Question

Let us assume that a wire can be broken by applying a load of $20kgwt$. What will be the force needed to break the wire of twice the diameter?
$\begin{aligned} & A.20kgwt \\\ & B.5kgwt \\\ & C.80kgwt \\\ & D.160kgwt \\\ \end{aligned}$

Answer 1

Breaking stress will be the characteristic of the material. Therefore for two wires of identical material which is having different thickness. Find out the ratio of the force to the square of the radius of the wire. This will be equal in both situations. Substitute the values in the equation and rearrange the equation. This will help you in answering this question.

Complete step by step answer:
Breaking stress will be the characteristic of the material. Therefore for two wires of identical material which is having different thickness. The ratio of the force to the square of the radius of the wire. That is we can write that,
$\dfrac{{{F}_{1}}}{{{r}_{1}}^{2}}=\dfrac{{{F}_{2}}}{{{r}_{2}}^{2}}$
The radius of the wire can be written as,
$\begin{aligned} & {{r}_{1}}^{2}={{r}^{2}} \\\ & {{r}_{2}}^{2}={{\left( 2r \right)}^{2}} \\\ \end{aligned}$
Substituting the values in it will give,
$\dfrac{{{F}_{1}}}{{{r}^{2}}}=\dfrac{{{F}_{2}}}{{{\left( 2r \right)}^{2}}}$
That is,
$\dfrac{20}{{{r}^{2}}}=\dfrac{{{F}_{2}}}{4{{r}^{2}}}$
Cancelling the common terms in it and rearranging this will give,
${{F}_{2}}=80kgwt$
Therefore the force needed to break the wire of twice the diameter has been obtained.

So, the correct answer is “Option C”.

Note: The stress on a material is defined as the force per unit area which is applied on the material. The breaking stress of a material will be the maximum amount of the tensile stress that a material can overcome before failure like the breaking or permanent deformation. Breaking stress will be the limiting state of the tensile stress which is leading to the tensile failure. The stress can be expressed in the unit of newton per square metre. Maximum amount of tensile stress can be defined as the elastic limit.