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(C) According to Hooke's Law,the extension in a wire is proportional to the applied force: $F = k(l - l_0)$,where $l_0$ is the original length and $k$

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$\frac{l_1 F_2-l_2 F_1}{F_2-F_1}$

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(C) According to Hooke's Law,the extension in a wire is proportional to the applied force: $F = k(l - l_0)$,where $l_0$ is the original length and $k$ is the force constant.For the first case: $F_1 = k(l_1 - l_0)$ -- $(1)$For the second case: $F_2 = k(l_2 - l_0)$ -- $(2)$Dividing $(1)$ by $(2)$,we get: $\frac{F_1}{F_2} = \frac{l_1 - l_0}{l_2 - l_0}$Cross-multiplying: $F_1(l_2 - l_0) = F_2(l_1 - l_0)$$F_1 l_2 - F_1 l_0 = F_2 l_1 - F_2 l_0$$F_2 l_0 - F_1 l_0 = F_2 l_1 - F_1 l_2$$l_0(F_2 - F_1) = F_2 l_1 - F_1 l_2$$l_0 = \frac{F_2 l_1 - F_1 l_2}{F_2 - F_1}$

The length of a metal wire is $l_1$ when the tension in it is $F_1$ and $l_2$ when the tension is $F_2$. Then,the original length of the wire is

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