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(A) Let the natural length of the wire be $L_0$. According to Hooke's law, the extension $\Delta l$ produced by a tension $T$ is given by $\Delta l =

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$\frac{L(\Delta T)-(\Delta L) T}{\Delta T}$

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(A) Let the natural length of the wire be $L_0$. According to Hooke's law, the extension $\Delta l$ produced by a tension $T$ is given by $\Delta l = \frac{T L_0}{A Y}$, where $A$ is the cross-sectional area and $Y$ is Young's modulus.In the first case, the total length is $L = L_0 + \Delta l_1 = L_0 + \frac{T L_0}{A Y}$. Thus, $L - L_0 = \frac{T L_0}{A Y}$ (Equation $1$).In the second case, the total length is $L + \Delta L = L_0 + \Delta l_2 = L_0 + \frac{(T + \Delta T) L_0}{A Y}$. Thus, $(L + \Delta L) - L_0 = \frac{(T + \Delta T) L_0}{A Y}$ (Equation $2$).Dividing Equation $1$ by Equation $2$:$\frac{L - L_0}{L + \Delta L - L_0} = \frac{T}{T + \Delta T}$.Cross-multiplying gives: $(L - L_0)(T + \Delta T) = T(L + \Delta L - L_0)$.$LT + L \Delta T - L_0 T - L_0 \Delta T = TL + T \Delta L - L_0 T$.Simplifying the terms: $L \Delta T - L_0 \Delta T = T \Delta L$.$L_0 \Delta T = L \Delta T - T \Delta L$.$L_0 = \frac{L \Delta T - T \Delta L}{\Delta T}$.

The length of a metal wire is $L$, when it is subjected to tension $T$. If the tension is increased to $T+\Delta T$, the length becomes $L+\Delta L$. The natural length of the wire is

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