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⚛️ Physics  ·  Mechanical Properties of Solids  ·  NEET & JEE

Two wires A and B of the same material and same length, but radius of A is twice that of B, are stretched by the same force. The ratio of elastic potential energy stored in A to that in B is:

Answer: 1:4.

  • A 4:1
  • B 1:4
  • C 2:1
  • D 1:2

Correct answer: B. 1:4

Explanation: Elastic PE = F²L/(2AY). Since A ∝ r², energy ratio = A<sub>B</sub>/A<sub>A</sub> = (r<sub>B</sub>/r<sub>A</sub>)² = (1/2)² = 1/4, so ratio of A:B is 1:4.

Stress-Strain Curve for a Ductile Metal WireStrainStresselastic limityield pointUTS (max stress)fractureSlope of the straight (elastic) portion = Young's modulus Y; the curve beyond elastic limit shows permanent (plastic) deformation

A typical stress-strain curve: the initial straight line (Hooke's Law region, slope = Young's modulus) ends at the elastic limit; beyond the yield point, deformation becomes permanent, peaking at the ultimate tensile strength before the wire finally fractures.

Concept context

Stress, strain, Hookes law, and the elastic moduli that describe how solids deform and recover under load.

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