Water flows through a pipe of diameter 8.0 cm with a speed of 10.0 m/s . It then enters a smaller pipe of diameter 3.0 cm . What is the speed of the water as it flows through the smaller pipe ?

Options

A. 1.4 m/s
B. 71 m/s
C. 2.8 m/s
D. 54 m/s

Answer

To solve the problem of water flowing through pipes of different diameters, we will use the principle of conservation of mass, specifically the equation of continuity for incompressible fluids. The equation states that the mass flow rate must remain constant throughout the flow.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Diameter of the larger pipe (D1) = 8.0 cm
- Speed of water in the larger pipe (V1) = 10.0 m/s
- Diameter of the smaller pipe (D2) = 3.0 cm

2. **Convert Diameters to Meters:**
- D1 = 8.0 cm = 0.08 m
- D2 = 3.0 cm = 0.03 m

3. **Calculate Cross-Sectional Areas:**
- The cross-sectional area (A) of a circular pipe is given by the formula:
\[ A = \frac{\pi}{4} D^2 \]
- For the larger pipe (A1):
\[ A1 = \frac{\pi}{4} (0.08)^2 = \frac{\pi}{4} \times 0.0064 = 0.0050265 \, \text{m}^2 \]
- For the smaller pipe (A2):
\[ A2 = \frac{\pi}{4} (0.03)^2 = \frac{\pi}{4} \times 0.0009 = 0.0007069 \, \text{m}^2 \]

4. **Apply the Equation of Continuity:**
- According to the equation of continuity:
\[ A1 \cdot V1 = A2 \cdot V2 \]
- Rearranging to find V2:
\[ V2 = \frac{A1 \cdot V1}{A2} \]

5. **Substitute the Values:**
- Substitute A1, V1, and A2 into the equation:
\[ V2 = \frac{0.0050265 \cdot 10.0}{0.0007069} \]

6. **Calculate V2:**
- Performing the calculation:
\[ V2 = \frac{0.050265}{0.0007069} \approx 71.1 \, \text{m/s} \]

7. **Conclusion:**
- The speed of the water as it flows through the smaller pipe is approximately **71.1 m/s**.