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(C) The given resonance frequencies are $f_1 = 425 \, Hz$,$f_2 = 595 \, Hz$,and $f_3 = 765 \, Hz$.Taking the ratio of these frequencies: $425 : 595 :

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(C) The given resonance frequencies are $f_1 = 425 \, Hz$,$f_2 = 595 \, Hz$,and $f_3 = 765 \, Hz$.Taking the ratio of these frequencies: $425 : 595 : 765 = 5 : 7 : 9$.Since the ratio of successive resonance frequencies is in the ratio of odd integers $(5:7:9)$,this indicates that the pipe is a closed organ pipe (closed at one end).The resonance frequencies for a closed organ pipe are given by $f_n = \frac{n V}{4 \ell}$,where $n$ is an odd integer $(1, 3, 5, \dots)$.For the first observed frequency $f_1 = 425 \, Hz$,we have $n = 5$ (since $425/85 = 5$).Thus,$\frac{5 V}{4 \ell} = 425 \, Hz$.Given $V = 340 \, m/s$,we substitute the value: $\frac{5 	imes 340}{4 \ell} = 425$.$\frac{1700}{4 \ell} = 425 \Rightarrow \frac{425}{\ell} = 425$.Therefore,$\ell = 1 \, m$.

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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