A stretched wire of length 110 , cm is divide | vedclass

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(B) For a stretched wire under constant tension and linear mass density, the frequency $n$ is inversely proportional to the length $l$, i.e., $n \prop

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

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(B) For a stretched wire under constant tension and linear mass density, the frequency $n$ is inversely proportional to the length $l$, i.e., $n \propto 1/l$ or $nl = 	ext{constant}$.Given the total length $L = l_1 + l_2 + l_3 = 110 \, cm$.The frequencies are in the ratio $n_1 : n_2 : n_3 = 1 : 2 : 3$.Since $n_1 l_1 = n_2 l_2 = n_3 l_3 = k$, we have $l_1 = k/n_1, l_2 = k/n_2, l_3 = k/n_3$.Thus, $l_1 : l_2 : l_3 = 1/1 : 1/2 : 1/3 = 6 : 3 : 2$.Let the lengths be $6x, 3x,$ and $2x$.Then $6x + 3x + 2x = 110 \, cm \Rightarrow 11x = 110 \, cm \Rightarrow x = 10 \, cm$.Therefore, 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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