A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is 0.1 J.
Elastic energy = (1/2) F × x
Here, F = 200N and x = 1mm = 1 × 10-3
E = (1/2)×200×10-3
E = 0.1 J.