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.

Options

A. `50 ms^-1`
B. `5 ms^-1`
C. `0.65 ms^-1`
D. `0.5 ms^-1`

Explanation

To solve the problem, we can use the principle of conservation of mass, which is expressed through the continuity equation. The continuity equation states that the mass flow rate must be constant throughout the flow of an incompressible fluid.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Cross-sectional area of the tube, \( A_1 = 8 \, \text{cm}^2 = 8 \times 10^{-4} \, \text{m}^2 \)
- Number of holes, \( n = 40 \)
- Area of each hole, \( A_h = 10^{-8} \, \text{m}^2 \)
- Speed of liquid 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 Total Area of the Holes:**
- Total area of the holes, \( A_2 = n \times A_h = 40 \times 10^{-8} \, \text{m}^2 = 4 \times 10^{-7} \, \text{m}^2 \)

3. **Apply the Continuity Equation:**
- According to the continuity equation, \( A_1 \cdot v_1 = A_2 \cdot v_2 \)
- Rearranging gives us \( v_2 = \frac{A_1 \cdot v_1}{A_2} \)

4. **Substitute the Values:**
- Substitute \( A_1 \), \( v_1 \), and \( A_2 \) into the equation:
\[ v_2 = \frac{(8 \times 10^{-4} \, \text{m}^2) \cdot (0.0025 \, \text{m/s})}{(4 \times 10^{-7} \, \text{m}^2)} \]

5. **Calculate \( v_2 \):**
- Calculate the numerator:
\[ 8 \times 10^{-4} \cdot 0.0025 = 2 \times 10^{-6} \, \text{m}^3/\text{s} \]
- Now divide by the area of the holes:
\[ v_2 = \frac{2 \times 10^{-6}}{4 \times 10^{-7}} = 5 \, \text{m/s} \]

6. **Final Result:**
- The speed with which the liquid is ejected through the holes is \( v_2 = 5 \, \text{m/s} \).