A highly viscous liquid of viscosity coefficient $\eta$ flows through a fixed horiwntal cylindrical tube (fixed from outer surface) of internal radius $r$, thickness $t (t << r)$ and length $l$. Volume of liquid flowing per;second is $Q$ and pressure difference across the tube is $P$. Modulus of rigidity of material of tube is $\beta$. Shear strain in the tube will be

Q: A highly viscous liquid of viscosity coefficient $\eta$ flows through a fixed horiwntal cylindrical tube (fixed from outer surface) of internal radius $r$, thickness $t (t << r)$ and length $l$. Volume of liquid flowing per;second is $Q$ and pressure difference across the tube is $P$. Modulus of rigidity of material of tube is $\beta$. Shear strain in the tube will be

${\beta=\frac{\left(\frac{F}{2 \pi r \ell}\right)}{\phi} \Rightarrow \phi=\frac{F}{2 \pi r \ell \beta}}$ $ \Rightarrow \phi = \frac{{P \times \pi {r^2}}}{{2\pi r\ell \beta }} = \frac{{Pr}}{{2\beta \ell }}$ ........$(1)$ Now rate of flow $\Rightarrow Q=\frac{\pi \operatorname{Pr}^{4}}{8 \eta \ell}$ ...........$(2)$ $\therefore \frac{\phi}{Q}=\frac{\operatorname{Pr}}{2 \beta \ell} \times \frac{8 \eta l}{\pi p r^{4}}=\frac{4 \eta}{\pi \beta r^{3}}$ $\therefore \phi=\frac{4 \eta Q}{\pi \beta r^{3}}$

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Solution

${\beta=\frac{\left(\frac{F}{2 \pi r \ell}\right)}{\phi} \Rightarrow \phi=\frac{F}{2 \pi r \ell \beta}}$

$ \Rightarrow \phi = \frac{{P \times \pi {r^2}}}{{2\pi r\ell \beta }} = \frac{{Pr}}{{2\beta \ell }}$ ........$(1)$

Now rate of flow

$\Rightarrow Q=\frac{\pi \operatorname{Pr}^{4}}{8 \eta \ell}$ ...........$(2)$

$\therefore \frac{\phi}{Q}=\frac{\operatorname{Pr}}{2 \beta \ell} \times \frac{8 \eta l}{\pi p r^{4}}=\frac{4 \eta}{\pi \beta r^{3}}$

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

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