Two rods,one of copper (Cu) and the other of | vedclass

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(C) Given that the increase in length for both rods is the same,we have $\Delta L_1 = \Delta L_2$.Using the formula for linear expansion $\Delta L = L

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$\frac{\alpha_i-\alpha_c}{\alpha_c+\alpha_i}$

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(C) Given that the increase in length for both rods is the same,we have $\Delta L_1 = \Delta L_2$.Using the formula for linear expansion $\Delta L = L \alpha \Delta T$,we get:$L_1 \alpha_c t = L_2 \alpha_i t$Dividing both sides by $t$,we get $L_1 \alpha_c = L_2 \alpha_i$,which implies $L_1 = \frac{\alpha_i}{\alpha_c} L_2$.Now,we need to find the ratio $\frac{L_1-L_2}{L_1+L_2}$.Substituting $L_1 = \frac{\alpha_i}{\alpha_c} L_2$ into the expression:$\frac{L_1-L_2}{L_1+L_2} = \frac{(\frac{\alpha_i}{\alpha_c}) L_2 - L_2}{(\frac{\alpha_i}{\alpha_c}) L_2 + L_2}$$= \frac{L_2 (\frac{\alpha_i}{\alpha_c} - 1)}{L_2 (\frac{\alpha_i}{\alpha_c} + 1)}$$= \frac{\frac{\alpha_i - \alpha_c}{\alpha_c}}{\frac{\alpha_i + \alpha_c}{\alpha_c}}$$= \frac{\alpha_i - \alpha_c}{\alpha_c + \alpha_i}$.

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