"A closed organ and an open organ tube 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 :" (1) 20 / 7 (2) 15 / 7 (3) 20 / 9 (4) 15 / 9

1. Wave speed in a gas

For any gas, the speed of sound is

$$v = \sqrt{\frac{B}{\rho}}\,$$

where $B$ is the bulk modulus and $\rho$ is the density.
If $B$ is the same for two gases but their densities differ, the speed ratio becomes the square-root of the inverse density ratio.

2. Harmonic frequencies

Closed pipe (one end closed)

$$\text{Allowed frequencies:}\quad f_{n}^{(\text{closed})} = (2m-1)\,\frac{v_1}{4L_1},\qquad m = 1,2,3,\dots$$

Only odd multiples appear.

Open pipe (both ends open)

$$\text{Allowed frequencies:}\quad f_{n}^{(\text{open})} = n\,\frac{v_2}{2L_2},\qquad n = 1,2,3,\dots$$

All integer multiples appear.

3. Setting up the equality

The question says

$$\text{9th harmonic of closed pipe} \; = \; \text{4th harmonic of open pipe}.$$

That translates to

$$9\,\frac{v_1}{4L_1} \;=\; 4\,\frac{v_2}{2L_2}.$$

4. Using density information

Given $\rho_1 : \rho_2 = 1 : 16$,

$$\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2}} = \sqrt{\frac{1}{16}} = \frac14.$$

So $v_2 = \dfrac{v_1}{4}$.

5. Solving for the unknown length

Insert $v_2 = v_1/4$:

$$9\,\frac{v_1}{4L_1} = 4\,\frac{\tfrac{v_1}{4}}{2L_2} \;\Rightarrow\; 9\,\frac{v_1}{4L_1} = \frac{v_1}{2L_2}.$$

Cancel $v_1$ and cross-multiply:

$$\frac{9}{4L_1} = \frac{1}{2L_2} \;\Longrightarrow\; 18L_2 = 4L_1 \;\Longrightarrow\; L_2 = \frac{4}{18}L_1 = \frac{2}{9}L_1.$$

With $L_1 = 10\,\text{cm}$,

$$L_2 = \frac{2}{9} \times 10\,\text{cm} = \frac{20}{9}\,\text{cm} \;\approx\; 2.22\,\text{cm}.$$

Hence the correct option is (3) 20 / 9 cm.

What is the question?

We have two pipes full of gases that can make musical notes.

• Pipe-1 is closed at one end (like a flute blocked by a thumb).
• Pipe-2 is open at both ends (like a normal flute).

They are blown with different gases but the gases are equally ‘stiff’ (same bulk modulus). One gas is 16 times heavier (denser) than the other.

The problem tells us that the 9th sound of the closed pipe matches (same pitch) the 4th sound of the open pipe.
We know the closed pipe is 10 cm long and we must find the length of the open pipe.

How to think of it like a kid

1. A note’s pitch depends on how fast sound travels in the gas and how long the pipe is.
2. Heavier gas ⟹ sound travels slower (like running through water vs air).
   If a gas is 16× heavier, sound becomes 4× slower (because $\sqrt{16}=4$).
3. A closed pipe only sings the odd-numbered notes (1st, 3rd, 5th …!) but an open pipe can sing every note (1st, 2nd, 3rd …).
4. We match the formulas of these notes and solve for the unknown length.

Step-by-step solution

1. Write the formula for harmonics

   Closed pipe (one end sealed):

   $$f_{9}^{(c)} = 9\,\frac{v_1}{4L_1}$$

   Open pipe (both ends open):

   $$f_{4}^{(o)} = 4\,\frac{v_2}{2L_2} = 2\,\frac{v_2}{L_2}$$

2. Set them equal (given condition)

   $$9\,\frac{v_1}{4L_1} = 2\,\frac{v_2}{L_2}$$

3. Insert velocity ratio
   Same bulk modulus but $\rho_1: \rho_2 = 1:16$ ⇒ $v_2 = v_1/4$.

   $$9\,\frac{v_1}{4L_1} = 2\,\frac{\tfrac{v_1}{4}}{L_2}$$

4. Simplify

   $$\frac{9}{4L_1} = \frac{1}{2L_2}$$

   Cross-multiplying:

   $$18L_2 = 4L_1$$

5. Solve for $L_2$

   $$L_2 = \frac{4}{18}L_1 = \frac{2}{9}L_1$$

6. Insert $L_1 = 10\,\text{cm}$

   $$L_2 = \frac{2}{9} \times 10\,\text{cm} = \frac{20}{9}\,\text{cm}$$

7. Answer

   $$L_2 = \boxed{\dfrac{20}{9}\;\text{cm}}\qquad \text{(Option 3)}$$