Q126 / 150 · MHT CET 2026 (19 April Shift 2) · Physics
In a resonance tube open at one end, the end correction is $1.1$ cm. If the shortest length of resonating air column with a tuning fork is $18$ cm, the next resonating length will be
- A$45.9$ cm
- B$49.6$ cm
- C$51.3$ cm
- Correct answerD$56.2$ cm
Solution
For a resonance tube closed at one end, the resonating lengths $l_1, l_2, l_3, \dots$ are related to the wavelength $\lambda$ and end correction $e$ by:
$l_1 + e = \dfrac{\lambda}{4}$
$l_2 + e = \dfrac{3\lambda}{4}$
Given $l_1 = 18$ cm and $e = 1.1$ cm.
$\dfrac{\lambda}{4} = 18 + 1.1 = 19.1$ cm
For the next resonating length $l_2$:
$l_2 + e = 3 \left( \dfrac{\lambda}{4} \right)$
$l_2 + 1.1 = 3 \times 19.1$
$l_2 + 1.1 = 57.3$
$l_2 = 57.3 - 1.1 = 56.2$ cm
Answer: $56.2$ cm
$l_1 + e = \dfrac{\lambda}{4}$
$l_2 + e = \dfrac{3\lambda}{4}$
Given $l_1 = 18$ cm and $e = 1.1$ cm.
$\dfrac{\lambda}{4} = 18 + 1.1 = 19.1$ cm
For the next resonating length $l_2$:
$l_2 + e = 3 \left( \dfrac{\lambda}{4} \right)$
$l_2 + 1.1 = 3 \times 19.1$
$l_2 + 1.1 = 57.3$
$l_2 = 57.3 - 1.1 = 56.2$ cm
Answer: $56.2$ cm
Hence, the correct answer is option: 'D'.