Water from a pipe is coming at a rate of 100 liters per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of: (density of water = 100 kg/m$$^{3}$$, coefficient of viscosity of water = 1 mPa s)

We are asked to estimate the Reynolds number for water flowing through a pipe. The Reynolds number for flow in a circular pipe is defined by the formula

$$\displaystyle \text{Re} \;=\;\frac{\rho\,v\,D}{\eta},$$

where $$\rho$$ is the density of the fluid, $$v$$ is the average speed of flow, $$D$$ is the internal diameter of the pipe and $$\eta$$ is the dynamic (shear) viscosity.

First we turn every given quantity into S.I. units. The volumetric flow rate is given as 100 litres per minute. Because $$1\;\text{litre}=10^{-3}\;\text{m}^3$$ and $$1\;\text{min}=60\;\text{s}$$, we have

$$Q \;=\;100\times10^{-3}\,\text{m}^3\ \text{per}\ 60\ \text{s} \;=\;\frac{0.100}{60}\,\text{m}^3\!\big/\!\text{s} \;=\;1.6667\times10^{-3}\,\text{m}^3\!\big/\!\text{s}.$$

The radius of the pipe is 5 cm, that is $$r=5\;\text{cm}=0.05\;\text{m}.$$ Hence the diameter is

$$D \;=\;2r \;=\;2\times0.05\;\text{m}=0.10\;\text{m}.$$

Next we find the cross-sectional area of the pipe. Using $$A=\pi r^{2},$$

$$A \;=\;\pi\,(0.05\;\text{m})^{2} \;=\;\pi\,(2.5\times10^{-3}\,\text{m}^2) \;=\;7.8539\times10^{-3}\,\text{m}^2.$$

The average flow speed $$v$$ is the volume flow rate divided by the cross-sectional area, so

$$v \;=\;\frac{Q}{A} \;=\;\frac{1.6667\times10^{-3}}{7.8539\times10^{-3}} \;\text{m}\!/\!\text{s} \;=\;0.212\;\text{m/s}\;(\text{approximately}).$$

Now we substitute every value into the Reynolds number formula. For ordinary water the density is practically $$\rho=1000\;\text{kg/m}^{3}$$ and the viscosity is $$\eta=1\;\text{mPa·s}=1\times10^{-3}\;\text{Pa·s}$$ (remember $$1\;\text{Pa·s}=1\;\text{kg·m}^{-1}\text{s}^{-1}$$). Therefore

$$\text{Re} \;=\;\frac{\rho\,v\,D}{\eta} \;=\;\frac{(1000\;\text{kg/m}^3)(0.212\;\text{m/s})(0.10\;\text{m})} {1\times10^{-3}\;\text{Pa·s}} \;=\;\frac{21.2\;\text{kg·m}^{-1}\text{s}^{-1}}{10^{-3}\;\text{kg·m}^{-1}\text{s}^{-1}} \;=\;2.12\times10^{4}.$$

The number $$2.12\times10^{4}$$ is clearly of the order of $$10^{4}.$$ Hence, the correct answer is Option B.