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JEE Main 2019 (12 Apr, Shift 2) · Thermal Stress & Breaking Stress

Thermal Stress & Breaking Stress question from JEE Main 2019 (12 Apr, Shift 2)

Q14JEE Main 2019 · 12 APR · SHIFT 2Medium

Thermal Stress & Breaking Stress

A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to:

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Answer: (a)

  1. Length unchanged, so the compressive stress equals the thermal stress: Fπr2=YαT\dfrac{F}{\pi r^2} = Y\alpha T.
  2. α=Fπr2YT\alpha = \dfrac{F}{\pi r^2 YT}
  3. Coefficient of volume expansion γ=3α=3Fπr2YT\gamma = 3\alpha = \dfrac{3F}{\pi r^2 YT}
  4. So the answer is (a).
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