Question

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 $\frac{5}{4}\frac{\sigma }{\rho g}$ 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.

• A.
  $D$
• B.
  $C$
• C.
  $B$
• D.
  $A$

Answer 1

The correct option is A

$D$
Given,
Length of wire $=\frac{5}{4}\frac{\sigma }{\rho g}$
Breaking stress $=\sigma$
Density of material $=\rho$
Therefore,
$\text{stress}=\text{weight}/\text{area}$
$\sigma =AL\rho g/A$
$L=\sigma /\rho g$
Length of wire $=5\sigma /4\rho g$
From $O$ to $D$,
Length $=4\sigma /4\rho g=\sigma /\rho g$
Hence wire will break at point $D$.