An organ pipe `P_(1)`, closed at one end and containing a gas of density `rho_(1)` is vibrating in its first harmonic. Another organ pipe `P_(2)`, open at both ends and containing a gas of density `rho_(2)`, is vibrating in its third harmonic. Both the pipes are in resonance with a given tuning fork. If the compressibility of gases is equal in both pipes, the ratio of the lengths of `P_(1)` and `P_(2)` is (assume the given gases to be monoatomic)

Options

A. `(1)/(3)`
B. `3`
C. `(1)/(6)sqrt((rho_(1))/(rho_(2)))`
D. `(1)/(6)sqrt((rho_(2))/(rho_(1)))`

Explanation

To solve the problem, we need to determine the ratio of the lengths of the two organ pipes, \( P_1 \) and \( P_2 \), based on the information provided about their harmonics and the gases they contain.

### Step-by-Step Solution:

1. **Understanding the Harmonics**:
- For pipe \( P_1 \) (closed at one end), the first harmonic (fundamental frequency) has a wavelength \( \lambda_1 \) given by:
\[ \lambda_1 = 4L_1 \]
where \( L_1 \) is the length of pipe \( P_1 \).
- For pipe \( P_2 \) (open at both ends), the third harmonic has a wavelength \( \lambda_2 \) given by:
\[ \lambda_2 = \frac{2L_2}{3} \]
where \( L_2 \) is the length of pipe \( P_2 \).

2. **Relating Frequency and Wavelength**:
- The speed of sound \( v \) in a gas is given by:
\[ v = \sqrt{\frac{B}{\rho}} \]
where \( B \) is the bulk modulus and \( \rho \) is the density of the gas.
- Since the compressibility is equal for both gases, we can express the speed of sound for each pipe as:
\[ v_1 = k \sqrt{\rho_1} \quad \text{and} \quad v_2 = k \sqrt{\rho_2} \]
where \( k \) is a constant related to compressibility.

3. **Finding Frequencies**:
- The frequency \( f_1 \) of the first harmonic in pipe \( P_1 \) is given by:
\[ f_1 = \frac{v_1}{\lambda_1} = \frac{v_1}{4L_1} \]
- The frequency \( f_2 \) of the third harmonic in pipe \( P_2 \) is given by:
\[ f_2 = \frac{v_2}{\lambda_2} = \frac{v_2}{\frac{2L_2}{3}} = \frac{3v_2}{2L_2} \]

4. **Setting Frequencies Equal**:
- Since both pipes are in resonance with the same tuning fork, we have:
\[ f_1 = f_2 \]
- Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{v_1}{4L_1} = \frac{3v_2}{2L_2} \]

5. **Substituting Speeds of Sound**:
- Substituting \( v_1 \) and \( v_2 \):
\[ \frac{k \sqrt{\rho_1}}{4L_1} = \frac{3k \sqrt{\rho_2}}{2L_2} \]
- The \( k \) cancels out:
\[ \frac{\sqrt{\rho_1}}{4L_1} = \frac{3\sqrt{\rho_2}}{2L_2} \]

6. **Rearranging for Length Ratio**:
- Rearranging gives:
\[ \frac{L_1}{L_2} = \frac{3\sqrt{\rho_2}}{2\sqrt{\rho_1}} \cdot 4 = \frac{6\sqrt{\rho_2}}{\sqrt{\rho_1}} \]

7. **Final Ratio**:
- Thus, the ratio of the lengths of the pipes is:
\[ \frac{L_1}{L_2} = \frac{6\sqrt{\rho_2}}{\sqrt{\rho_1}} \]