Water flows at a speed 5 cm `s^(-1)` through a pipe of radius 2 cm. The visosity of water is 0.001 PI. The Reynolds number and the nature of flow are respectively

Options

A. 2000, unsteady
B. 1500, turbulent
C. 1000, turbulent
D. 2500, laminar

Answer

To solve the problem, we need to calculate the Reynolds number (Re) using the formula:

\[ Re = \frac{\rho V d}{\eta} \]

Where:
- \(\rho\) = density of water
- \(V\) = velocity of water
- \(d\) = diameter of the pipe
- \(\eta\) = viscosity of water

### Step 1: Identify the given values
- Velocity \(V = 5 \, \text{cm/s} = 5 \times 10^{-2} \, \text{m/s}\)
- Radius of the pipe \(r = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}\)
- Diameter of the pipe \(d = 2r = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m}\)
- Density of water \(\rho = 1000 \, \text{kg/m}^3\)
- Viscosity of water \(\eta = 0.001 \, \text{poise} = 0.001 \times 0.1 \, \text{kg/(m·s)} = 10^{-4} \, \text{kg/(m·s)}\)

### Step 2: Substitute the values into the Reynolds number formula
Substituting the values we have into the Reynolds number formula:

\[ Re = \frac{(1000 \, \text{kg/m}^3)(5 \times 10^{-2} \, \text{m/s})(4 \times 10^{-2} \, \text{m})}{10^{-4} \, \text{kg/(m·s)}} \]

### Step 3: Calculate the numerator
Calculating the numerator:

\[ 1000 \times 5 \times 10^{-2} \times 4 \times 10^{-2} = 1000 \times 5 \times 4 \times 10^{-4} = 20000 \times 10^{-4} = 2 \, \text{kg m/s}^2 \]

### Step 4: Divide by the viscosity
Now, divide by the viscosity:

\[ Re = \frac{2 \, \text{kg m/s}^2}{10^{-4} \, \text{kg/(m·s)}} = 2 \times 10^{4} = 20000 \]

### Step 5: Conclusion about the nature of flow
The Reynolds number is \(20000\). According to fluid dynamics:
- If \(Re < 2000\), the flow is laminar.
- If \(Re > 2000\), the flow is turbulent.

Since \(20000 > 2000\), the flow is turbulent.

### Final Answer
The Reynolds number is \(20000\) and the nature of flow is turbulent.

---