JEE Physics 2025 Question with Solution

Given: A closed tube of length $L_1 = 10 \, \text{cm}$ contains a gas of density $\rho_1$, and an open tube of length $L_2$ contains a gas of density $\rho_2$. Both gases have the same bulk modulus. The 9th harmonic of the closed tube has the same frequency as the 4th harmonic of the open tube.

Find: The length $L_2$ of the open tube.

For the closed organ pipe,

$$f_{9} = \frac{9v_1}{4L_1}$$

and for the open organ pipe,

$$f_{4} = \frac{4v_2}{2L_2} = \frac{2v_2}{L_2}$$

Since these frequencies are identical,

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

So,

$$L_2 = \frac{8L_1v_2}{9v_1}$$

The speed of sound in a gas is

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

Since the bulk modulus $B$ is the same for both gases,

$$\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}}$$

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

$$\frac{v_1}{v_2} = \sqrt{\frac{16}{1}} = 4$$

Thus,

$$\frac{v_2}{v_1} = \frac{1}{4}$$

Substitute into the expression for $L_2$:

$$L_2 = \frac{8 \times 10 \, \text{cm} \times \frac{1}{4}}{9}$$ $$L_2 = \frac{20}{9} \, \text{cm}$$

Therefore, the length of the open tube is $\frac{20}{9} \, \text{cm}$. The correct option is C.

Given: Same bulk modulus for both gases and density ratio $\rho_1 : \rho_2 = 1 : 16$.

Find: The open tube length.

From

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

with the same $B$,

$$v \propto \frac{1}{\sqrt{\rho}}$$

Hence,

$$\frac{v_1}{v_2} = 4$$

and

$$\frac{v_2}{v_1} = \frac{1}{4}$$

Now use frequency equality:

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

Rearranging,

$$L_2 = \frac{8L_1v_2}{9v_1}$$

Substitute $L_1 = 10 \, \text{cm}$ and $\frac{v_2}{v_1} = \frac{1}{4}$:

$$L_2 = \frac{8 \times 10}{9} \times \frac{1}{4} = \frac{20}{9} \, \text{cm}$$

Therefore, the correct option is C.