Q 11-08-007JEE MainMedium
A steel wire ($Y = 2 \times 10^{11}$ N m$^{-2}$) is under a tensile stress of $2 \times 10^8$ N m$^{-2}$. Find the elastic energy stored per unit volume of the wire, in kJ m$^{-3}$.
Numerical value type. Enter your answer.
Answer: 100
Concept. Elastic energy per unit volume is $u = \dfrac12 \times \text{stress} \times \text{strain} = \dfrac{\text{stress}^2}{2Y}$.
Step 1. Substitute.
$$u = \frac{(2 \times 10^8)^2}{2 \times 2 \times 10^{11}} = \frac{4 \times 10^{16}}{4 \times 10^{11}} = 10^5\ \text{J m}^{-3}$$
Step 2. Convert. $10^5\ \text{J m}^{-3} = 100\ \text{kJ m}^{-3}$.
Answer. $100$.
Solution by Sreeraj P, M.Sc Physics