Water is flowing steadily through a horizontal pipe of non - uniform cross - section . If the velocity of water cross - section is ` 0.02 m^(2)` is 2m/s , what is the velocity of water at another point where the cross - section is ` 0.01 m^(2)` ?

Options

A. 1 m/s
B. 2 m/s
C. 3 m/s
D. 4 m/s

Answer

To solve the problem, we will use the principle of continuity for fluid flow, which states that the product of the cross-sectional area (A) and the velocity (v) of the fluid remains constant along a streamline. This can be expressed mathematically as:

\[ A_1 \cdot v_1 = A_2 \cdot v_2 \]

Where:
- \( A_1 \) is the cross-sectional area at the first point,
- \( v_1 \) is the velocity at the first point,
- \( A_2 \) is the cross-sectional area at the second point,
- \( v_2 \) is the velocity at the second point.

### Step-by-Step Solution:

1. **Identify the given values:**
- Cross-sectional area at point 1, \( A_1 = 0.02 \, m^2 \)
- Velocity at point 1, \( v_1 = 2 \, m/s \)
- Cross-sectional area at point 2, \( A_2 = 0.01 \, m^2 \)

2. **Apply the equation of continuity:**
\[ A_1 \cdot v_1 = A_2 \cdot v_2 \]

3. **Substitute the known values into the equation:**
\[ 0.02 \, m^2 \cdot 2 \, m/s = 0.01 \, m^2 \cdot v_2 \]

4. **Calculate the left side:**
\[ 0.04 \, m^3/s = 0.01 \, m^2 \cdot v_2 \]

5. **Solve for \( v_2 \):**
\[ v_2 = \frac{0.04 \, m^3/s}{0.01 \, m^2} \]
\[ v_2 = 4 \, m/s \]

### Final Answer:
The velocity of water at the point where the cross-section is \( 0.01 \, m^2 \) is \( 4 \, m/s \).

---