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The length of metallic wire is $\ell_1$ when tension in it is $T_1$. It is $\ell_2$ when the tension is $T_2$. The original length of the wire will be -
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Verified Answer
The correct answer is:
$\frac{\mathrm{T}_2 \ell_1-\mathrm{T}_1 \ell_2}{\mathrm{~T}_2-\mathrm{T}_1}$
Assuming Hooke's law to be valid.
$\begin{aligned} & \mathrm{T} \propto(\Delta \ell) \\ & \mathrm{T}=\mathrm{k}(\Delta \ell)\end{aligned}$
Let, $\ell_0=$ natural length $($ original length)
$\begin{aligned} & \Rightarrow \mathrm{T}=\mathrm{k}\left(\ell-\ell_0\right) \\ & \text { so, } \mathrm{T}_1=\mathrm{k}\left(\ell_1-\ell_0\right) \& \mathrm{~T}_2=\mathrm{k}\left(\ell_2-\ell_0\right) \\ & \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\ell_1-\ell_0}{\ell_2-\ell_0} \Rightarrow \ell_0=\frac{\mathrm{T}_2 \ell_1-\mathrm{T}_1 \ell_2}{\mathrm{~T}_2-\mathrm{T}_1}\end{aligned}$
$\begin{aligned} & \mathrm{T} \propto(\Delta \ell) \\ & \mathrm{T}=\mathrm{k}(\Delta \ell)\end{aligned}$
Let, $\ell_0=$ natural length $($ original length)
$\begin{aligned} & \Rightarrow \mathrm{T}=\mathrm{k}\left(\ell-\ell_0\right) \\ & \text { so, } \mathrm{T}_1=\mathrm{k}\left(\ell_1-\ell_0\right) \& \mathrm{~T}_2=\mathrm{k}\left(\ell_2-\ell_0\right) \\ & \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\ell_1-\ell_0}{\ell_2-\ell_0} \Rightarrow \ell_0=\frac{\mathrm{T}_2 \ell_1-\mathrm{T}_1 \ell_2}{\mathrm{~T}_2-\mathrm{T}_1}\end{aligned}$
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