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(B) The frequency produced by a stretched wire is given by $f = \frac{p}{2l} \sqrt{\frac{T}{m}}$.Here,$p$ is the number of loops,$l$ is the length,$T$

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$120 \ cm, 80 \ cm, 60 \ cm$

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(B) The frequency produced by a stretched wire is given by $f = \frac{p}{2l} \sqrt{\frac{T}{m}}$.Here,$p$ is the number of loops,$l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.Since $T$ and $m$ are constant for the same wire,we have $f \propto \frac{1}{l}$.Given the ratio of frequencies $f_1 : f_2 : f_3 = 2 : 3 : 4$,the ratio of lengths is $l_1 : l_2 : l_3 = \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$.Multiplying by the least common multiple $(12)$,we get $l_1 : l_2 : l_3 = 6 : 4 : 3$.The sum of the ratio parts is $6 + 4 + 3 = 13$.Given the total length $L = 260 \ cm$,the individual lengths are:$l_1 = \frac{6}{13} 	imes 260 = 120 \ cm$$l_2 = \frac{4}{13} 	imes 260 = 80 \ cm$$l_3 = \frac{3}{13} 	imes 260 = 60 \ cm$.

$A$ stretched wire of length $260 \ cm$ is set into vibrations. It is divided into three segments whose frequencies are in the ratio $2:3:4$. Their lengths must be

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