To solve this problem, we need to understand that the increase in length of both rods is independent of temperature.$\\$This implies that the thermal expansion of both rods is the same.$\\$The formula for linear expansion is:$\\ \Delta L = \alpha L \Delta T \\$Where:$\\$• $\Delta L$ is the change in length,$\\$• $\alpha$ is the coefficient of linear expansion,$\\$• $L$ is the original length, and$\\$• $\Delta T$ is the change in temperature.$\\$Given that the increase in length is the same for both rods, we have:$\\ \alpha_{Cu} L_{Cu} \Delta T = \alpha_{Al} L_{Al} \Delta T \\$We can cancel $\Delta T$ from both sides since it is common:$\\ \alpha_{Cu} L_{Cu} = \alpha_{Al} L_{Al} \\$Rearranging for $L_{Al},$ we get:$\\ L_{Al} = \frac{\alpha_{Cu}}{\alpha_{Al}} L_{Cu} \\$The coefficients of linear expansion are:$\\$• $\alpha_{Cu} = 1.7 \times 10^{-5} \,$per $^\circ C \\$• $\alpha_{Al} = 2.2 \times 10^{-5} \,$per $^\circ C \\$Substitute these values and the length of the copper rod:$\\ L_{Al} = \frac{1.7 \times 10^{-5}}{2.2 \times 10^{-5}} \times 88 \,$cm$\\ L_{Al} = \frac{1.7}{2.2} \times 88 \\$Calculate the ratio:$\\ \frac{1.7}{2.2} \approx 0.7727 \\$Now, calculate $L_{Al}: \\ L_{Al} = 0.7727 \times 88 \\ L_{Al} \approx 68 \,$cm$\\$Therefore, the length of the aluminium rod is $68 \,$cm$. \\$This corresponds to Option 4.