In a resonance tube the first resonance with a tuning fork occurs at $16 cm$ and second at $49 cm.$ If the velocity of sound is $330 m/s,$ the frequency of tuning fork is
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(a) For closed pipe ${l_1} = \frac{v}{{4n}}$;
${l_2} = \frac{{3v}}{{4n}}$==>$v = 2n\,({l_2} - {l_1})$
==> $n = \frac{v}{{2({l_2} - {l_1})}} = \frac{{330}}{{2 \times (0.49 - 0.16)}} = 500\,Hz$
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