A stretched wire of length 110 cm is divided | vedclass

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(B) For a stretched wire,the frequency $n$ is inversely proportional to the length $l$ (i.e.,$n \propto 1/l$),assuming tension and mass per unit lengt

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$60 \ cm, 30 \ cm, 20 \ cm$

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(B) For a stretched wire,the frequency $n$ is inversely proportional to the length $l$ (i.e.,$n \propto 1/l$),assuming tension and mass per unit length remain constant.Therefore,$n_1 l_1 = n_2 l_2 = n_3 l_3 = k$.Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 2 : 3$,we have:$l_1 : l_2 : l_3 = 1/1 : 1/2 : 1/3 = 6 : 3 : 2$.Let the lengths be $6x, 3x, 2x$.The total length is $6x + 3x + 2x = 110 \ cm$.$11x = 110 \ cm \Rightarrow x = 10 \ cm$.Thus,the lengths are $l_1 = 6(10) = 60 \ cm$,$l_2 = 3(10) = 30 \ cm$,and $l_3 = 2(10) = 20 \ cm$.

$A$ stretched wire of length $110 \ cm$ is divided into three segments whose frequencies are in the ratio $1 : 2 : 3$. Their lengths must be

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