A brass wire of length 2 m and radius 1 mm at 27 is held taut between two rigid supports. Initially it was cooled to a temperature of -43 creating a tension T in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to 1.4T, is. (a) -71 (b) -80 (c) -65 (d) -86

Answer: (a). Tension in a wire held between rigid supports: T = YAαΔθ, so T ∝ Δθ. First cooling: Δθ = 27 - (-43) = 70 gives T. For 1.4T: Δθ = 1.4 × 70 = 98, so final temperature = 27 - 98 = -71. So the answer is (a) -71.

A brass wire of length 2 m and radius 1 mm at 27 is held taut between… | JEE Main 2026 (24 Jan, Shift 1)

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Thermal Stress & Breaking Stress

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

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Answer: (a)

Tension in a wire held between rigid supports: $T = YA\alpha\Delta\theta$ , so $T \propto \Delta\theta$ .
First cooling: $\Delta\theta = 27 - (-43) = 70^\circ\text{C}$ gives $T$ .
For $1.4T$ : $\Delta\theta = 1.4 \times 70 = 98^\circ\text{C}$ , so final temperature $= 27 - 98 = -71^\circ\text{C}$ .
So the answer is (a) $-71^\circ\text{C}$ .

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