NTA Abhyas JEE Main2020PhysicsMechanical Properties of SolidsPractice
A uniform rod of length l is acted upon by a force F in a gravity-free region, as shown in the figure. If the area of cross-section of the rod is A and it's Young's modulus is Y , then the elastic potential energy stored in the rod due to elongation is
Options
- AU = F 2 l 6 A 2 Y
- BU = F 2 l 3 A 2 Y
- CU = F 2 l 2 4 A 2 Y
- DU = F 2 l 2 2 A 2 Y
Correct answer
A. U = F 2 l 6 A 2 Y
Step-by-step solution
The acceleration of the rod is a = F m So, the tension in the rod at a distance x from the free end is T x = m x l × F m = F x l The longitudinal stress developed inside the rod at a distance x from the free end is σ = T x A = F x l A The strain energy density (energy per unit volume) is d U d V = 1 2 σ 2 Y = 1 2 F 2 x 2 Y A 2 l 2 , where d V = A d x is the volume of a small part of the rod So the energy stored in the rod is U = 1 2 F 2 Y A 2 l 2 ∫ 0 l x 2 d x U = F 2 l 6 A 2 Y