Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
In a wire stretched by hanging a weight from its end, the elastic potential energy per unit volume in terms of longitudinal strain σ and modulus of elasticity Y is
PhysicsMechanical Properties of SolidsNEET
Options:
  • A Yσ22
  • B Yσ2
  • C 2Yσ22
  • D Y2σ2
Solution:
2385 Upvotes Verified Answer
The correct answer is: Yσ22
Energy density=12×stress×strain,

Y=stressσ or stress=Yσ,

 Energy density=12 Yσ×σ=Yσ22

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.