Thermal Properties of Matter

JEE Main level, single correct, medium

14 of 126

Question 14 of 126

A brass wire of length 2 m and radius 1 mm at 27∘C27^{\circ} \mathrm{C} is held taut between two rigid supports. Initially it was cooled to a temperature of −43∘C-43^{\circ} \mathrm{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 1.4T1.4 T, is ____\_\_\_\_ ∘C{ }^{\circ} \mathrm{C}.

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Answer: (C) -71

Thermal stress in the wire is caused by contraction.
At -43°C, the temperature drop from 27°C is
ΔT1=70°C\Delta T_1 = 70°C
It creates tension
T=YAαΔT1T = YA\alpha\Delta T_1 where YY is Young's modulus and α\alpha is linear expansion coefficient.
For tension to increase to
1.4T1.4T, the new temperature drop must be: ΔT2=1.4×70=98°C\Delta T_2 = 1.4 \times 70 = 98°C.
Required temperature:
T3=27−98=−71°CT_3 = 27 - 98 = -71°C

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