A closed organ and an open organ tube filled by two different gases having the same bulk modulus but different densities \( \rho_1 \) and \( \rho_2 \), 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 \( \rho_1 : \rho_2 = 1 : 16 \), then the length of the open tube is:
Show Hint
- \( \frac{20}{7} \, \text{cm} \)
- \( \frac{15}{7} \, \text{cm} \)
- \( \frac{20}{9} \, \text{cm} \)
- \(\frac{15}{9} \, \text{cm} \)
The Correct Option is C
Solution and Explanation
Given Information: 9th harmonic of closed pipe: $\frac{9V_1}{4l_1}$ 4th harmonic of open pipe: $\frac{2V_2}{l_2}$ Step 1: Equate the frequencies The problem states that the 9th harmonic of the closed pipe is equal to the 4th harmonic of the open pipe. Therefore, we can write: \[ \frac{9V_1}{4l_1} = \frac{2V_2}{l_2} \]
Explanation: This equation sets the frequencies of the two harmonics equal to each other, which is a crucial step in relating the properties of the two pipes.
Step 2: Substitute the velocity expressions We know that the velocity of sound in a medium is given by $V = \sqrt{\frac{B}{\rho}}$, where $B$ is the bulk modulus and $\rho$ is the density. Substituting this into the previous equation, we get: \[ \frac{9}{4l_1} \sqrt{\frac{B}{\rho_1}} = \frac{2}{l_2} \sqrt{\frac{B}{\rho_2}} \]
Explanation: This step replaces the velocities $V_1$ and $V_2$ with their expressions in terms of bulk modulus and density, allowing us to relate the densities of the media in the pipes.
Step 3: Rearrange to find the ratio of lengths Rearranging the equation to find the ratio $\frac{l_2}{l_1}$, we get: \[ \frac{l_2}{l_1} = \frac{8}{9} \sqrt{\frac{\rho_1}{\rho_2}} \]
Explanation: This isolates the ratio of the lengths of the two pipes in terms of the ratio of the square roots of their densities.
Step 4: Substitute given values and solve for $l_2$ We are given that $l_2 = l_1 \times \frac{8}{9} \times \frac{1}{4}$. Substituting this into the equation, we get: \[ l_2 = l_1 \times \frac{8}{9} \times \frac{1}{4} = \frac{20}{9} \text{ cm} \]
Explanation: This step substitutes the given numerical values and simplifies the expression to find the value of $l_2$ in centimeters.
Final Result: The length of the open pipe $l_2$ is $\frac{20}{9}$ cm.
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