A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities $$\rho_1$$ and $$\rho_2$$, respectively. The frequency of $$9^{th}$$ harmonic of closed tube is identical with $$4^{th}$$ harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is $$\rho_1 : \rho_2=1:16$$, then the length of the open tube is :

We need to find the length of the open organ tube given that the 9th harmonic of the closed tube equals the 4th harmonic of the open tube.

For a closed organ pipe, the nth harmonic frequency (only odd harmonics exist, but the problem says 9th harmonic meaning the harmonic number is 9):

$$f_{closed} = \frac{n \cdot v_1}{4L_1}$$

For an open organ pipe, the nth harmonic frequency:

$$f_{open} = \frac{n \cdot v_2}{2L_2}$$

The speed of sound in a gas is $$v = \sqrt{\frac{B}{\rho}}$$ where B is the bulk modulus.

Since both gases have the same bulk modulus B:

$$\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{16}{1}} = 4$$

The 9th harmonic of the closed tube equals the 4th harmonic of the open tube:

$$\frac{9 v_1}{4 L_1} = \frac{4 v_2}{2 L_2}$$

Substituting $$L_1 = 10$$ cm and $$v_1 = 4v_2$$:

$$\frac{9 \times 4v_2}{4 \times 10} = \frac{4 v_2}{2 L_2}$$

$$\frac{36 v_2}{40} = \frac{4 v_2}{2 L_2}$$

$$\frac{9 v_2}{10} = \frac{2 v_2}{L_2}$$

$$L_2 = \frac{2 \times 10}{9} = \frac{20}{9} \text{ cm}$$

The correct answer is Option 4: $$\frac{20}{9}$$ cm.