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If $l_{1}$ are $l_{2}$ the lengths of an air column for the first and second resonance when a tuning fork of frequency n is sounded on a resonance tube, then the minimum distance of the anti-node from the top end of the resonance tube is

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$2 (l_{2} - l_{1})$

$\frac{1}{2} (2 l_{1} - l_{2})$

$\frac{l_{2} - 3 l_{1}}{2}$

$\frac{l_{2} + l_{1}}{2}$

answer is C.

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Detailed Solution

$l_{1} + x = λ / 4; l_{2} + x = \frac{3 λ}{4} \Rightarrow x = \frac{l_{2} - 3 l_{1}}{2}$

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