Step 1: Understanding the Question:
A closed pipe and an open pipe contain different gases (densities \(\rho_1\) and \(\rho_2\)) but same compressibility (hence same bulk modulus \(B\)). Both vibrate in their first overtone with identical frequency. We need \(L_o\) in terms of \(L_c\) and densities.
Step 2: Key Formula or Approach:
For a closed pipe, first overtone = third harmonic: \(f_c = \frac{3v_1}{4L_c}\).
For an open pipe, first overtone = second harmonic: \(f_o = \frac{v_2}{L_o}\).
Speed of sound in a gas: \(v = \sqrt{\frac{B}{\rho}}\). Since \(B\) is same, \(v \propto \frac{1}{\sqrt{\rho}}\).
Step 3: Detailed Explanation:
Let \(v_1 = \sqrt{B/\rho_1}\) and \(v_2 = \sqrt{B/\rho_2}\). Equate frequencies:
\[
\frac{3v_1}{4L_c} = \frac{v_2}{L_o} \quad \Rightarrow \quad L_o = \frac{4L_c}{3} \cdot \frac{v_2}{v_1}.
\]
Now \(\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2}}\). Thus:
\[
L_o = \frac{4L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}.
\]
In the given options, option (A) writes \(\sqrt{\frac{\rho}{\rho_2}}\) where \(\rho\) stands for \(\rho_1\). Hence (A) is correct.
Step 4: Final Answer:
The length of the open organ pipe is \(\frac{4L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}\), which corresponds to option (A).