A steel wire and a copper wire, both possessing equal length and...

Let $F$ be the tension applied to the combination of the wires. Since the wires are joined end to end, the tension $F$ is the same in both wires.

The cross-sectional area $A$ is also the same for both wires. Therefore, the stress $\sigma = \dfrac{F}{A}$ is equal for both the steel and copper wires.

Young's modulus is given by $Y = \dfrac{\text{Stress}}{\text{Strain}}$, which implies $\text{Strain} = \dfrac{\text{Stress}}{Y}$.

Let $\epsilon_c$ and $\epsilon_s$ be the strains developed in the copper and steel wires, respectively.

$\epsilon_c = \dfrac{\sigma}{Y_c}$

$\epsilon_s = \dfrac{\sigma}{Y_s}$

Taking the ratio of the strain in copper to the strain in steel:

$\dfrac{\epsilon_c}{\epsilon_s} = \dfrac{Y_s}{Y_c}$

Substituting the given values, $Y_s = 2 \times 10^{11} \text{ N/m}^2$ and $Y_c = 1.3 \times 10^{11} \text{ N/m}^2$:

$\dfrac{\epsilon_c}{\epsilon_s} = \dfrac{2 \times 10^{11}}{1.3 \times 10^{11}} = \dfrac{20}{13}$