A brass wire of length 2 m and radius 1 mm at $27^{\circ} \mathrm{C}$ is held taut between two rigid

Question

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 $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.

Professor PC Thomas Classes · Answer

The thermal tension developed in a clamped wire when cooled is $T = Y A \alpha \Delta 	heta$, where $\Delta 	heta$ is the change in temperature. Initially, the temperature drop is $\Delta 	heta_1 = 27^\circ	ext{C} - (-43^\circ	ext{C}) = 70^\circ	ext{C}$, yielding $T = YA\alpha(70)$. We want the new tension to be $1.4T$, so $1.4T = YA\alpha \Delta 	heta_2$. Dividing the two equations yields $\frac{\Delta 	heta_2}{70} = 1.4 \implies \Delta 	heta_2 = 98^\circ	ext{C}$. The final temperature required is $27^\circ	ext{C} - \Delta 	heta_2 = 27 - 98 = -71^\circ	ext{C}$.