Two metal rods A and B each of length 50 cm | vedclass

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(C) The change in length of the combined rod is the sum of the changes in lengths of the individual rods.$\Delta l = \Delta l_A + \Delta l_B$Using the

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$3$

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(C) The change in length of the combined rod is the sum of the changes in lengths of the individual rods.$\Delta l = \Delta l_A + \Delta l_B$Using the formula for linear expansion $\Delta l = l \alpha \Delta T$,where $l = 0.5 \ m$,$\Delta T = 230^{\circ} C - 30^{\circ} C = 200^{\circ} C$,$\alpha_A = 2.0 	imes 10^{-5} /^{\circ} C$,and $\alpha_B = 1.0 	imes 10^{-5} /^{\circ} C$:$\Delta l = (l_A \alpha_A \Delta T) + (l_B \alpha_B \Delta T)$$\Delta l = l \Delta T (\alpha_A + \alpha_B)$$\Delta l = 0.5 	imes 200 	imes (2.0 	imes 10^{-5} + 1.0 	imes 10^{-5})$$\Delta l = 100 	imes (3.0 	imes 10^{-5})$$\Delta l = 3.0 	imes 10^{-3} \ m = 3 \ mm$Therefore,the change in length is $3 \ mm$.

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