Question

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

Answer 1

Answer:

Explanation:

energy stored in per unit volume of the wire= 0.5*stress*strain

stress=50/πr² (r=0.5mm)

strain=ΔL/L=10⁻³/5

substitute values to get your answer.