A closed organ and an open organ tube 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 :

Step 1: Frequency Formula for Organ Pipes

• Closed Pipe (Odd Harmonics Only): $$f_n = \frac{n v}{4L_c}, \quad n = 1,3,5,7,\dots$$
• Open Pipe (All Harmonics): $$f_m = \frac{m v}{2L_o}, \quad m = 1,2,3,4,\dots$$
• Speed of Sound in a Gas: $$v = \sqrt{\frac{B}{\rho}}$$

For closed pipe (9th harmonic, $n = 9$):

$$f_9 = \frac{9 v_1}{4L_c}$$

For open pipe (4th harmonic, $m = 4$):

$$f_4 = \frac{4 v_2}{2L_o} = \frac{2 v_2}{L_o}$$

Since the two frequencies are equal:

$$\frac{9 v_1}{4L_c} = \frac{2 v_2}{L_o}$$

 Step 2: Relating Speeds Using Density Ratio

Since both gases have the same bulk modulus, the speed of sound ratio is:

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

Thus, $v_1 = 4 v_2$.

Substituting $v_1 = 4 v_2$ in the equation:

$$\frac{9 (4 v_2)}{4L_c} = \frac{2 v_2}{L_o}$$

$$L_o = \frac{2 L_c}{9}$$

Given $L_c = 10$ cm:

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