( {5\pi \times 0.12} \right) = - 14.56\;\)

Now,

\({\tau _{y = 0.05}} = \mu {\left. {\frac{{du}}{{dy}}} \right|_{y = 0.05}} = 5 \times {10^{ - 1}} \times 33.31 = 16.66\;N/{m^2}\)

\({\tau _{y = 0.12}} = \;\mu {\left. {\frac{{du}}{{dy}}} \right|_{y = 0.12}} = 5 \times {10^{ - 1}} \times \left( { - 14.56} \right) = - 7.28\;N/{m^2}\)