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(D) According to Newton's law of viscosity,the shear stress $	au$ is given by $	au = \eta \frac{du}{dy}$.Given the velocity profile $u = \alpha \lef

Vedclass · Accepted Answer

$\frac{\alpha \eta}{h}$

Vedclass · Answer

(D) According to Newton's law of viscosity,the shear stress $	au$ is given by $	au = \eta \frac{du}{dy}$.Given the velocity profile $u = \alpha \left[ \frac{y}{h} - 2\left( \frac{y}{h} ight)^2 ight]$.Differentiating $u$ with respect to $y$:$\frac{du}{dy} = \alpha \left[ \frac{1}{h} - 2 \cdot 2 \left( \frac{y}{h} ight) \cdot \frac{1}{h} ight] = \alpha \left[ \frac{1}{h} - \frac{4y}{h^2} ight]$.At the base of the plate,$y = 0$.Therefore,the velocity gradient at the base is $\left( \frac{du}{dy} ight)_{y=0} = \alpha \left[ \frac{1}{h} - 0 ight] = \frac{\alpha}{h}$.The magnitude of the shear stress at the base is $	au = \eta \left| \frac{du}{dy} ight|_{y=0} = \eta \left( \frac{\alpha}{h} ight) = \frac{\alpha \eta}{h}$.

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