Fifth harmonic of an closed organ pipe matches with first overtone of an open organ pipe. Now the pipes are joined to form a closed organ pipe. Then the ratio of the fundamental frequency of the open pipe to that of the composite pipe is

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9 : 2

11 : 2

7 : 2

2 : 7

answer is A.

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Detailed Solution

(1)

$\begin{array}{l} 5. \frac{V}{4 l_{C}} = 4. \frac{V}{4 l_{C}} \Rightarrow l_{0} = \frac{4}{5} l_{C} \\ ∴ L = l e n g t h o f c o m p o s i t e p i p e = l_{o} + l_{c} = l_{o} + \frac{5}{4} l_{o} = \frac{9}{4} l_{o} \\ ∴ Re q u i r e d r a t i o = \frac{2. \frac{V}{4 l_{o}}}{1. \frac{V}{4 L}} = \frac{V}{2 l_{o}} \times \frac{4 \times (\frac{9 l_{o}}{4})}{V} \\ = 9 : 2 \end{array}$

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