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JEE Main 2026 (24 Jan, Shift 1) · Thermal Stress & Breaking Stress

Thermal Stress & Breaking Stress question from JEE Main 2026 (24 Jan, Shift 1)

Q3JEE Main 2026 · 24 JAN · SHIFT 1Difficult

Thermal Stress & Breaking Stress

A brass wire of length 2 m and radius 1 mm at 27∘C27^\circ\text{C} is held taut between two rigid supports. Initially it was cooled to a temperature of −43∘C-43^\circ\text{C} creating a tension TT in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to 1.4T1.4T, is ‾ ∘C\underline{\qquad}\ {}^\circ\text{C}.

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Answer: (a)

  1. Tension in a wire held between rigid supports: T=YAαΔθT = YA\alpha\Delta\theta, so T∝ΔθT \propto \Delta\theta.
  2. First cooling: Δθ=27−(−43)=70∘C\Delta\theta = 27 - (-43) = 70^\circ\text{C} gives TT.
  3. For 1.4T1.4T: Δθ=1.4×70=98∘C\Delta\theta = 1.4 \times 70 = 98^\circ\text{C}, so final temperature =27−98=−71∘C= 27 - 98 = -71^\circ\text{C}.
  4. So the answer is (a) −71∘C-71^\circ\text{C}.
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