Two uniform strings X and Y made of same material made to vibrate...

The frequency of the $p$-th harmonic of a stretched string is given by $f = \dfrac{p}{2L} \sqrt{\dfrac{T}{\mu}}$

The linear mass density is $\mu = \pi r^2 \rho$, where $r$ is the radius and $\rho$ is the density of the material.

Substituting $\mu$, we get $f = \dfrac{p}{2Lr} \sqrt{\dfrac{T}{\pi \rho}}$

For string X, the second overtone corresponds to the third harmonic, so $p_X = 3$.

For string Y, the third overtone corresponds to the fourth harmonic, so $p_Y = 4$.

Since both strings are made of the same material and are under the same tension, $\rho$ and $T$ are constant.

Given that the frequencies are equal ($f_X = f_Y$):

$\dfrac{3}{2L_X r_X} \sqrt{\dfrac{T}{\pi \rho}} = \dfrac{4}{2L_Y r_Y} \sqrt{\dfrac{T}{\pi \rho}}$

$\dfrac{3}{L_X r_X} = \dfrac{4}{L_Y r_Y}$

$\dfrac{L_X}{L_Y} = \dfrac{3}{4} \times \dfrac{r_Y}{r_X}$

It is given that the radius of string X is thrice that of string Y, so $r_X = 3r_Y$, which means $\dfrac{r_Y}{r_X} = \dfrac{1}{3}$.

$\dfrac{L_X}{L_Y} = \dfrac{3}{4} \times \dfrac{1}{3} = \dfrac{1}{4}$

The ratio of the length of string X to that of Y is $1 : 4$.

Answer: $1 : 4$