Question

A force $F$ is needed to break a copper wire having a radius $R$. What will be the force needed to break a copper wire of radius $2R$. [assume $F$ is applied along the wire and the wire obeys Hooke’s law until it breaks]

Answer 1

The main hint for this problem is that the wire will obey Hooke’s law until it breaks. So we will use the concept of Hooke’s law or, say, the concept of Young’s modulus to solve our problem. Furthermore, we will think about how the required force changes with the variation of the radius of the wire.

Complete step by step solution:
In physics, Hooke’s law states that stress is directly proportional to strain within the elastic limit.
Hence, stress $\propto$ strain
When an object experiences a strain or deformation in its dimension, it produces a reaction force to gain its previous dimension. It is a fundamental mechanical property of any object. The stress developed inside the body, i.e., the reaction force developed per unit cross-sectional area while it undergoes longitudinal strain, is called longitudinal strain. For longitudinal strain, we have Young’s modulus. Within the elastic limit, longitudinal stress divided by longitudinal strain is called Young’s modulus.
Therefore,
$Y = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta L}}{L}} \right)}}$
$\Rightarrow \dfrac{{F.L}}{{A.\Delta L}}$
Here,
$\left( {\dfrac{F}{A}} \right)$ is the longitudinal stress, i.e., reaction force per unit area
$\left( {\dfrac{{\Delta L}}{L}} \right)$ is the longitudinal strain
From the above equation, we have-
$\Rightarrow F = Y.\dfrac{{A.\Delta L}}{L}$
$\Rightarrow F \propto A$
For a cylindrical wire,
$\Rightarrow F \propto \pi .{r^2}$…………..$(1)$
Where $r$ is the radius of the wire
We say $F'$ force is required to break the wire of radius $2R$. From $(1)$, we have-
$\Rightarrow \dfrac{F}{{\pi .{R^2}}} = \dfrac{{F'}}{{\pi .{{\left( {2R} \right)}^2}}}$
$\Rightarrow \dfrac{F}{{\pi .{R^2}}} = \dfrac{{F'}}{{\pi .4{R^2}}}$
We cancel $\pi .{R^2}$ from both sides and get-
$\Rightarrow F = \dfrac{{F'}}{4}$
$\Rightarrow F' = 4F$
Therefore, the required force will be four times the previous force.

Note: Like Young’s modulus, we have some other modules related to the elasticity of an object, such as Bulk modulus and Shear modulus. Bulk modulus is related to volume stress and strain, and shear modulus is related to shearing stress and strain.