A lawn sprinkler has `20` holes, each of cross-sectional area `2.0 x 10^(-2) cm^(2)`, and is connected to a hose pipe of cross sectional area `2.4 cm^(2)`. If the speed of water in the hose pipe is `1.5 ms^(-1)`, the speed of water as it emerges from the holes is

Options

A. `2.25ms^(-1)`
B. `4.5ms^(-1)`
C. `9ms^(-1)`
D. `18ms^(-1)`

Explanation

To find the speed of water as it emerges from the holes of the lawn sprinkler, we can use the principle of conservation of mass, which is expressed through the equation of continuity. The equation of continuity states that the product of the cross-sectional area (A) and the velocity (v) of a fluid must remain constant along a streamline.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of holes (n) = 20
- Cross-sectional area of each hole (A_hole) = \(2.0 \times 10^{-2} \, \text{cm}^2\)
- Cross-sectional area of the hose pipe (A_pipe) = \(2.4 \, \text{cm}^2\)
- Speed of water in the hose pipe (v_pipe) = \(1.5 \, \text{m/s}\)

2. **Convert Areas to Consistent Units:**
- Convert the area of the holes from cm² to m²:
\[ A_hole = 2.0 \times 10^{-2} \, \text{cm}^2 = 2.0 \times 10^{-2} \times 10^{-4} \, \text{m}^2 = 2.0 \times 10^{-6} \, \text{m}^2 \]
- Total cross-sectional area of all holes:
\[ A_{total\_holes} = n \times A_hole = 20 \times 2.0 \times 10^{-6} \, \text{m}^2 = 4.0 \times 10^{-5} \, \text{m}^2 \]

3. **Apply the Equation of Continuity:**
The equation of continuity can be written as:
\[ A_{pipe} \cdot v_{pipe} = A_{total\_holes} \cdot v_{holes} \]
Where \(v_{holes}\) is the speed of water emerging from the holes.

4. **Substitute the Known Values:**
\[ 2.4 \times 10^{-4} \, \text{m}^2 \cdot 1.5 \, \text{m/s} = 4.0 \times 10^{-5} \, \text{m}^2 \cdot v_{holes} \]

5. **Calculate the Left Side:**
\[ 2.4 \times 10^{-4} \cdot 1.5 = 3.6 \times 10^{-4} \, \text{m}^3/s \]

6. **Solve for \(v_{holes}\):**
\[ v_{holes} = \frac{3.6 \times 10^{-4}}{4.0 \times 10^{-5}} = 9 \, \text{m/s} \]

### Final Answer:
The speed of water as it emerges from the holes is \(9 \, \text{m/s}\).