A wire is stretched by a force 'F' which causes an extension ∆l, the energy stored in the wire is:
Correct answer: D. ½ F∆l
- A. F∆l
- B. 2F∆l
- C. ½ F∆l2
- D. ½ F∆l
Explanation
Given that, force applied = F . extension = ∆l and we assume the length of the wire to be = L.If the elastic limit is not exceeded then the stress is directly proportional to strain.Where stress is the amount of force applied per unit area (σ = F/A)And strain is extension per unit length (ε = ∆l/l)Hence,Energy stored = ½ x stress x strain x volumeEnergy stored = ½ x (F/A) x (∆l / L) x A x LEnergy stored = ½ F∆l
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Solids are compared through crystal structure, amorphous arrangement, elasticity, plastic deformation, stress, strain and Young’s modulus. Electrical behavior depends on available charge carriers and band structure, so conductors, insulators and semiconductors must be distinguished from mechanical properties such as strength, stiffness and brittleness.
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