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

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

(A)20/7 cm
(B)15/7 cm
(C)20/9 cm✓ Correct
(D)15/9 cm

Answer: (C)

Solution

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

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

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