Elastic PEMediumQuestion 6357
Question
Elastic PE stored in wire (Y=2×10¹¹ Pa, V=1 cm³, strain=10⁻³):
- A100 J
- B0.1 JCorrect
- C10 J
- D0.01 J
Correct answer
0.1 J
Explanation
u = ½Yε² = ½×2×10¹¹×10⁻⁶ = 10⁵ J/m³. PE = u×V = 10⁵×10⁻⁶ = 0.1 J.
Solve this quality-checked NEET multiple-choice question, then review the correct answer and explanation.
u = ½Yε² = ½×2×10¹¹×10⁻⁶ = 10⁵ J/m³. PE = u×V = 10⁵×10⁻⁶ = 0.1 J.