At $20^{\circ }C$ temperature in horizontal pipe pressure falls $60kPa/100m$ and water flow in pipe with rate $3lit/min$, what is approximate radius of pipe. (Coefficient of viscosity, $20^{o}C=10^{-3}Pa$)

Given,

Pressure difference, $\Delta P=60kPa$

Coefficient of viscosity, $\eta ={10}^{-3}Ns{m}^{-2}$

Length of pipe, $l=100m$

$Q=\frac{3\times {10}^{-3}}{60}=5\times {10}^{-5}{m}^3{s}^{-1}$

Using Poiseuille's law we can write

$Q=\frac{\Delta P\pi r^4}{8\eta l}$

$r^4=\frac{8\eta Q}{\pi }\times \frac{l}{\Delta P}$

$\frac{\Delta P}{l}=\frac{60kPa}{100}=\frac{60000}{100}=600Pa/m$

Substituting values

$r^4=\frac{8\times {10}^{-3}\times 5\times {10}^{-5}}{3.14}\times \frac{1}{600}$

$r=3.8\times {10}^{-3}m=0.38\text{cm}$

Hence, radius of pipe is $0.38cm$