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

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$40$

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(C) The fundamental frequency of a stretched string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.Taking the ratio of initial frequency $f_1$ to final frequency $f_2$:$\frac{f_1}{f_2} = \frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}}$Given $l_2 = l_1(1 - \frac{x}{100})$,$T_2 = T_1(1 + \frac{44}{100}) = 1.44 T_1$,and $\frac{f_1}{f_2} = \frac{1}{2}$.Substituting these values:$\frac{1}{2} = \frac{l_1(1 - \frac{x}{100})}{l_1} \sqrt{\frac{T_1}{1.44 T_1}}$$\frac{1}{2} = (1 - \frac{x}{100}) 	imes \frac{1}{\sqrt{1.44}}$$\frac{1}{2} = (1 - \frac{x}{100}) 	imes \frac{1}{1.2}$$0.6 = 1 - \frac{x}{100}$$\frac{x}{100} = 0.4$$x = 40$.

If the length of a stretched string is shortened by $x \%$ and the tension is increased by $44 \%$,then the ratio of the initial and final fundamental frequencies is $1:2$. The value of '$x$' is:

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