One end of a horizontal thick copper wire of length $2\mathrm{L}$and radius $2\mathrm{R}$ is welded to an end of another horizontal thin copper wire of length$\mathrm{L}$ and radius$\mathrm{R}$. When the arrangement stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is 

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

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