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(A) According to Newton's law of viscosity,the shear stress $	au$ is given by $	au = \eta \frac{dv}{dy}$.Given the velocity profile $v = k \left( \f

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$\frac{\eta k}{a}$

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(A) According to Newton's law of viscosity,the shear stress $	au$ is given by $	au = \eta \frac{dv}{dy}$.Given the velocity profile $v = k \left( \frac{2y^2}{a^2} - \frac{y^3}{a^3} ight)$.Differentiating with respect to $y$,we get $\frac{dv}{dy} = k \left( \frac{4y}{a^2} - \frac{3y^2}{a^3} ight)$.Substituting this into the stress formula,$	au = \eta k \left( \frac{4y}{a^2} - \frac{3y^2}{a^3} ight)$.At $y = a$,the shear stress is $	au = \eta k \left( \frac{4a}{a^2} - \frac{3a^2}{a^3} ight) = \eta k \left( \frac{4}{a} - \frac{3}{a} ight) = \frac{\eta k}{a}$.

Water is flowing on a horizontal fixed surface,such that its flow velocity varies with $y$ (vertical direction) as $v = k \left( \frac{2y^2}{a^2} - \frac{y^3}{a^3} \right)$. If the coefficient of viscosity for water is $\eta$,what will be the shear stress between the layers of water at $y = a$?

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