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An open pipe is resonance in its $2^{n d}$ harmonic with tuning fork of frequency $f_{1}$ _. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from $f_{1}$ then again a resonance is obtained with a frequency $f_{2}$ . If in this case the pipe vibrates $n^{t h}$ harmonics, then

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$n = 3, f_{2} = \frac{5}{4} f_{1}$

$n = 5, f_{2} = \frac{5}{4} f_{1}$

$n = 3, f_{2} = \frac{3}{4} f_{1}$

$n = 5, f_{2} = \frac{3}{4} f_{1}$

answer is C.

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Detailed Solution

$f_{1} = 2 (\frac{v}{2 l})$

$f_{2} = n (\frac{v}{4 l})$

$f_{2} = \frac{n}{4} f_{1}$ ; (Where $n$ is odd number)

As $f_{2} > f_{1}$ Therefore $n = 5$ .

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