Two narrow organ pipes, one open (length `l_(1)`) and the other closed (length `l_(2)`) are sounded in their respective fundamental modes. The beat frequency heard is `5 Hz`. If now the pipes are sounded in their first overtones, then also the beat frequency heard is `5 Hz`. Then:

Options

A. `(l_(1))/(l_(2))=1/2`
B. `(l_(1))/(l_(2))=1/1`
C. `(l_(1))/(l_(2))=3/2`
D. `(l_(1))/(l_(2))=2/3`

Answer

To solve the problem, we need to find the ratio of the lengths of two organ pipes (one open and one closed) based on the given beat frequencies in their fundamental modes and first overtones.

### Step-by-Step Solution:

1. **Identify the Frequencies for the Fundamental Modes:**
- For the open organ pipe (length \( l_1 \)), the frequency in the fundamental mode is given by:
\[ f_1 = \frac{V}{2l_1} \]
- For the closed organ pipe (length \( l_2 \)), the frequency in the fundamental mode is:
\[ f_2 = \frac{V}{4l_2} \]

2. **Set Up the Beat Frequency Equation for Fundamental Modes:**
- The beat frequency is given as \( 5 \, \text{Hz} \). Therefore, we can write:
\[ \left| f_1 - f_2 \right| = 5 \]
- Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \left| \frac{V}{2l_1} - \frac{V}{4l_2} \right| = 5 \]

3. **Simplify the Equation:**
- We can express the equation as:
\[ \frac{V}{2l_1} - \frac{V}{4l_2} = 5 \]
- Rearranging gives:
\[ \frac{V}{2l_1} = 5 + \frac{V}{4l_2} \]

4. **Identify the Frequencies for the First Overtones:**
- For the open organ pipe, the frequency in the first overtone (n=2) is:
\[ f_1' = \frac{V}{l_1} \]
- For the closed organ pipe, the frequency in the first overtone (n=2) is:
\[ f_2' = \frac{3V}{4l_2} \]

5. **Set Up the Beat Frequency Equation for First Overtones:**
- The beat frequency is still \( 5 \, \text{Hz} \):
\[ \left| f_1' - f_2' \right| = 5 \]
- Substituting the expressions for \( f_1' \) and \( f_2' \):
\[ \left| \frac{V}{l_1} - \frac{3V}{4l_2} \right| = 5 \]

6. **Simplify the Second Equation:**
- We can express this as:
\[ \frac{V}{l_1} - \frac{3V}{4l_2} = 5 \]
- Rearranging gives:
\[ \frac{V}{l_1} = 5 + \frac{3V}{4l_2} \]

7. **Solve the Two Equations:**
- We now have two equations:
1. \( \frac{V}{2l_1} = 5 + \frac{V}{4l_2} \) (Equation 1)
2. \( \frac{V}{l_1} = 5 + \frac{3V}{4l_2} \) (Equation 2)
- Multiply Equation 1 by 2:
\[ \frac{V}{l_1} = 10 + \frac{V}{2l_2} \]
- Set this equal to Equation 2:
\[ 10 + \frac{V}{2l_2} = 5 + \frac{3V}{4l_2} \]

8. **Rearranging and Solving for Lengths:**
- Rearranging gives:
\[ 5 = \frac{3V}{4l_2} - \frac{V}{2l_2} \]
- Simplifying this leads to:
\[ 5 = \frac{3V - 2V}{4l_2} = \frac{V}{4l_2} \]
- Thus, we find:
\[ l_2 = \frac{V}{20} \]

9. **Finding the Ratio \( \frac{l_1}{l_2} \):**
- From the first equation, substituting \( l_2 \):
\[ \frac{V}{2l_1} = 5 + \frac{20}{4V} \]
- Solving this will yield \( l_1 \) and ultimately the ratio:
\[ \frac{l_1}{l_2} = 1 \]

### Final Result:
The ratio of the lengths of the two pipes is:
\[ \frac{l_1}{l_2} = 1 \]