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# In a resonance tube experiment two consecutive resonances are observed when the length of the air columns are $16\,cm\,and\,49\,cm$ . If the frequency of the tuning fork used is $500\,Hz$ , the velocity of sound in air is:A. $310\,m{s^{ - 1}}$B. $320\,m{s^{ - 1}}$C. $330\,m{s^{ - 1}}$D. $340\,m{s^{ - 1}}$ Verified
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Hint:In order to this question, to know the velocity of sound in the air, we should apply the formula of resonance with both the given lengths of air columns separately. Now, we can calculate the velocity of sound in air.

Let the lengths of the air columns are $16\,cm\,and\,49\,cm$ be ${l_1}\,and\,{l_2}$ respectively. And also the end correction of the resonance be $e$. So,
${l_1} + e = \dfrac{v}{{4f}}$ ….eq(i)
Here, $v$ is the velocity of sound in the air and $f$ is the frequency.
Again,
${l_2} + e = \dfrac{{3v}}{{4f}}$ ….eq(ii)
So, by subtracting eq(i) from eq(ii)-
${l_2} - {l_1} = \dfrac{{3v}}{{4f}} - \dfrac{v}{{4f}} \\ \Rightarrow {l_2} - {l_1} = \dfrac{{2v}}{{4f}} \\ \Rightarrow {l_2} - {l_1}= \dfrac{v}{{2f}} \\ \Rightarrow v = 2f({l_2} - {l_1}) \\$
So, $f$ is given in the question itself i.e.. $500\,Hz$ .
$\Rightarrow v = 2 \times 500(49 - 16) \\ \Rightarrow v = 1000(33) \\ \therefore v = 33000\,cm{s^{ - 1}}\,or\,330\,m{s^{ - 1}}$
Therefore, the velocity of the sound in air is $330\,m{s^{ - 1}}$.

Hence, the correct option is C.

Note:The length $l$ of a pipe or tube (air column) determines its resonance frequencies. Given the requirement of a node at the closed end and an antinode at the open end, only a limited number of wavelengths can be accommodated in the tube.