Water is moving with a speed of `5.18 ms^(-1)` through a pipe with a cross-sectional area of `4.20 cm^2`. The water gradually descends `9.66 m` as the pipe increase in area to `7.60 cm^2`. The speed of flow at the lower level is

Options

A. `2.86 ms^(-1)`
B. `3.0 ms^(-1)`
C. `3.82 ms^(-1)`
D. `5.7 ms^(-1)`

Explanation

To solve the problem, we will use the principle of continuity for fluids, which states that the product of the cross-sectional area and the velocity of the fluid must remain constant along a streamline. This can be expressed mathematically as:

\[ A_1 V_1 = A_2 V_2 \]

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. **Convert the Cross-sectional Areas to Square Meters**:
- Given \( A_1 = 4.20 \, \text{cm}^2 \) and \( A_2 = 7.60 \, \text{cm}^2 \).
- Convert these areas to square meters:
\[ A_1 = 4.20 \, \text{cm}^2 = 4.20 \times 10^{-4} \, \text{m}^2 \]
\[ A_2 = 7.60 \, \text{cm}^2 = 7.60 \times 10^{-4} \, \text{m}^2 \]

2. **Identify the Given Values**:
- The initial velocity \( V_1 = 5.18 \, \text{m/s} \).

3. **Use the Continuity Equation to Find \( V_2 \)**:
- Rearranging the continuity equation gives:
\[ V_2 = \frac{A_1 V_1}{A_2} \]
- Substitute the known values:
\[ V_2 = \frac{(4.20 \times 10^{-4} \, \text{m}^2)(5.18 \, \text{m/s})}{7.60 \times 10^{-4} \, \text{m}^2} \]

4. **Calculate \( V_2 \)**:
- First, calculate the numerator:
\[ 4.20 \times 10^{-4} \times 5.18 = 2.1736 \times 10^{-3} \, \text{m}^3/\text{s} \]
- Now, divide by \( A_2 \):
\[ V_2 = \frac{2.1736 \times 10^{-3}}{7.60 \times 10^{-4}} \approx 2.86 \, \text{m/s} \]

5. **Final Result**:
- The speed of flow at the lower level is approximately \( V_2 \approx 2.86 \, \text{m/s} \).