Two rods,one made of aluminium and the other | vedclass

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(C) The change in length $\Delta L$ for a rod of initial length $L_0$ due to a temperature change $\Delta T$ is given by $\Delta L = L_0 \alpha \Delta

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$\frac{\alpha_2}{(\alpha_1 + \alpha_2)}$

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(C) The change in length $\Delta L$ for a rod of initial length $L_0$ due to a temperature change $\Delta T$ is given by $\Delta L = L_0 \alpha \Delta T$.Given that the increase in length for both rods is the same,we have:$\Delta L_1 = \Delta L_2$$l_1 \alpha_1 t = l_2 \alpha_2 t$Dividing both sides by $t$ (assuming $t 
eq 0$):$l_1 \alpha_1 = l_2 \alpha_2$From this,we can express $l_2$ in terms of $l_1$:$l_2 = l_1 \frac{\alpha_1}{\alpha_2}$We want to find the ratio $\frac{l_1}{l_1 + l_2}$. Substituting $l_2$ into the expression:$\frac{l_1}{l_1 + l_1 \frac{\alpha_1}{\alpha_2}} = \frac{l_1}{l_1 (1 + \frac{\alpha_1}{\alpha_2})} = \frac{1}{\frac{\alpha_2 + \alpha_1}{\alpha_2}} = \frac{\alpha_2}{\alpha_1 + \alpha_2}$

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