JEE Main 2023 — Mechanical Properties of Solids Question with Solution

The data given is 

$$\begin{gathered} Y=7\times {10}^{10} \mathrm{N} {\mathrm{m}}^{-2} \text{and} \\ \frac{∆\mathrm{l}}{\mathrm{l}}=\frac{0.04}{100} \end{gathered}$$

Young's modulus is given by the ratio of stress and strain,

$Y=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆l}{l}}}=\frac{Fl}{A∆l}$

The energy stored is given as 

$$\begin{gathered} E=\left(\frac{YA}{2l}\right)∆l^2 \\ \Rightarrow E=\left(\frac{YA}{2}\right){\left(\frac{∆l}{l}\right)}^2\times l ...\left(i\right) \end{gathered}$$

The energy per unit volume can be written from equation (i)

$\frac{E}{V}=\frac{Y}{2}\times {\left(\frac{{\displaystyle ∆l}}{l}\right)}^2 ...\left(ii\right)$

Substituting the values in equation (ii)

$\frac{E}{V}=\frac{1}{2}\times 7\times {10}^{10}\times \frac{0.04\times 0.04}{{10}^4}=56\times {10}^2$