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

We need to find the elastic energy stored per unit volume in an aluminium rod.

Formula for energy per unit volume. The energy stored per unit volume in an elastically deformed material is:

$$u = \frac{1}{2} \times Y \times (\text{strain})^2$$

Convert strain to decimal. Strain = 0.04% = $$\frac{0.04}{100} = 4 \times 10^{-4}$$

Calculate energy per unit volume: $$u = \frac{1}{2} \times 7.0 \times 10^{10} \times (4 \times 10^{-4})^2$$

$$= \frac{1}{2} \times 7.0 \times 10^{10} \times 16 \times 10^{-8}$$

$$= \frac{1}{2} \times 7.0 \times 16 \times 10^{2}$$

$$= \frac{1}{2} \times 11200 = 5600 \text{ J m}^{-3}$$

The correct answer is Option C: 5600.