A) 0.3 N
B) 3 N
C) 10 N
D) 16 N
Correct Answer: B
Solution :
\[A=0.1\,{{m}^{2}},\] \[U=30\,\text{cm/s},\] \[b=0.01\,\,\text{cm},\mu =0.01\,\,\text{N}\text{.s/}{{\text{m}}^{\text{2}}}\] \[\tau =\frac{\mu U}{b}-\frac{1}{2}\,\,\frac{dp}{dx}\,(b-2y)\] \[q=\frac{Ub}{2}-\frac{{{b}^{3}}}{12\mu }.\frac{dp}{dx}\] For \[q=0,\,\,\frac{dp}{dx}=\frac{6\mu U}{{{b}^{2}}}\] \[\tau =\frac{\mu U}{b}-\frac{1}{2}\times \frac{6\mu U}{{{b}^{2}}}\,(b-2y)\] \[\tau =\frac{\mu U}{b}-\frac{3\mu U}{{{b}^{2}}}\,(b-2y)\] Average shear stress \[{{\tau }_{avg}}=\frac{1}{b}\,\int_{0}^{b}{\left[ \frac{\mu U}{b}-\frac{3\mu U}{{{b}^{2}}}(b-2y) \right]}\,\,dy\] \[=\frac{1}{b}\,\,\left[ \frac{\mu Uy}{b}=\frac{3\mu U}{{{b}^{2}}}\,(by-{{y}^{2}}) \right]_{0}^{b}\] \[=\frac{1}{b}\,\,{{\left[ \mu U-3\mu U\,({{b}^{2}}-{{b}^{2}}) \right]}^{0}}=\frac{\mu U}{b}\] \[=\frac{0.01\times 0.3}{0.01\times {{10}^{-2}}}=30\,\text{N/}{{\text{m}}^{\text{2}}}\] Force required \[={{t}_{avg}}\times A=30\times 0.1=3N\]
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