When a wire of length 5m and diameter 1mm was stretched by a load of 5kg the elongation produced in the wire was 1mm. Find the energy stored in per unit volume of the wire?

Explanation

To find the energy stored per unit volume of the wire, we can follow these steps:

### Step 1: Gather Given Data
- Length of the wire, \( L = 5 \, \text{m} \)
- Diameter of the wire, \( D = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Radius of the wire, \( R = \frac{D}{2} = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \)
- Load applied, \( F = 5 \, \text{kg} \)
- Elongation, \( \Delta L = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
...