Q70 / 75 · JEE Main 2026 (24 January Shift 1) · Physics
A brass wire of length 2 m and radius 1 mm at $27^{\circ} \mathrm{C}$ is held taut between two rigid supports. Initially it was cooled to a temperature of $-43^{\circ} \mathrm{C}$ 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 $1.4 T$, is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.
- A-86
- B-65
- Correct answerC-71
- D-80
Solution
Thermal stress in the wire is caused by contraction.
At -43°C, the temperature drop from 27°C is $\Delta T_1 = 70°C$
It creates tension $T = YA\alpha\Delta T_1$ where $Y$ is Young's modulus and $\alpha$ is linear expansion coefficient.
For tension to increase to $1.4T$, the new temperature drop must be: $\Delta T_2 = 1.4 \times 70 = 98°C$.
Required temperature: $T_3 = 27 - 98 = -71°C$
At -43°C, the temperature drop from 27°C is $\Delta T_1 = 70°C$
It creates tension $T = YA\alpha\Delta T_1$ where $Y$ is Young's modulus and $\alpha$ is linear expansion coefficient.
For tension to increase to $1.4T$, the new temperature drop must be: $\Delta T_2 = 1.4 \times 70 = 98°C$.
Required temperature: $T_3 = 27 - 98 = -71°C$
Hence, the correct answer is option: 'C'.