Waves

JEE Main level, single correct, medium

84 of 142

Question 84 of 142

A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities ρ1\rho_1 and ρ2′\rho_{2^{\prime}}, respectively. The frequency of 9th 9^{\text {th }} harmonic of closed tube is identical with 4th 4^{\text {th }} harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is ρ1:ρ2=1:16\rho_1: \rho_2=1: 16, then the length of the open tube is :

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Answer: (D) 209 cm\frac{20}{9} \mathrm{~cm}

9th  harmonic of closed pipe =9 V14ℓ14th  harmonic of open pipe =2 V2ℓ2∴9 V14ℓ1=2 V2ℓ2∴94ℓ1 Bρ1=2ℓ2 Bρ2⇒ℓ2ℓ1=89ρ1ρ2ℓ2=ℓ1×89×14=209 cm\begin{aligned} & 9^{\text {th }} \text { harmonic of closed pipe }=\frac{9 \mathrm{~V}_1}{4 \ell_1} \\ & 4^{\text {th }} \text { harmonic of open pipe }=\frac{2 \mathrm{~V}_2}{\ell_2} \\ & \therefore \frac{9 \mathrm{~V}_1}{4 \ell_1}=\frac{2 \mathrm{~V}_2}{\ell_2} \\ & \therefore \frac{9}{4 \ell_1} \sqrt{\frac{\mathrm{~B}}{\rho_1}}=\frac{2}{\ell_2} \sqrt{\frac{\mathrm{~B}}{\rho_2}} \Rightarrow \frac{\ell_2}{\ell_1}=\frac{8}{9} \sqrt{\frac{\rho_1}{\rho_2}} \\ & \ell_2=\ell_1 \times \frac{8}{9} \times \frac{1}{4}=\frac{20}{9} \mathrm{~cm}\end{aligned}

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