Question

In a resonance tube, the first and second resonance are heard when water level is \(24.1\,\text{cm}\) and \(74.1\,\text{cm}\) respectively below the open end of the tube. The inner diameter of the tube is

• A. \(5\,\text{cm}\)
• B. \(3\,\text{cm}\)
• C. \(4\,\text{cm}\)
• D. \(2\,\text{cm}\)

Hint:

In resonance tubes, end correction must always be included while calculating wavelength.

Answer 1

The Correct Option is B

Step 1: Effective length condition in resonance tube.
For first resonance:
\[ L_1 + e = \frac{\lambda}{4} \] For second resonance:
\[ L_2 + e = \frac{3\lambda}{4} \]
Step 2: Subtract the two equations.
\[ L_2 - L_1 = \frac{\lambda}{2} \]
Step 3: Substitute given values.
\[ L_2 - L_1 = 74.1 - 24.1 = 50\,\text{cm} \] \[ \Rightarrow \lambda = 100\,\text{cm} \]
Step 4: Calculate end correction.
\[ e = \frac{\lambda}{4} - L_1 = 25 - 24.1 = 0.9\,\text{cm} \]
Step 5: Relation between end correction and radius.
\[ e = 0.6r \Rightarrow r = \frac{0.9}{0.6} = 1.5\,\text{cm} \]
Step 6: Find inner diameter.
\[ d = 2r = 3\,\text{cm} \]
Step 7: Conclusion.
The inner diameter of the tube is \(3\,\text{cm}\).