Question

A steel wire of mass 3.16 Kg is stretched to a tensile strain of 1 x${10}^{-3}$. What is the elastic deformation energy if density p=7.9g/cc and Y=2x${10}^{11}N/m^2$

• A. 4 KJ
• B. 0.4 KJ
• C. 0.04 KJ
• D. 40 J

Answer 1

Answer: (D)

40 J

$\begin{array}{c}\text{Given}\\ \text{mass}=3.16kg\\ \text{Strain}\left(\frac{\Delta L}{L}\right)=1\times {10}^{-3}\end{array}$

$\rho \left(\text{density}\right)=7.8g/cc$

$Y\left(\text{young modulus}\right)=2\times {10}^{11}$

$f=\frac{7.9gram}{{cm}^3}=\frac{7.9\times {10}^{-3}kg}{{10}^{-6}{m}^3}$

$=7.9\times {10}^{3}kg{m}^3$

$\begin{array}{c}\text{Now,}\\ \text{Energy stored}=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}\end{array}$

$\begin{array}{c}\therefore Y=\frac{\text{stress}}{\text{strain}}\\ \text{Stress}=Y\text{strain}\end{array}$

$E=\frac{1}{2}\times Y\times (\mathrm{strain}{)}^2\times \frac{\left(\rho V\right)}{\rho }$

$\therefore \rho V=m$

$=\frac{1\times 2\times {10}^{11}\times {(1\times {10}^{-3})}^2\times 3.16}{2\times 7.9\times {10}^{+3}}$

$\begin{array}{c}=0.4\times {10}^2\\ E=40J\end{array}$