Let the lengths of the first and second wires be $L_1$ and $L_2$.

Let their radii be $r_1$ and $r_2$.

Given:

$r_1 = 2r_2$

Both wires are of same material and are under same tension.

Frequency of $n^\text{th}$ overtone (i.e., $(n+1)^\text{th}$ harmonic) of a stretched string:

$$ f_n = (n+1)\frac{1}{2L}\sqrt{\frac{T}{\mu}} $$

where

$L$ = length of wire
$T$ = tension (same for both)
$\mu =$ mass per unit length $= \pi r^2 \rho$ (as $\rho$ is density and $r$ is radius)

Given:

$1^{\text{st}}$ overtone of $1^{\text{st}}$ wire = $2^{\text{nd}}$ overtone of $2^{\text{nd}}$ wire

So,

$1^{\text{st}}$ overtone means $n=1$; frequency $= 2\frac{1}{2L_1}\sqrt{\frac{T}{\mu_1}} = \frac{1}{L_1}\sqrt{\frac{T}{\mu_1}}$
$2^{\text{nd}}$ overtone means $n=2$; frequency $= 3\frac{1}{2L_2}\sqrt{\frac{T}{\mu_2}} = \frac{3}{2L_2}\sqrt{\frac{T}{\mu_2}}$

Set them equal:

$$ \frac{1}{L_1}\sqrt{\frac{T}{\mu_1}} = \frac{3}{2L_2}\sqrt{\frac{T}{\mu_2}} $$

Tension $T$ is same, so cancel it:

$$ \frac{1}{L_1}\frac{1}{\sqrt{\mu_1}} = \frac{3}{2L_2}\frac{1}{\sqrt{\mu_2}} $$

Rewriting:

$$ \frac{1}{L_1} \cdot \frac{\sqrt{\mu_2}}{\sqrt{\mu_1}} = \frac{3}{2L_2} $$

So,

$$ \frac{L_1}{L_2} = \frac{2}{3} \cdot \frac{\sqrt{\mu_2}}{\sqrt{\mu_1}} $$

We know:

$\mu = \pi r^2 \rho$

So,

$$ \frac{\sqrt{\mu_2}}{\sqrt{\mu_1}} = \frac{\sqrt{\pi r_2^2 \rho}}{\sqrt{\pi r_1^2 \rho}} = \frac{r_2}{r_1} $$

Given: $r_1 = 2r_2 \implies r_2 = \frac{r_1}{2}$

Therefore,

$$ \frac{r_2}{r_1} = \frac{\frac{r_1}{2}}{r_1} = \frac{1}{2} $$

Plug this back:

$$ \frac{L_1}{L_2} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} $$

So, the ratio of lengths $L_1 : L_2 = 1:3$

Answer:

Option A: $1:3$

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