Properties of Matter Part 3

Concept:When two rods of the same length and cross-section are joined end-to-end, the total expansion is the sum of individual expansions.Use the linear expansion formula $\Delta L = L_0 \alpha \Delta T$.Explanation:The total initial length at $30^\circ \text{C}$ is $120\ \text{cm}$.Since the rods are equal in length, each rod initially measures $L_0 = \frac{120}{2}=60\ \text{cm}$.The temperature change is $\Delta T = 100^\circ\text{C} - 30^\circ\text{C}=70^\circ\text{C}$.Coefficient for aluminium: $\alpha_{\text{Al}} = 24\times10^{-6}\ /^\circ\text{C}$.Coefficient for steel: $\alpha_{\text{Steel}} = 1.2\times10^{-5}\ /^\circ\text{C} = 12\times10^{-6}\ /^\circ\text{C}$.Expansion of aluminium:$\Delta L_{\text{Al}} = 60 \times (24\times10^{-6}) \times 70 = 60\times70\times24\times10^{-6} = 4200\times24\times10^{-6} = 100800\times10^{-6} = 0.1008\ \text{cm}$.Expansion of steel:$\Delta L_{\text{Steel}} = 60 \times (12\times10^{-6}) \times 70 = 4200\times12\times10^{-6} = 50400\times10^{-6} = 0.0504\ \text{cm}$.Total expansion: $\Delta L_{\text{total}} = 0.1008 + 0.0504 = 0.1512\ \text{cm}$.Final length: $L_{\text{final}} = 120 + 0.1512 = 120.1512\ \text{cm} \approx 120.15\ \text{cm}$.Answer:The length of the composite rod is $120.15\ \text{cm}$.Hence, the correct option is (C).