The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of $$5 \times 10^{-4}$$ is ______ kJ m$$^{-3}$$. Assume that material is elastic upto the linear strain of $$5 \times 10^{-4}$$.

From the graph, at a stress of $$20\text{ Pa}$$, the strain is $$1 \times 10^{-10}$$.

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{20}{1 \times 10^{-10}} = 2 \times 10^{11} \text{ Pa}$$

The energy density ($$u$$) is the energy stored per unit volume, calculated as: $$u = \frac{1}{2} \times Y \times (\text{Strain})^2$$

$$u = \frac{1}{2} \times (2 \times 10^{11}) \times (5 \times 10^{-4})^2$$

$$u = 10^{11} \times (25 \times 10^{-8})$$

$$u = 25 \times 10^3 \text{ J m}^{-3}$$

Since $$10^3 \text{ J} = 1 \text{ kJ}$$: $$u = 25 \text{ kJ m}^{-3}$$

The value of $$n$$ is 25.