1. Given data

• Flow rate: $100$ litres/minute
• Radius of pipe: $r = 5\text{ cm} = 0.05\text{ m}$
• Density of water: $\rho = 1000\,\text{kg m}^{-3}$
• Coefficient of viscosity: $\eta = 1\,\text{mPa s} = 10^{-3}\,\text{Pa s}$

We need Reynolds number:

$Re = \frac{\rho v D}{\eta}$

where $D = 2r$ is the diameter.

2. Convert flow rate into SI units

Since

$100\text{ litres} = 100 \times 10^{-3} = 0.1\text{ m}^3$

So flow rate is

$Q = \frac{0.1}{60}\text{ m}^3/\text{s} = 1.667 \times 10^{-3}\text{ m}^3/\text{s}$

3. Find cross-sectional area of the pipe

$A = \pi r^2 = \pi (0.05)^2 = \pi \times 2.5 \times 10^{-3}$

$A \approx 7.85 \times 10^{-3}\text{ m}^2$

4. Find average speed of flow

Using $Q = Av$,

$v = \frac{Q}{A} = \frac{1.667 \times 10^{-3}}{7.85 \times 10^{-3}}$

$v \approx 0.212\text{ m/s}$

5. Calculate Reynolds number

Diameter:

$D = 2r = 0.1\text{ m}$

Now,

$Re = \frac{\rho v D}{\eta} = \frac{1000 \times 0.212 \times 0.1}{10^{-3}}$

$Re = \frac{21.2}{10^{-3}} = 2.12 \times 10^4$

So Reynolds number is of the order of

$10^4$

6. Evaluate options

• A: $10^6$ ❌
• B: $10^4$ ✅
• C: $10^3$ ❌
• D: $10^2$ ❌

Therefore, the correct option is B.