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(C) The given frequencies are $425 \, Hz$,$595 \, Hz$,and $765 \, Hz$.Calculate the ratio of these frequencies: $425:595:765 = 5:7:9$.Since the freque

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(C) The given frequencies are $425 \, Hz$,$595 \, Hz$,and $765 \, Hz$.Calculate the ratio of these frequencies: $425:595:765 = 5:7:9$.Since the frequencies are in the ratio of odd integers $(5, 7, 9)$,the pipe must be a closed organ pipe.For a closed organ pipe,the resonance frequencies are given by $f_n = n \cdot f_1$,where $n$ takes odd values $(1, 3, 5, \dots)$.Here,the $5^{th}$ harmonic is $425 \, Hz$,so $5 \cdot f_1 = 425 \, Hz$,which gives the fundamental frequency $f_1 = 85 \, Hz$.The fundamental frequency of a closed pipe is $f_1 = \frac{v}{4L}$.Substituting the values: $85 = \frac{340}{4L}$.$L = \frac{340}{4 	imes 85} = \frac{340}{340} = 1 \, m$.

For a certain organ pipe,three successive resonance frequencies are observed at $425 \, Hz$,$595 \, Hz$,and $765 \, Hz$ respectively. If the speed of sound in air is $340 \, m/s$,then the length of the pipe is .... $m$

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