The fundamental frequency of a sonometer wire increases by 6 Hz if its tension is increased by 44%, keeping the length constant. The fundamental frequency of the sonometer wire is $x\times 10Hz.$The value of $x$ is ____

Fundamental frequency of vibrations of string

            $n=\frac{1}{2l}\sqrt{\frac{T}{m}}$.

            At constant length,

            $n\propto \sqrt{T}$ or $\frac{n}{\sqrt{T}}=$ constant.--------(1)

            As constant tension,

            $n\propto \frac{1}{l}$ or $nl=$ constant ---------- (2)

            New tension, $T^{\text{'}}=T+\frac{44}{100}T=1.44T$

            If n’ is new frequency, then

            $\frac{n^{\text{'}}}{\sqrt{T^{\text{'}}}}=\frac{n}{T}\Rightarrow n^{\text{'}}=\sqrt{\left(\frac{T^{\text{'}}}{T}\right)n}$.--------------(3)

            As $n\propto \sqrt{T}$ the increases in tension results in an increase in frequency

            $n^{\text{'}}=n+6$.

            From (3) $n+6=\left(\sqrt{\frac{1.44T}{T}}\right)n$

            Or $n+6=1.2n\Rightarrow 0.2n=6$

            Or $n=\frac{6}{0.2}=30Hz$.