Two uniform wires of same material are vibrating under the same tension. If the first overtone of first wire is equal to the \(2^{\text {nd }}\) overtone of \(2^{\text {nd }}\) wire and radius of the first wire is twice the radius of the \(2^{\text {nd }}\) wire then the ratio of length of first wire to \(2^{\text {nd }}\) wire is

Fundamental frequency of the first wire is
\(\mathrm{n}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\pi \mathrm{r}_1^2 \rho}}=\frac{1}{2 l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}\)
The first overtone \(\mathrm{n}_1=2 \mathrm{n}=\frac{1}{l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}\)
Similarly, the second overtone of the second wire will be,
\(\mathrm{n}_2=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi \rho}}\)
Given that \(\mathrm{n}_1=\mathrm{n}_2\)
\(\begin{aligned}
& \therefore \quad \frac{1}{l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi \rho}} \\
& \therefore \quad 3 l_1 \mathrm{r}_1=2 l_2 \mathrm{r}_2 \\
& \\
& \quad \frac{l_1}{l_2}=\frac{2 \mathrm{r}_2}{3 \mathrm{r}_1}
\end{aligned}\)
\(=\frac{2 r_2}{3\left(2 r_2\right)} \quad \ldots .\left(\because r_1=2 r_2\right)\)
\(=\frac{1}{3}\)