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(C) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear

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$16/9$

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(C) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Let the initial tension be $T$. Then,$256 = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.When the tension is increased by $1 \,kg \,wt$,the new tension is $(T+1) \,kg \,wt$. The new frequency is $320 = \frac{1}{2l} \sqrt{\frac{T+1}{\mu}}$.Dividing the two equations: $\frac{320}{256} = \sqrt{\frac{T+1}{T}}$.Simplifying the ratio: $\frac{320}{256} = \frac{5}{4}$.Squaring both sides: $\frac{25}{16} = \frac{T+1}{T}$.$25T = 16(T+1) \implies 25T = 16T + 16$.$9T = 16 \implies T = \frac{16}{9} \,kg \,wt$.

$A$ wire of length $1 \,m$ under a certain initial tension emits a sound of fundamental frequency $256 \,Hz$. When the tension is increased by $1 \,kg \,wt$,the frequency of the fundamental note increases to $320 \,Hz$. The initial tension is ........... $kg \,wt$.

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