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(B) The shear stress $	au$ on the inner surface of the tube is given by $	au = \frac{F}{A} = \frac{P \cdot \pi r^2}{2 \pi r l} = \frac{Pr}{2l}$.By t

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$\frac{4\eta Q}{\pi \beta r^3}$

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(B) The shear stress $	au$ on the inner surface of the tube is given by $	au = \frac{F}{A} = \frac{P \cdot \pi r^2}{2 \pi r l} = \frac{Pr}{2l}$.By the definition of modulus of rigidity $\beta$,the shear strain $\phi$ is $\phi = \frac{	au}{\beta} = \frac{Pr}{2\beta l}$  ........$(1)$According to Poiseuille's law,the rate of flow $Q$ is given by $Q = \frac{\pi P r^4}{8 \eta l}$.From this,we can express the pressure difference $P$ as $P = \frac{8 \eta l Q}{\pi r^4}$  ........$(2)$Substituting equation $(2)$ into equation $(1)$:$\phi = \frac{r}{2 \beta l} \cdot \left( \frac{8 \eta l Q}{\pi r^4} ight) = \frac{4 \eta Q}{\pi \beta r^3}$.

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