Two uniform wires of same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and radius of first wire is twice the radius of second wire then the ratio of length of first to second wire is

The frequency of the $p$-th overtone of a stretched string is given by $f = \dfrac{(p+1)v}{2L}$, where $v$ is the wave speed and $L$ is the length of the wire.

For the first wire, the first overtone ($p=1$) frequency is:
$f_1 = \dfrac{2v_1}{2L_1} = \dfrac{v_1}{L_1}$

For the second wire, the second overtone ($p=2$) frequency is:
$f_2 = \dfrac{3v_2}{2L_2}$

Given that $f_1 = f_2$, we have:
$\dfrac{v_1}{L_1} = \dfrac{3v_2}{2L_2}$

$\Rightarrow \dfrac{L_1}{L_2} = \dfrac{2v_1}{3v_2}$

The velocity of a transverse wave in a wire is $v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{T}{\pi r^2 \rho}}$, where $T$ is tension, $r$ is radius, and $\rho$ is density.

Since $T$ and $\rho$ are the same for both wires, $v \propto \dfrac{1}{r}$.

$\Rightarrow \dfrac{v_1}{v_2} = \dfrac{r_2}{r_1}$

Given that the radius of the first wire is twice the radius of the second wire ($r_1 = 2r_2$), we get:
$\dfrac{v_1}{v_2} = \dfrac{r_2}{2r_2} = \dfrac{1}{2}$

Substituting this into the length ratio equation:
$\dfrac{L_1}{L_2} = \dfrac{2}{3} \times \dfrac{1}{2} = \dfrac{1}{3}$

Answer: $\dfrac{1}{3}$