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 area `10^-8 m^2`. If the liquid flows inside the tube with a speed of `0.15 m min^-1`, the speed with which the liquid is ejected through the holes is.

Explanation

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

### Step-by-step Solution:

1. **Identify Given Data:**
- Cross-sectional area of the tube, \( A = 8 \, \text{cm}^2 = 8 \times 10^{-4} \, \text{m}^2 \)
- Number of holes, \( n = 40 \)
- Area of each hole, \( a = 10^{-8} \, \text{m}^2 \)
- Speed of liquid inside the tube, \( v = 0.15 \, \text{m/min} \)

2. **Convert Speed to Standard Units:**
- Convert speed from meters per minute to meters per second:
\[ v = 0.15 \, \text{m/min} = \frac{0.15}{60} \, \text{m/s} = 0.0025 \, \text{m/s} \]

3. **Calculate Total Area of Holes:**
- Total area of the holes, \( A' \):
\[ A' = n \times a = 40 \times 10^{-8} \, \text{m}^2 = 4 \times 10^{-7} \, \text{m}^2 \]

4. **Apply the Equation of Continuity:**
- According to the equation of continuity:
\[ A \cdot v = A' \cdot v' \]
- Where \( v' \) is the speed of the liquid ejected through the holes.

5. **Substitute Known Values:**
- Substitute \( A \), \( v \), and \( A' \) into the equation:
\[ (8 \times 10^{-4} \, \text{m}^2) \cdot (0.0025 \, \text{m/s}) = (4 \times 10^{-7} \, \text{m}^2) \cdot v' \]

6. **Calculate \( v' \):**
- Calculate the left-hand side:
\[ 8 \times 10^{-4} \cdot 0.0025 = 2 \times 10^{-6} \, \text{m}^3/\text{s} \]
- Set the equation:
\[ 2 \times 10^{-6} = (4 \times 10^{-7}) \cdot v' \]
- Solve for \( v' \):
\[ v' = \frac{2 \times 10^{-6}}{4 \times 10^{-7}} = 5 \, \text{m/s} \]

### Final Answer:
The speed with which the liquid is ejected through the holes is \( 5 \, \text{m/s} \).

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