The length of a metal wire is l₁, when the tension in it is T₁ and is l₂ when the tension is T₂. The natural length of the wire is

According to question, length of metal wire is l₁ when the tension in it is T₁ and l₂ when the tension is T₂.

Let initial length of metal wire be l₀ and area of cross-section A.

We know that,

$\frac{\mathrm{T}}{\mathrm{A}}=\frac{\mathrm{Y\Delta l}}{\mathrm{I}}$                                ...(i)

where, T = tension in the metal wire,

A = area of cross-section of wire,

Y = Young’s modulus of the material of wire

and $\mathrm{\Delta l}$ = change in length of wire

Considering Eq. (i), we can write

$\frac{{\mathrm{T}}_1}{\mathrm{A}}=\frac{\mathrm{Y}\left({\mathrm{l}}_1-{\mathrm{l}}_0\right)}{{\mathrm{l}}_0}$                             ...(ii)

and $\frac{{\mathrm{T}}_2}{\mathrm{A}}=\frac{\mathrm{Y}\left({\mathrm{l}}_2-{\mathrm{l}}_0\right)}{{\mathrm{l}}_0}$                      ...(iii)

On dividing Eq. (ii) by Eq. (iii), we get

$\frac{\frac{{\mathrm{T}}_1}{\mathrm{A}}}{\frac{{\mathrm{T}}_2}{\mathrm{A}}}=\frac{\frac{\mathrm{Y}\left({\mathrm{l}}_1-{\mathrm{l}}_0\right)}{{\mathrm{l}}_0}}{\frac{\mathrm{Y}\left({\mathrm{l}}_2-{\mathrm{l}}_0\right)}{{\mathrm{l}}_0}}\Rightarrow \frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{{\mathrm{l}}_1-{\mathrm{l}}_0}{{\mathrm{l}}_2-{\mathrm{l}}_0}$

$\begin{array}{l}\Rightarrow {\mathrm{T}}_1{\mathrm{l}}_2-{\mathrm{T}}_1{\mathrm{l}}_0={\mathrm{T}}_2{\mathrm{l}}_1-{\mathrm{T}}_2{\mathrm{l}}_0\\ \Rightarrow {\mathrm{T}}_1{\mathrm{l}}_0-{\mathrm{T}}_2{\mathrm{l}}_0={\mathrm{T}}_1{\mathrm{l}}_2-{\mathrm{T}}_2{\mathrm{l}}_1\\ \Rightarrow {\mathrm{l}}_0=\frac{{\mathrm{T}}_1{\mathrm{I}}_2-{\mathrm{T}}_2{\mathrm{l}}_1}{{\mathrm{T}}_1-{\mathrm{T}}_2}{\text{ or l}}_0=\frac{{\mathrm{l}}_1{\mathrm{T}}_2-{\mathrm{l}}_2{\mathrm{T}}_1}{{\mathrm{T}}_2-{\mathrm{T}}_1}\end{array}$