JEE Main2003PhysicsMechanical Properties of SolidsActual
A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 ~N to the lower end. The weight stretches the wire by 1 ~mm . Then the elastic energy stored in the wire is
Options
- A0.2 ~J
- B10 ~J
- C20 ~J
- D0.1 ~J
Correct answer
D. 0.1 ~J
Step-by-step solution
Elastic energy = 1 2 F x F =200 ~N , x =1 ~mm =10⁻³ ~m E = 1 2 200 1 10⁻³=0.1 ~J