A copper wire of length 2.4 m and an aluminu | vedclass

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(C) The total elongation $\Delta l$ is the sum of the elongations of the copper wire and the aluminum wire: $\Delta l = \Delta l_c + \Delta l_a$. Sinc

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) The total elongation $\Delta l$ is the sum of the elongations of the copper wire and the aluminum wire: $\Delta l = \Delta l_c + \Delta l_a$. Since the wires are connected in series, the same load $F$ acts on both. Using the formula $\Delta l = \frac{F l}{Y A}$, where $A = \pi r^2 = \pi (10^{-3} \ m)^2 = \pi 	imes 10^{-6} \ m^2$: $\Delta l = \frac{F l_c}{Y_c A} + \frac{F l_a}{Y_a A} = \frac{F}{A} \left( \frac{l_c}{Y_c} + \frac{l_a}{Y_a} ight)$. Substituting the given values: $0.6 	imes 10^{-3} = \frac{F}{\pi 	imes 10^{-6}} \left( \frac{2.4}{1.2 	imes 10^{11}} + \frac{0.7}{0.7 	imes 10^{11}} ight)$. $0.6 	imes 10^{-3} = \frac{F}{\pi 	imes 10^{-6}} \left( 2 	imes 10^{-11} + 1 	imes 10^{-11} ight)$. $0.6 	imes 10^{-3} = \frac{F}{\pi 	imes 10^{-6}} 	imes 3 	imes 10^{-11}$. $0.6 	imes 10^{-3} = F 	imes \frac{3 	imes 10^{-11}}{\pi 	imes 10^{-6}} = F 	imes \frac{3 	imes 10^{-5}}{\pi}$. $F = \frac{0.6 	imes 10^{-3} 	imes \pi}{3 	imes 10^{-5}} = \frac{0.6 	imes 10^2 	imes \pi}{3} = 0.2 	imes 100 	imes \pi = 20 \pi \ N$.

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