The length of wire OE is divided into five equal parts OA, AB, BC, CD and DE. The wire is hanging from E and its length is given by $\frac{5}{4}\frac{\mathrm{\sigma }}{\mathrm{\rho g}}$, where $\mathrm{\sigma }$ is the breaking stress and $\mathrm{\rho }$ is the density of the material of the wire. Find the point at which wire will break:

$\mathrm{Braking} \mathrm{Stress} = \frac{\mathrm{Weight}}{\mathrm{Area}}$

     $\sigma = \frac{\left(\mathrm{AL}\right)\mathrm{\rho g}}{\mathrm{A}} \mathrm{or} \mathrm{L} = \frac{\mathrm{\sigma }}{\mathrm{\rho g}}$

So, at length $\mathrm{L} = \frac{\mathrm{\sigma }}{\mathrm{\rho g}} \mathrm{wire} \mathrm{will} \mathrm{break}.$

Given, length of wire = $\frac{5\mathrm{\sigma }}{4\mathrm{\rho g}}$

                             = $\frac{\mathrm{\sigma }}{\mathrm{\rho g}}+\frac{1}{4}\frac{\mathrm{\sigma }}{\mathrm{\rho g}} = \mathrm{L}+\frac{1}{4}\mathrm{L} = \frac{5}{4}\mathrm{L}$

At point D, length of wire ($\mathrm{L} = \frac{\mathrm{\sigma }}{\mathrm{\rho g}}$) is completed when measured from the bottom and at that point the wire will break under its own weight, because four parts, each of length $\left(\frac{1}{4}\frac{\mathrm{\sigma }}{\mathrm{\rho g}}\right)$ are completed up to point D.