Question

A metallic rod of length l and cross section area A is made of a material of Young’s modulus Y. If the rod is elongated by an amount y, then the work done is proportional to
(A). Y
(B). $\dfrac{1}{y}$
(C). ${y^2}$
(D). $\dfrac{1}{{{y^2}}}$

Answer 1

Hint: 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 a uniaxial deformation.

Complete step-by-step solution -
Consider a rod of length, l and cross section area A being pulled by equal and opposite forces. The length of the rod increases from its natural length$L{\text{ to }}L + \Delta L$. The fractional change $\dfrac{{\Delta L}}{L}$is called the longitudinal strain.
If the length increases from its natural length, the longitudinal strain is tensile strain. If it decreases it is called a compressive strain.
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.
In the situation described above, the Young’s modulus is
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}} = \dfrac{{FL}}{{A\Delta L}}$
So, for the given question:
$\eqalign{
  & Volume = A \times l \cr
  & {\text{or }}V = Al \cr
  & {\text{Strain = }}\dfrac{{{\text{Elongation}}}}{{{\text{Natural length}}}} = \dfrac{y}{l} \cr
  & {\text{We know that Work done, }}W = \dfrac{1}{2} \times stress \times strain \times volume \cr
  & \Rightarrow W = \dfrac{1}{2} \times Y \times {\left( {strain} \right)^2} \times Al \cr
  & \Rightarrow W = \dfrac{1}{2} \times Y \times {\left[ {\dfrac{y}{l}} \right]^2} \times Al \cr
  & \Rightarrow W = \dfrac{1}{2}\left[ {\dfrac{{YA}}{l}} \right]{y^2} \cr
  & \Rightarrow W \propto {y^2} \cr} $
So, the correct answer is C. i.e. , $W \propto {y^2}$

Note: Students need to first define the physical quantities being discussed about properly and then use the understanding and mathematical expression of these quantities such as stress, strain and Young’s modulus to solve the question. Additionally, students try to skip steps when solving these types of questions that can result in silly mistakes. So avoid skipping steps.