Q. 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 :
A) $\frac{15}7$ cm
B) $\frac{20}7$ cm
C) $\frac{15}9$ cm
D) $\frac{20}9$ cm
Explanation
[Solution by Dr. Abubacker] $v=\sqrt{B/\rho}$, so $\frac{v_1}{v_2}=\sqrt{\rho_2/\rho_1}=4$. Closed tube, 9th harmonic: $f=\frac{9v_1}{4(10)}$; open tube, 4th harmonic: $f=\frac{4v_2}{2L}$. Equating, $\frac{9v_1}{40}=\frac{2v_2}{L}\Rightarrow L=\frac{80}{9}\cdot\frac{v_2}{v_1}=\frac{20}9$ cm. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.