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Step-by-Step Solution
Step 1: Identify the Given Data
• Flow rate of water, $Q = 100 \text{ litres per minute}$.
• Radius of the pipe, $r = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$.
• Density of water, $\rho = 1000 \text{ kg/m}^3$.
• Coefficient of viscosity of water, $\eta = 1 \text{ mPa·s} = 1 \times 10^{-3} \text{ Pa·s}$.
• We need to find the Reynolds number, $R_e$.
Step 2: Convert the Flow Rate into SI Units
1 litre $= 10^{-3} \text{ m}^3$. Therefore,
$100 \text{ litres} = 100 \times 10^{-3} \text{ m}^3 = 0.1 \text{ m}^3.$
Since the flow rate is per minute, convert to seconds:
$Q = \dfrac{0.1 \text{ m}^3}{60 \text{ s}} = \dfrac{1}{600} \text{ m}^3/\text{s} = 1.6667 \times 10^{-3} \text{ m}^3/\text{s}.$
It is often convenient to keep it as a fraction:
$Q = \dfrac{5}{3} \times 10^{-3} \text{ m}^3/\text{s}.$
Step 3: Calculate the Cross-Sectional Area of the Pipe
The cross-sectional area $A$ of a circular pipe of radius $r$ is given by
$A = \pi r^2.$
Here, $r = 5 \times 10^{-2} \text{ m}$, so
$A = \pi \times \bigl(5 \times 10^{-2}\bigr)^2 = \pi \times 25 \times 10^{-4} \text{ m}^2 = 2.5 \pi \times 10^{-3} \text{ m}^2.$
Step 4: Compute the Flow Velocity
The flow velocity $v$ is given by the volumetric flow rate divided by the cross-sectional area:
$ v = \dfrac{Q}{A}.$
Using the values,
$ v = \dfrac{\dfrac{5}{3} \times 10^{-3} \text{ m}^3/\text{s}}{2.5 \pi \times 10^{-3} \text{ m}^2}
= \dfrac{5}{3} \times \dfrac{1}{2.5 \pi} \text{ m/s}
= \dfrac{2}{3 \pi} \text{ m/s (approximately)}
\approx 0.2 \text{ m/s}.$
Step 5: Calculate the Reynolds Number
The Reynolds number $R_e$ for flow in a pipe is given by
$ R_e = \dfrac{\rho \, v \, D}{\eta}, $
where $D$ is the diameter of the pipe $(D = 2r = 10 \text{ cm} = 0.1 \text{ m}).$
Substitute the known values:
$ R_e = \dfrac{(1000 \text{ kg/m}^3) \times (0.2 \text{ m/s}) \times (0.1 \text{ m})}{1 \times 10^{-3} \text{ Pa·s}}. $
Simplifying,
$ R_e = \dfrac{1000 \times 0.2 \times 0.1}{10^{-3}}
= \dfrac{20}{10^{-3}}
= 2 \times 10^4. $
Step 6: Conclude the Order of Magnitude
Since $R_e \approx 2 \times 10^4$, the order of the Reynolds number is $10^4$.
Hence, the correct answer is $10^4$.
