A pipe of length `l_(1)`, closed at one end is kept in a chamber of gas of density `rho_(1)`. A second pipe open at both ends is placed in a second chamber of gas of density `rho_(2)`. The compressibility of both the gases is equal. Calculate the length of the second pipe if frquency of first overtone in both the cases is equal.

Options

A. `(4)/(3) l_1 sqrt((rho_2)/(rho_(1))`
B. `(4)/(3) l_1 sqrt(rho_1)/(rho_(2))`
C. `l_1 sqrt((rho_2)/(rho_(1))`
D. `l_1 sqrt(rho_1)/(rho_(2))`

Explanation

To solve the problem, we need to analyze the frequencies of the first overtone for both the closed and open pipes and set them equal to each other.

### Step-by-step Solution:

1. **Understand the Frequencies**:
- For a closed pipe (closed at one end), the frequency of the first overtone (n=2) is given by:
\[ f_1 = \frac{3}{4} \frac{V}{L_1} \]
- For an open pipe (open at both ends), the frequency of the first overtone (n=3) is given by:
\[ f_2 = \frac{4}{L_2} V \]

2. **Identify the Speed of Sound**:
- The speed of sound \( V \) in both gases can be expressed in terms of the compressibility \( \beta \) and density \( \rho \):
\[ V = \sqrt{\frac{\beta}{\rho}} \]
- Since the compressibility of both gases is equal, we can denote it as \( \beta \).

3. **Substituting the Speed of Sound**:
- For the closed pipe:
\[ f_1 = \frac{3}{4} \frac{1}{L_1} \sqrt{\frac{\beta}{\rho_1}} \]
- For the open pipe:
\[ f_2 = \frac{4}{L_2} \sqrt{\frac{\beta}{\rho_2}} \]

4. **Setting Frequencies Equal**:
- Since the frequencies of the first overtone are equal:
\[ \frac{3}{4} \frac{1}{L_1} \sqrt{\frac{\beta}{\rho_1}} = \frac{4}{L_2} \sqrt{\frac{\beta}{\rho_2}} \]

5. **Canceling Common Terms**:
- We can cancel \( \sqrt{\beta} \) from both sides:
\[ \frac{3}{4} \frac{1}{L_1 \sqrt{\rho_1}} = \frac{4}{L_2 \sqrt{\rho_2}} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 3 L_2 \sqrt{\rho_2} = 16 L_1 \sqrt{\rho_1} \]

7. **Solving for \( L_2 \)**:
- Rearranging the equation to find \( L_2 \):
\[ L_2 = \frac{16 L_1 \sqrt{\rho_1}}{3 \sqrt{\rho_2}} \]

### Final Result:
The length of the second pipe \( L_2 \) is given by:
\[ L_2 = \frac{16 L_1 \sqrt{\rho_1}}{3 \sqrt{\rho_2}} \]