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NEET 2024 · Thermal Stress & Breaking Stress

Thermal Stress & Breaking Stress question from NEET 2024

Q4NEET 2024Easy

Thermal Stress & Breaking Stress

A metallic bar of Young's modulus, 0.5×1011 N m−20.5 \times 10^{11}\text{ N m}^{-2} and coefficient of linear thermal expansion 10−5 ∘C−110^{-5}\,{}^\circ\text{C}^{-1}, length 1 m1\text{ m} and area of cross-section 10−3 m210^{-3}\text{ m}^2 is heated from 0∘C0^\circ\text{C} to 100∘C100^\circ\text{C} without expansion or bending. The compressive force developed in it is:

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Answer: (b)

  1. Thermal force when expansion is prevented: F=YAαΔTF = YA\alpha\Delta T.
  2. F=0.5×1011×10−3×10−5×100=5×104 NF = 0.5 \times 10^{11} \times 10^{-3} \times 10^{-5} \times 100 = 5 \times 10^4\text{ N}
  3. So the answer is (b) 50×103 N50 \times 10^3\text{ N}.
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