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

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For a resonance tube closed at one end, the resonating lengths $l_{1}, l_{2}, l_{3}, \dots$ are related to the wavelength $λ$ and end correction $e$ by:

$l_{1} + e = \frac{λ}{4}$

$l_{2} + e = \frac{3 λ}{4}$

Given $l_{1} = 18$ cm and $e = 1.1$ cm.

$\frac{λ}{4} = 18 + 1.1 = 19.1$ cm

For the next resonating length $l_{2}$ :

$l_{2} + e = 3 (\frac{λ}{4})$

$l_{2} + 1.1 = 3 \times 19.1$

$l_{2} + 1.1 = 57.3$

$l_{2} = 57.3 - 1.1 = 56.2$ cm

Answer: $56.2$ cm

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