• # question_answer When a fait plate of $0.1\,{{m}^{2}}$ area is pulled at a constant velocity of 30 cm/s parallel to another stationary plate pocaled at a distance 0.01 cm from it and the space in between filled with a fluid of dynamic viscosity $=\,\,0.01\,\,\text{Ns/}{{\text{m}}^{\text{2,}}}$ the force required to be applied is A) 0.3 N               B) 3 NC) 10 N                            D) 16 N

$A=0.1\,{{m}^{2}},$ $U=30\,\text{cm/s},$ $b=0.01\,\,\text{cm},\mu =0.01\,\,\text{N}\text{.s/}{{\text{m}}^{\text{2}}}$ $\tau =\frac{\mu U}{b}-\frac{1}{2}\,\,\frac{dp}{dx}\,(b-2y)$ $q=\frac{Ub}{2}-\frac{{{b}^{3}}}{12\mu }.\frac{dp}{dx}$ 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 $={{t}_{avg}}\times A=30\times 0.1=3N$