Water flows through a horizontal tube of diameter 8cm and length 1km at the rate of 40 liters/s. Find the pressure difference required if ${\mathrm{\eta }}_{\mathrm{water}}=0.01 poise$

$\begin{array}{l}\mathrm{d}=8\mathrm{cm}=8\times {10}^{-2}\mathrm{m}\Rightarrow \mathrm{r}=\frac{\mathrm{d}}{2}=4\times {10}^{-2}\mathrm{m}; L=1\mathrm{km}=1000\mathrm{m};\\ \mathrm{Q}=\frac{40\text{ lit }}{\sec }=\frac{40\times {10}^{-3}{\mathrm{m}}^3}{\sec }; \mathrm{\eta }=0.01\text{ poise }=\left(0.01\times {10}^{-1}\right)Pa-s\end{array}$
We know that, $\mathrm{Q}=\text{ flow rate }=\left(\frac{\pi P{r}^4}{8\eta L}\right)$
$\therefore$Pressure difference, $\mathrm{P}=\frac{\mathrm{Q}\left(8\eta L\right)}{{\mathrm{\pi r}}^4}$ 

On substituting the values, $\mathrm{P}=3.98\times {10}^5\frac{\mathrm{dyne}}{{\mathrm{cm}}^2}$