A tuning fork of frequency n is held near the open end of a tube which is closed at the other end and the length of the tube is adjusted until resonance occurs. If the two shortest lengths that produce resonance are ,l₁ and 1₂, the speed of the sound is

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$n (l_{2} â l_{1})$

$2 n (l_{2} + l_{1})$

$2 n (l_{2} â l_{1})$

$n (l_{2} â l_{1}) / 2$

answer is D.

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Detailed Solution

$n = \frac{v}{4 (l_{1} + x)} and n = \frac{3 v}{4 (l_{2} + x)}, h e r e x = e n d c o r r e c t i o n . V = 4 {nl}_{1} and 3 v = 4 {nl}_{2} or 2 v = 4 n (l_{2} â l_{1}) V = 2 n (l_{2} â l_{1})$

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