1. Wave speed in a gas

For sound in a gas, $v=\sqrt{\frac{1}{\rho \beta}}$ where $\rho$ is density and $\beta$ is compressibility.

Since compressibility is same in both pipes, $v \propto \frac{1}{\sqrt{\rho}}$ So, $\frac{v_1}{v_2}=\sqrt{\frac{\rho_2}{\rho_1}}$ where:

• $v_1$ = speed in closed pipe gas
• $v_2$ = speed in open pipe gas

2. First overtone frequencies

(a) Closed organ pipe of length $L$

Allowed harmonics are odd harmonics only.

• Fundamental: $f_{c,1}=\frac{v_1}{4L}$
• First overtone = 3rd harmonic: $f_c=\frac{3v_1}{4L}$

(b) Open organ pipe of length $\ell$

Let the open pipe length be $\ell$.

• Fundamental: $f_{o,1}=\frac{v_2}{2\ell}$
• First overtone = 2nd harmonic: $f_o=\frac{v_2}{\ell}$

3. Given same frequency in first overtone

So, $\frac{3v_1}{4L}=\frac{v_2}{\ell}$

Hence, $\ell=\frac{4L}{3}\cdot \frac{v_2}{v_1}$

Using $\frac{v_2}{v_1}=\sqrt{\frac{\rho_1}{\rho_2}}$ we get $\ell=\frac{4L}{3}\sqrt{\frac{\rho_1}{\rho_2}}$

4. Compare with given form

Given, $\ell=\frac{x}{3}L\sqrt{\frac{\rho_1}{\rho_2}}$

Comparing, $\frac{x}{3}L\sqrt{\frac{\rho_1}{\rho_2}}=\frac{4}{3}L\sqrt{\frac{\rho_1}{\rho_2}}$

So, $x=4$

5. Final answer

The required integer is $\boxed{4}$