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 $9^{\text {th }}$ harmonic of closed tube is identical with $4^{\text {th }}$ harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is $\rho_{1}: \rho_{2}=1: 16$, then the length of the open tube is:

(C) $\frac{20}{9} \mathrm{~cm}$

(A) $\frac{20}{7} \mathrm{~cm}$

(B) $\frac{15}{7} \mathrm{~cm}$

(D) $\frac{15}{9} \mathrm{~cm}$

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Waves & Sound

A closed organ and an open organ tube filled by two different gases having same bulk modulus but different densities

ρ_{1}

and

ρ_{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

ρ_{1} : ρ_{2} = 1 : 16

, then the length of the open tube is:

Answer: (C)

The 9th harmonic of a closed pipe of length

ℓ_{1}

has frequency

f_{c l ose d} = \frac{9 v _{1}}{4 ℓ _{1}},

and the 4th harmonic of an open pipe of length

ℓ_{2}

has frequency

f_{o p e n} = \frac{2 v _{2}}{ℓ _{2}} .

Setting these equal:

\frac{9 v _{1}}{4 ℓ _{1}} = \frac{2 v _{2}}{ℓ _{2}} \Rightarrow \frac{ℓ _{2}}{ℓ _{1}} = \frac{8}{9} \cdot \frac{v _{2}}{v _{1}} .

Since both gases have the same bulk modulus

B

, the sound speed in each is

v = B / ρ

, so

\frac{v _{2}}{v _{1}} = \frac{ρ _{1}}{ρ _{2}} = \frac{1}{16} = \frac{1}{4} .

Therefore

ℓ_{2} = ℓ_{1} \times \frac{8}{9} \times \frac{1}{4} = 10 \times \frac{2}{9} = \frac{20}{9} cm .

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option