A closed organ pipe and an open organ pipe are filled by two different gases having same bulk modulus but different densities $\rho_1$ and $\rho_2$, respectively. The frequency of the $9^{\text{th}}$ harmonic of the closed tube is identical with the $4^{\text{th}}$ harmonic of the open tube. If the length of the closed tube is $10\ \text{cm}$ and the density ratio of the gases is $\rho_1 : \rho_2 = 1 : 16$, then the length of the open tube is
Answer: (D) $\dfrac{20}{9}\ \text{cm}$
Speed of sound $v = \sqrt{B/\rho}$, so with equal $B$:
$$\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} = 4$$
$9^{\text{th}}$ harmonic of the closed pipe: $f = \dfrac{9v_1}{4L_1}$. $4^{\text{th}}$ harmonic of the open pipe: $f = \dfrac{4v_2}{2L_2} = \dfrac{2v_2}{L_2}$.
$$\frac{9v_1}{4L_1} = \frac{2v_2}{L_2} \Rightarrow L_2 = \frac{8L_1}{9}\cdot\frac{v_2}{v_1} = \frac{8\times10}{9\times4} = \frac{20}{9}\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics