Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small holes. Consider one such fountain which is produced by a pipe of internal diameter ? Cm in which water flows at a rate `30 ms^(-1)`. The enclosure has `100` holes of diameter `0.5 cm` The velocity of water coming out of the holes is `("in ms"^(-1))`

Options

A. `0.48`
B. 96
C. 24
D. 48

Answer

To solve the problem, we will use the principle of conservation of mass, specifically the equation of continuity, which states that the mass flow rate must be constant throughout a fluid system.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Internal diameter of the pipe, \( D_1 = ? \) cm (not specified in the question, but we will assume it as 1 cm for calculation).
- Velocity of water in the pipe, \( v_1 = 30 \, \text{m/s} \).
- Number of holes, \( n = 100 \).
- Diameter of each hole, \( D_2 = 0.5 \, \text{cm} \).

2. **Convert Units:**
- Convert the diameters from cm to meters:
- \( D_1 = 1 \, \text{cm} = 0.01 \, \text{m} \)
- \( D_2 = 0.5 \, \text{cm} = 0.005 \, \text{m} \)

3. **Calculate the Areas:**
- Area of the pipe (wide end):
\[ A_1 = \pi \left( \frac{D_1}{2} \right)^2 = \pi \left( \frac{0.01}{2} \right)^2 = \pi \left( 0.005 \right)^2 = \pi \times 0.000025 \approx 7.85 \times 10^{-5} \, \text{m}^2 \]
- Area of one hole:
\[ A_2 = \pi \left( \frac{D_2}{2} \right)^2 = \pi \left( \frac{0.005}{2} \right)^2 = \pi \left( 0.0025 \right)^2 = \pi \times 0.00000625 \approx 1.96 \times 10^{-5} \, \text{m}^2 \]
- Total area of all holes:
\[ A_{total} = n \times A_2 = 100 \times 1.96 \times 10^{-5} \approx 1.96 \times 10^{-3} \, \text{m}^2 \]

4. **Apply the Equation of Continuity:**
- According to the equation of continuity:
\[ A_1 v_1 = A_{total} v_2 \]
- Rearranging gives us:
\[ v_2 = \frac{A_1 v_1}{A_{total}} \]

5. **Substituting the Values:**
- Substitute the values we calculated:
\[ v_2 = \frac{(7.85 \times 10^{-5}) \times 30}{1.96 \times 10^{-3}} \]
- Calculate \( v_2 \):
\[ v_2 \approx \frac{2.355 \times 10^{-3}}{1.96 \times 10^{-3}} \approx 1.20 \, \text{m/s} \]

6. **Final Result:**
- The velocity of water coming out of the holes is approximately \( 1.20 \, \text{m/s} \).