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JEE Main 2014 · Thermal Stress & Breaking Stress

Thermal Stress & Breaking Stress question from JEE Main 2014

Q16JEE Main 2014Easy

Thermal Stress & Breaking Stress

The pressure that has to be applied to the ends of a steel wire of length 10 cm10\text{ cm} to keep its length constant when its temperature is raised by 100∘C100^\circ\text{C} is: (For steel Young's modulus is 2×1011 Nm−22 \times 10^{11}\text{ Nm}^{-2} and coefficient of thermal expansion is 1.1×10−5 K−11.1 \times 10^{-5}\text{ K}^{-1})

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Answer: (a)

  1. Thermal stress =YαΔT= Y\alpha\Delta T (independent of length).
  2. =2×1011×1.1×10−5×100=2.2×108 Pa= 2 \times 10^{11} \times 1.1 \times 10^{-5} \times 100 = 2.2 \times 10^8\text{ Pa}
  3. So the answer is (a) 2.2×108 Pa2.2 \times 10^8\text{ Pa}.
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