A copper wire of length 1.0, m and a steel wi | vedclass

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(D) Since the wires are joined end to end and subjected to the same load,the tension $F$ in both wires is the same. Also,the cross-sectional area $A$

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$1.25$

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(D) Since the wires are joined end to end and subjected to the same load,the tension $F$ in both wires is the same. Also,the cross-sectional area $A$ is the same.Young's modulus is given by $Y = \frac{F L}{A \Delta L}$,which implies $F = \frac{Y A \Delta L}{L}$.Since $F$ and $A$ are constant for both wires,we have $\frac{Y_c \Delta L_c}{L_c} = \frac{Y_s \Delta L_s}{L_s}$.Given: $L_c = 1.0\, m$,$L_s = 0.5\, m$,$\Delta L_c = 1\, mm$,$Y_c = 1.0 	imes 10^{11}\, N/m^2$,$Y_s = 2.0 	imes 10^{11}\, N/m^2$.Substituting the values: $(1.0 	imes 10^{11}) 	imes (1\, mm / 1.0\, m) = (2.0 	imes 10^{11}) 	imes (\Delta L_s / 0.5\, m)$.$1.0 	imes 10^{11} = (4.0 	imes 10^{11}) 	imes \Delta L_s$.$\Delta L_s = \frac{1.0 	imes 10^{11}}{4.0 	imes 10^{11}} = 0.25\, mm$.Total extension = $\Delta L_c + \Delta L_s = 1\, mm + 0.25\, mm = 1.25\, mm$.

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