Question

The work done per unit volume to stretch the length of area of cross-section $2{\text{mm}}^2$ by $2\%$ will be
Take the young's modulus of material as $[Y=8\times {10}^{10}{\text{N/m}}^2]$

• A. $40{\text{MJ/m}}^3$
• B. $16{\text{MJ/m}}^3$
• C. $64{\text{MJ/m}}^3$
• D. $32{\text{MJ/m}}^3$

Answer 1

The correct option is B $16{\text{MJ/m}}^3$

Given, elongation of length $=2\%$ of original length
$\text{Strain}=\frac{\Delta L}{L}$
$\therefore \text{Strain}=\frac{2}{100}$
Work done per unit volume in stretching is given by:
$W=\frac{1}{2}\times \text{Stress}\times \text{Strain}...\left(i\right)$
$\because \text{Stress}=Y\times \text{Strain}...\left(ii\right)$
From Eq. $\left(i\right)$ and $\left(ii\right)$
$W=\frac{1}{2}\times Y\times {\text{Strain}}^2$
$W=\frac{1}{2}\times 8\times {10}^{10}\times {\left(\frac{2}{100}\right)}^2=16\times {10}^6{\text{J/m}}^3$
$\therefore W=16{\text{MJ/m}}^3$