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(C) The total length of the system at $20^{\circ}C$ is $L = 50 \ cm + 100 \ cm = 150 \ cm$.Let the temperature change be $\Delta T = 100^{\circ}C - 20

Vedclass · Accepted Answer

$20 	imes 10^{-6} {^{\circ}C}^{-1}$

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(C) The total length of the system at $20^{\circ}C$ is $L = 50 \ cm + 100 \ cm = 150 \ cm$.Let the temperature change be $\Delta T = 100^{\circ}C - 20^{\circ}C = 80^{\circ}C$.The new length of the iron rod at $100^{\circ}C$ is $L_{Fe}' = 50(1 + \alpha_{Fe} \Delta T) = 50(1 + 12 	imes 10^{-6} 	imes 80) = 50(1 + 0.00096) = 50.048 \ cm$.The new length of the aluminum rod at $100^{\circ}C$ is $L_{Al}' = 100(1 + \alpha_{Al} \Delta T) = 100(1 + 24 	imes 10^{-6} 	imes 80) = 100(1 + 0.00192) = 100.192 \ cm$.The total length at $100^{\circ}C$ is $L' = 50.048 + 100.192 = 150.24 \ cm$.The change in length is $\Delta L = L' - L = 150.24 - 150 = 0.24 \ cm$.The effective coefficient of linear expansion $\alpha_{eff}$ is given by $\Delta L = L \alpha_{eff} \Delta T$.$\alpha_{eff} = \frac{\Delta L}{L \Delta T} = \frac{0.24}{150 	imes 80} = \frac{0.24}{12000} = 20 	imes 10^{-6} {^{\circ}C}^{-1}$.

$A$ $50 \ cm$ iron rod is connected to a $100 \ cm$ aluminum rod at $20^{\circ}C$. If $\alpha_{Fe} = 12 \times 10^{-6} {^{\circ}C}^{-1}$ and $\alpha_{Al} = 24 \times 10^{-6} {^{\circ}C}^{-1}$,what is the effective coefficient of linear expansion of the system?

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