The cylindrical tube of a spray pump has a cross-section of ` 8 cm^(2)`. one end of which has 40 fine holes each of diameter 10 mm. If the liquid flow inside the tube is 0.15 m per minute, what is the speed of ejection of the liquid through the holes ?

Explanation

To solve the problem, we will use the principle of continuity, which states that the product of the cross-sectional area and the velocity of flow is constant along a streamline.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Cross-sectional area of the tube, \( A_1 = 8 \, \text{cm}^2 = 8 \times 10^{-4} \, \text{m}^2 \)
- Diameter of each hole, \( d = 10 \, \text{mm} = 0.01 \, \text{m} \)
- Number of holes, \( n = 40 \)
- Flow rate inside the tube, \( V_1 = 0.15 \, \text{m/min} = \frac{0.15}{60} \, \text{m/s} = 0.0025 \, \text{m/s} \)

2. **Calculate the Area of One Hole:**
- The area of one hole, \( A_h \), can be calculated using the formula for the area of a circle:
\[ A_h = \frac{\pi d^2}{4} = \frac{\pi (0.01)^2}{4} = \frac{\pi \times 0.0001}{4} \approx 7.854 \times 10^{-5} \, \text{m}^2 \]

3. **Calculate the Total Area of All Holes:**
- The total area of the 40 holes, \( A_2 \):
\[ A_2 = n \times A_h = 40 \times 7.854 \times 10^{-5} \approx 3.1416 \times 10^{-3} \, \text{m}^2 \]

4. **Apply the Equation of Continuity:**
- According to the equation of continuity:
\[ A_1 V_1 = A_2 V_2 \]
- Rearranging to find \( V_2 \):
\[ V_2 = \frac{A_1 V_1}{A_2} \]

5. **Substituting the Values:**
- Substitute \( A_1 \), \( V_1 \), and \( A_2 \) into the equation:
\[ V_2 = \frac{(8 \times 10^{-4}) \times (0.0025)}{3.1416 \times 10^{-3}} \]

6. **Calculate \( V_2 \):**
- Performing the calculation:
\[ V_2 \approx \frac{(2 \times 10^{-6})}{3.1416 \times 10^{-3}} \approx 0.000636 \, \text{m/s} \approx 0.636 \, \text{cm/s} \]

### Final Answer:
The speed of ejection of the liquid through the holes is approximately \( 0.636 \, \text{cm/s} \).