One end of a horizontal thick copper wire of length 2L and radius...

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

${F}_{1} = {F}_{2}$

$⇒ Y \times \dfrac{∆ {l}_{1}}{{l}_{1}} \times {A}_{1} = Y \times \dfrac{∆ {l}_{2}}{{l}_{2}} \times {A}_{2}$

Where $Y$ is Young's modulus, $∆ {l}_{1} and ∆ {l}_{2}$ is increase in length, ${l}_{1} and {l}_{2}$ are lengths, ${A}_{1} and {A}_{2}$ are areas, $2R and R$ are radii of thick and thin wires respectively.

\left[\text{∵ F} = \text{Y} \times \text{strain} \times \text{Area}\right]$</p><p>$⇒ \dfrac{∆ {l}_{1}}{{l}_{1}} \times {A}_{1} = \dfrac{∆ {l}_{2}}{{l}_{2}} \times {A}_{2}$</p><p>Here,${A}_{1} = \pi{\left(2R\right)}^{2}, {l}_{1} = 2L$</p><p>${A}_{2} = {\pi R}^{2}, {l}_{2} = L$<br >$⇒ \dfrac{∆ {l}_{1}}{2L} \times \pi{\left(2R\right)}^{2} = \dfrac{∆ {l}_{2}}{L} \times {\pi R}^{2}$<br >$⇒ \dfrac{∆ {l}_{2}}{∆ {l}_{1}} = 2