Explanation

(d) $L$ be original length of the wire
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When a mass $M_{1}$ is suspended from the wire, change in length of wire is $\mathrm{\Delta}L_{1} = L_{1} - L$
When a mass $M_{2}$ is suspended from it, change in length of wire is $\mathrm{\Delta}L_{2} = L_{2} - L$
From figure (b), $T_{1} = M_{1}g$ …(i)
From figure (c), $T_{2} = M_{2}g$ …(ii)
As young’s modulus, $Y = \frac{T_{1}L}{A\mathrm{\Delta}L_{1}} = \frac{T_{2}L}{A\mathrm{\Delta}L_{2}}$
$\frac{T_{1}}{\mathrm{\Delta}L_{1}} = \frac{T_{2}}{\mathrm{\Delta}L_{2}} \Rightarrow \frac{T_{1}}{L_{1} - L} = \frac{T_{2}}{L_{2} - L}$
$\frac{M_{1}g}{L_{1} - L} = \frac{M_{2}g}{L_{2} - L}$ [Using (i) and (ii)]
$M_{1}\left( L_{2} - L \right) = M_{2}(L_{1} - L)$
$M_{1}L_{2} - M_{1}L = M_{2}L_{1} - M_{2}L$
$L\left( M_{2} - M_{1} \right) = L_{1}M_{2} - L_{2}M_{1} \Rightarrow L = \frac{L_{1}M_{2} - L_{2}M_{1}}{M_{2} - M_{1}}$