Question

An elastic material of Young’s modulus Y is subjected to a stress S. The elastic energy stored per unit volume of the material is:
A. $\dfrac{{SY}}{2}$
B. $\dfrac{{{S^2}}}{{2Y}}$
C. $\dfrac{S}{{2Y}}$
D. $\dfrac{{2S}}{Y}$

Answer 1

Hint: The elastic energy stored in a material is nothing but its potential energy. This potential energy per unit volume is mathematically the product of half times the stress and strain. And Young’s modulus is nothing but the ratio of linear stress over strain. Using these concepts we can find the required answer.

Formula Used:
Young’s Modulus, $Y = \dfrac{{{\text{Stress}}}}{{{\text{Strain}}}}$
Elastic Energy per unit volume, $E = \dfrac{1}{2} \times {\text{stress}} \times {\text{strain}}$

Complete step by step answer:
Young's modulus is a mechanical property of a solid material that measures the stiffness of that material. It can be defined as the relationship between stress (force per unit area) and strain (proportional deformation) in a material in the linear elasticity regime of uniaxial deformation.
Now, Young’s modulus is derived from Hooke’s law for small deformations i.e., if the deformation is small, the stress in a body is proportional to the corresponding strain.
Mathematically,
 $\dfrac{{{\text{Tensile stress}}}}{{{\text{Tensile strain}}}} = Y$
where Y is a constant for a given material. This ratio of tensile stress over tensile strain is called Young’s modulus for the material.
Given:
Young’s Modulus of the material = Y
Stress on the material = S
Elastic Energy per unit volume on this material will be:
$E = \dfrac{1}{2} \times {\text{stress}} \times {\text{strain}} \cdots \cdots \cdots \left( 1 \right)$
But we know that:
$\eqalign{
  & Y = \dfrac{{{\text{Stress}}}}{{{\text{Strain}}}} \cr
  & \Rightarrow S = Y \times {\text{strain }}\left[ {\because {\text{Stress}} = S\left( {given} \right)} \right] \cr
  & \therefore {\text{ strain}} = \dfrac{S}{Y} \cr} $
Substituting the value of strain in equation (1) we get:
$\eqalign{
  & E = \dfrac{1}{2} \times {\text{S}} \times \dfrac{S}{Y} \cr
  & E = \dfrac{{{S^2}}}{{2Y}} \cr} $
Therefore, the correct option is B i.e., the elastic energy stored per unit volume of the material is $\dfrac{{{S^2}}}{{2Y}}$

Note: The elastic energy is also known as elastic potential energy of a strained body because this energy is in the form of potential energy and will be converted into some other form of energy such as kinetic or sound energy, when the object is allowed to return to its original shape.