⚛️ Physics · Mechanical Properties of Solids · NEET & JEE

A rod (Y = 2×1011 Pa, α = 1.2×10-5 /K) is clamped rigidly between two walls and heated through 50 K. The thermal stress developed is:

Answer: 1.2×10 8 Pa.

A 0.6×108 Pa
B 1.2×106 Pa
C 2.4×108 Pa
D 1.2×108 Pa

Correct answer: D. 1.2×108 Pa

Explanation: Thermal stress = YαΔT = 2×1011 × 1.2×10-5 × 50 = 1.2×108 Pa.

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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A rod (Y = 2×10<sup>11</sup> Pa, α = 1.2×10<sup>-5</sup> /K) is clamped rigidly between two walls and heated through 50 K. The thermal stress developed is:

Concept context

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