To solve this problem it is necessary to apply the concepts related to thermal linear expansion. Mathematically this concept can be expressed under the Equation:
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
Where,
[tex]L_0 =[/tex] Initial Length
[tex]\alpha =[/tex]Thermal Expanssion constant
[tex]\Delta T =[/tex]Change in Lenght
Our values are given as
[tex]L_0 = 2.99m[/tex]
[tex]\alpha = 5.50*10^{-7} ( \∘C )^{-1}[/tex]
[tex]\Delta T = T_2-T_1 = 213-20[/tex]
Replacing at our equation we have,
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
[tex]\Delta L = (2.99) (5.5 x 10^{-7}) (213 - 20)[/tex]
[tex]\Delta L = 0.317*10^{-3}m\approx 0.317mm[/tex]
Therefore the expantion of the bar is 0.317mm
