To solve the problem, we will use the principle of continuity in fluid mechanics, which states that the product of the cross-sectional area (A) and the velocity (V) of a fluid is constant along a streamline. This can be expressed mathematically as:
\[ A_1 V_1 = A_2 V_2 \]
Where:
- \( A_1 \) = cross-sectional area at section 1
- \( V_1 \) = velocity at section 1
- \( A_2 \) = cross-sectional area at section 2
- \( V_2 \) = velocity at section 2
### Step-by-Step Solution:
1. **Identify the given values:**
- Area at section 1, \( A_1 = 2 \, \text{cm}^2 \)
- Velocity at section 1, \( V_1 = 5 \, \text{cm/s} \)
- Area at section 2, \( A_2 = 4 \, \text{cm}^2 \)
2. **Write the equation of continuity:**
\[
A_1 V_1 = A_2 V_2
\]
3. **Substitute the known values into the equation:**
\[
2 \, \text{cm}^2 \times 5 \, \text{cm/s} = 4 \, \text{cm}^2 \times V_2
\]
4. **Calculate the left-hand side:**
\[
10 \, \text{cm}^3/s = 4 \, \text{cm}^2 \times V_2
\]
5. **Solve for \( V_2 \):**
\[
V_2 = \frac{10 \, \text{cm}^3/s}{4 \, \text{cm}^2} = 2.5 \, \text{cm/s}
\]
6. **Conclusion:**
The velocity at section 2, \( V_2 \), is \( 2.5 \, \text{cm/s} \).
