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

Solution

Physics sir · 4 steps

Same bulk modulus, so the lighter gas is faster. Find the speed ratio, then match the frequencies.

1. Speed ratio. In a gas v = √(B/ρ). With the same B, v₁/v₂ = √(ρ₂/ρ₁) = √16 = 4, so v₁ = 4v₂.
2. Write both harmonics. A closed tube has harmonics f = (n v)/4L with n odd, so its 9th harmonic is (9 v₁)/4L₁. An open tube has f = (n v)/2L for every n, so its 4th harmonic is (4 v₂)/2L₂.

   (9 v₁)/4L₁ = (4 v₂)/2L₂

3. Put in numbers. Use v₁ = 4v₂ and L₁ = 10 cm: (9 × 4 v₂)/40 = (2 v₂)/L₂. The v₂ cancels, giving 36/40 = 2/L₂.
4. So. L₂ = 80/36 = 20/9 cm, which is option (b).

The idea to keep. When two pipes hold different gases, the speed changes too. Find v₁/v₂ from √(B/ρ) first, then equate nv/4L for the closed pipe with nv/2L for the open one.

Answer from the NTA final answer key. Solution by PrepWiser.