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(B) Let the original length be $l$ and the cross-sectional area be $A$. The Young's modulus $Y$ is given by $Y = \frac{T}{A} \cdot \frac{l}{\Delta l}$

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$\frac{l_{1} T_{2}-l_{2} T_{1}}{T_{2}-T_{1}}$

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(B) Let the original length be $l$ and the cross-sectional area be $A$. The Young's modulus $Y$ is given by $Y = \frac{T}{A} \cdot \frac{l}{\Delta l}$,where $\Delta l$ is the change in length.For tension $T_1$,the length is $l_1$,so $\Delta l_1 = l_1 - l$. Thus,$Y = \frac{T_1}{A} \cdot \frac{l}{l_1 - l}$.For tension $T_2$,the length is $l_2$,so $\Delta l_2 = l_2 - l$. Thus,$Y = \frac{T_2}{A} \cdot \frac{l}{l_2 - l}$.Since the material is the same,$Y$ is constant. Equating the two expressions:$\frac{T_1}{A} \cdot \frac{l}{l_1 - l} = \frac{T_2}{A} \cdot \frac{l}{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 metallic wire is $l$. The tension in the wire is $T_1$ for length $l_1$ and the tension in the wire is $T_2$ for length $l_2$. Find the original length $l$.

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