Two uniform strings A and B made of steel are made to vibrate under the same tension. If first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B , the ratio of the length of string \(B\) to that of \(A\) is

\(\mathrm{R}_2=2 \mathrm{R}_1, \mathrm{~T}_1=\mathrm{T}_2\)
As the first overtone of A is equal to second overtone of B, we get
\(\begin{aligned}
2 n_A=3 n_B \\
\ldots(\because \text { first overtone }=\text { second }
\end{aligned}\)
harmonic)...(i)
Also, \(\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}}}\).
Where, \(\mathrm{m}=\) mass per unit length \(=\frac{\left(\pi \mathrm{R}^2 l\right) \rho}{l} . .(\mathrm{ii})\)
\(\begin{aligned}
& \therefore \quad 2 \times \frac{1}{2 l_{\mathrm{A}}} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}_{\mathrm{A}}}}=3 \times \frac{1}{2 l_B} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}_{\mathrm{B}}}} ...(from(i))\\
& \therefore \quad \frac{l_{\mathrm{A}}}{l_{\mathrm{B}}}=\frac{2}{3} \sqrt{\frac{\mathrm{~m}_{\mathrm{B}}}{\mathrm{~m}_{\mathrm{A}}}}=\frac{2}{3} \sqrt{\frac{\pi \mathrm{r}_{\mathrm{B}}^2}{\pi \mathrm{r}_{\mathrm{A}}^2}} \\
& \therefore \quad \frac{l_{\mathrm{A}}}{l_{\mathrm{B}}}=\frac{2}{3} \sqrt{\frac{\mathrm{r}_{\mathrm{B}}^2}{\left(2 \mathrm{r}_{\mathrm{B}}\right)^2}}=\frac{2}{3} \sqrt{\frac{1}{4}}=\frac{1}{3} \\
& \therefore \quad \frac{l_{\mathrm{B}}}{l_{\mathrm{A}}}=\frac{3}{1}
\end{aligned}\)
...(from (ii))