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(C) Let the unstretched length of the metal wire be $L$ and its cross-sectional area be $A$.According to Hooke's Law, the Young's modulus $Y$ is given

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$\frac{T_2 L_1-T_1 L_2}{T_2-T_1}$

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(C) Let the unstretched length of the metal wire be $L$ and its cross-sectional area be $A$.According to Hooke's Law, the Young's modulus $Y$ is given by $Y = \frac{T/A}{\Delta L/L}$, where $\Delta L$ is the change in length.For tension $T_1$, the length is $L_1$, so the extension is $\Delta L_1 = L_1 - L$. Thus, $Y = \frac{T_1 L}{A(L_1 - L)}$.For tension $T_2$, the length is $L_2$, so the extension is $\Delta L_2 = L_2 - L$. Thus, $Y = \frac{T_2 L}{A(L_2 - L)}$.Equating the two expressions for $Y$:$\frac{T_1 L}{A(L_1 - L)} = \frac{T_2 L}{A(L_2 - L)}$$\frac{T_1}{L_1 - L} = \frac{T_2}{L_2 - L}$$T_1(L_2 - L) = T_2(L_1 - L)$$T_1 L_2 - T_1 L = T_2 L_1 - T_2 L$$T_2 L - T_1 L = T_2 L_1 - T_1 L_2$$L(T_2 - T_1) = T_2 L_1 - T_1 L_2$$L = \frac{T_2 L_1 - T_1 L_2}{T_2 - T_1}$

The length of a metal wire is $L_1$ when the tension is $T_1$ and $L_2$ when the tension is $T_2$. The unstretched length of the wire is:

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