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
`(5)/(4)(sigma)/(rhog)`, where `sigma` is the breaking stress and `rho` is the density of the material of the wire. Find the point at which wire will break

Options

A. `A`
B. `B`
C. `C`
D. `D`

Answer

D

Explanation

Stress`=(weight)/(area)`
`alpha((AL)rhog)/(A)` or `L=(sigma)/(rhog)`
So, at length `L=(sigma)/(rhog)` wire will break
Given length of wire`=(5sigma)/(4rhog)`
`=(sigma)/(rhog)+(1)/(4)(sigma)/(rhog)=L+(1)/(4)L=(5)/(4)L`
At point D, length of wire `(L=(sigma)/(rhog))` 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 `((1)/(4)(sigma)/(rhog))` are completed up to point D.