An aluminium rod with Young's modulus $Y=7.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$ undergoes elastic strain of $0.04 \%$. The energy per unit volume stored in the rod in SI unit is:

<p>The strain energy stored per unit volume in a material under stress can be calculated using the following formula:</p> <p>$U = \frac{1}{2} \sigma \epsilon$</p> <p>where $\sigma$ is the stress and $\epsilon$ is the strain. </p> <p>For an elastic material, stress is proportional to strain (Hooke&#39;s law), and the constant of proportionality is the Young&#39;s modulus (Y). So we can write:</p> <p>$\sigma = Y \epsilon$</p> <p>Substituting this into the energy density equation we get:</p> <p>$U = \frac{1}{2} Y \epsilon^2$</p> <p>The strain given in the problem is 0.04%, which needs to be converted to a decimal for use in this formula. Therefore, $\epsilon = 0.04/100 = 0.0004$.</p> <p>Substituting the values into the equation gives:</p> <p>$U = \frac{1}{2} \times 7.0 \times 10^{10} N/m^2 \times (0.0004)^2 = 5600 ~J/m^3$</p>