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In a resonance tube, the first resonance with 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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500 Hz

165 Hz

330 Hz

300 Hz

answer is A.

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Detailed Solution

For closed pipe, $l_{1} = \frac{v}{4 f}, l_{2} = \frac{3 v}{4 f}$

Now, $v = 2 f (l_{2} - l_{1})$

$\Rightarrow f = \frac{v}{2 (l_{2} - l_{1})} = \frac{330}{2 (0.49 - 0.16)} = 500 Hz$

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