A sonometer wire has a length of 114 ~cm betw | vedclass

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(D) The fundamental frequency $n$ of a sonometer wire of length $l$ is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.Since $T$ and $m$ are constant f

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$l_1, l_2, l_3 = 72, 24, 18$

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(D) The fundamental frequency $n$ of a sonometer wire of length $l$ is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.Since $T$ and $m$ are constant for the wire,$n \propto \frac{1}{l}$,which implies $l \propto \frac{1}{n}$.Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 3 : 4$,the ratio of the lengths of the segments must be $l_1 : l_2 : l_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$.To simplify this ratio,multiply by the least common multiple of $1, 3, 4$,which is $12$: $l_1 : l_2 : l_3 = 12 : 4 : 3$.The sum of the parts is $12 + 4 + 3 = 19$.The total length is $114 ~cm$.Thus,$l_1 = \frac{12}{19} 	imes 114 = 12 	imes 6 = 72 ~cm$.$l_2 = \frac{4}{19} 	imes 114 = 4 	imes 6 = 24 ~cm$.$l_3 = \frac{3}{19} 	imes 114 = 3 	imes 6 = 18 ~cm$.Therefore,the lengths are $72 ~cm, 24 ~cm, 18 ~cm$.

$A$. $Pa \cdot s$	$(i)$ $[L^2 \ T^{-2} \ K^{-1}]$
$B$. $N \cdot m \cdot K^{-1}$	$(ii)$ $[M \ L \ T^{-3} \ K^{-1}]$
$C$. $J \cdot kg^{-1} \cdot K^{-1}$	$(iii)$ $[M \ L^{-1} \ T^{-1}]$
$D$. $W \cdot m^{-1} \cdot K^{-1}$	$(iv)$ $[M \ L^2 \ T^{-2} \ K^{-1}]$

$A$ sonometer wire has a length of $114 ~cm$ between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $cm$) whose fundamental frequencies are in the ratio $1:3:4$?

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