1. Wave speed in gases

For a gas, speed of sound is $v = \sqrt{\frac{B}{\rho}}$ where $B$ is bulk modulus and $\rho$ is density.

Since both gases have the same bulk modulus, their speeds are: $v_1 = \sqrt{\frac{B}{\rho_1}}, \qquad v_2 = \sqrt{\frac{B}{\rho_2}}$

Given $\rho_1 : \rho_2 = 1:16$ So, $\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{16} = 4$ Hence, $v_1 = 4v_2$

2. Frequency of harmonics

Closed organ tube

For a closed tube, only odd harmonics are present, and the $n$th allowed harmonic has frequency: $f_n = \frac{n v}{4L} \quad (n=1,3,5,\dots)$

The 9th harmonic of the closed tube is therefore: $f_c = \frac{9v_1}{4L_c}$

Given: $L_c = 10\text{ cm}$

So, $f_c = \frac{9v_1}{4\times 10}$

Open organ tube

For an open tube, all harmonics are present: $f_n = \frac{n v}{2L}$

The 4th harmonic of the open tube is: $f_o = \frac{4v_2}{2L_o} = \frac{2v_2}{L_o}$

3. Given condition: frequencies are equal

$\frac{9v_1}{4L_c} = \frac{4v_2}{2L_o}$ Substitute $L_c=10$ cm: $\frac{9v_1}{40} = \frac{2v_2}{L_o}$

Now use $v_1 = 4v_2$: $\frac{9(4v_2)}{40} = \frac{2v_2}{L_o}$ $\frac{36v_2}{40} = \frac{2v_2}{L_o}$ $\frac{9v_2}{10} = \frac{2v_2}{L_o}$

Cancel $v_2$: $\frac{9}{10} = \frac{2}{L_o}$

Therefore, $L_o = \frac{20}{9}\text{ cm}$

4. Option check

The correct option is: $\boxed{\frac{20}{9}\text{ cm}}$ which is Option B.