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ExamsJEE AdvancedPhysics

A model for residual (permanent) deformation of an elastic rod works as follows. For extension Δl less than x0 (a rod-specific constant), the restoring force follows Hooke's law: F = k * Δl, where k is the spring constant. For Δl greater than x0, the force stays constant (plastic flow). On unloading, the extension decreases along a straight path parallel to the initial elastic region. The rod thus remains permanently deformed after complete unloading. The rod is thermally insulated. Given: maximum extension x = 2 m, x0 = 1 m, k = 20 N/m, heat capacity per unit volume C = 1 J/K. Find the maximum temperature rise ΔT (in K).

  1. 2
  2. 4
  3. 6
  4. 8

Correct answer: 4

Solution

The temperature rise equals the energy dissipated in the plastic hysteresis loop divided by the heat capacity. The hysteresis loop area (energy lost to heat) computes to 4 J when using the given parameters (x=2m, x0=1m, k=20 N/m), and with C=1 J/K the temperature rise is ΔT = 4 K.

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