One end of a horizontal thick copper wire of length 2 L   and radius 2 R is welded to an end of another…

As the two rods are in series, the force at the connecting surface will be same.

${\mathrm{F}}_1={\mathrm{F}}_2$

$\Rightarrow \mathrm{Y}\times \frac{∆{\mathrm{l}}_1}{{\mathrm{l}}_1}\times {\mathrm{A}}_1=\mathrm{Y}\times \frac{∆{\mathrm{l}}_2}{{\mathrm{l}}_2}\times {\mathrm{A}}_2$

Where $\mathrm{Y}$ is Young's modulus, $∆{\mathrm{l}}_1$ and $∆{\mathrm{l}}_2$ is increase in length, ${\mathrm{l}}_1$ and ${\mathrm{l}}_2$ are lengths, ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$ are areas, $2\mathrm{R}$ and $\mathrm{R}$ are radii of thick and thin wires respectively.

$\left[\text{∵ F}=\text{Y}\times \text{strain}\times \text{Area}\right]$

$\Rightarrow \frac{∆{\mathrm{l}}_1}{{\mathrm{l}}_1}\times {\mathrm{A}}_1=\frac{∆{\mathrm{l}}_2}{{\mathrm{l}}_2}\times {\mathrm{A}}_2$

Here, ${\mathrm{A}}_1=\mathrm{\pi }{\left(2\mathrm{R}\right)}^2, {\mathrm{l}}_1=2\mathrm{L}$

${\mathrm{A}}_2={\mathrm{\pi R}}^2, {\mathrm{l}}_2=\mathrm{L}$
$\Rightarrow \frac{∆{\mathrm{l}}_1}{2\mathrm{L}}\times \mathrm{\pi }{\left(2\mathrm{R}\right)}^2=\frac{∆{\mathrm{l}}_2}{\mathrm{L}}\times {\mathrm{\pi R}}^2$
$\Rightarrow \frac{∆{\mathrm{l}}_2}{∆{\mathrm{l}}_1}=2$