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(D) Total length of the wire,$L = 114\, cm$.The fundamental frequency $n$ of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,wh

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At $72\, cm$ and $96\, cm$ from one end

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(D) Total length of the wire,$L = 114\, cm$.The fundamental frequency $n$ of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $n \propto \frac{1}{L}$.Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 3 : 4$.Therefore,the ratio of the lengths of the three segments is $L_1 : L_2 : L_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4} = 12 : 4 : 3$.The sum of the ratio parts is $12 + 4 + 3 = 19$.Calculating the lengths:$L_1 = \frac{12}{19} 	imes 114 = 12 	imes 6 = 72\, cm$.$L_2 = \frac{4}{19} 	imes 114 = 4 	imes 6 = 24\, cm$.$L_3 = \frac{3}{19} 	imes 114 = 3 	imes 6 = 18\, cm$.To divide the wire into these segments,the first bridge should be placed at $72\, cm$ from one end,and the second bridge should be placed at $72 + 24 = 96\, cm$ from the same end.

$A$ sonometer wire of length $114\, cm$ is fixed at both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio $1 : 3 : 4$?

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