Initial temperature, $T_{1}=27^{\circ} C$

Length of the brass wire at $T_{1}, l=1.8\; m$

Final temperature, $T_{2}=-39^{\circ} C$

Diameter of the wire, $d=2.0 mm =2 \times 10^{-3} m$

Tension developed in the wire $=F$

Coefficient of linear expansion of brass, $\alpha=2.0 \times 10^{-5} K ^{-1}$

Young's modulus of brass, $Y=0.91 \times 10^{11} Pa$ p://wuw tiwariacademy.com/

Young's modulus is given by the relation:

$Y=\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$

$\Delta L=\frac{F \times L}{A \times Y}$

$F=$ Tension developed in the wire

$A=$ Area of cross-section of the wire.

$\Delta L=$ Change in the length, given by the relation:

$\Delta L=\alpha L\left(T_{2}-T_{1}\right)$

$\alpha L\left(T_{2}-T_{1}\right)=\frac{F L}{\pi\left(\frac{d}{2}\right)^{2} \times Y}$

$F=\alpha\left(T_{2}-T_{1}\right) \pi Y\left(\frac{d}{2}\right)^{2}$

$F=2 \times 10^{-5} \times(-39-27) \times 3.14 \times 0.91 \times 10^{11} \times\left(\frac{2 \times 10^{-3}}{2}\right)^{2}$

$=-3.8 \times 10^{2} N$

(The negative sign indicates that the tension is directed inward.)

Hence, the tension developed in the wire is $3.8 \times 10^{2}\; N$