If the length of a stretched string is shortened by x % and the tension is increased by 44 % then the ratio…

Recall the formula of fundamental frequency in terms of tension and mass per unit length,

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

it means,

$\frac{f_1}{f_2}=\frac{\frac{1}{2l_1}\sqrt{\frac{T_1}{\mu }}}{f=\frac{1}{2l_2}\sqrt{\frac{T_2}{\mu }}}=\frac{l_2}{l_1}\times \sqrt{\frac{T_1}{T_2}}$

Put given data,

$$\begin{gathered} \frac{1}{2}=\frac{l\left(1-{\displaystyle \frac{x}{100}}\right)}{l}\times \sqrt{\frac{T}{T\left(1+{\displaystyle \frac{44}{100}}\right)}} \\ \Rightarrow \frac{1}{2}=\frac{\left(100-{\displaystyle x}\right)}{10}\times \frac{1}{12} \\ \Rightarrow 60=100-x \\ \Rightarrow x=40 \end{gathered}$$