Answer:
[tex]4.17\times 10^{-5}\ /^{\circ}C[/tex]
Explanation:
[tex]L_0[/tex] = Original length of rod
[tex]\alpha[/tex] = Coefficient of linear expansion = [tex]11\times 10^{-6}\ /^{\circ}C[/tex]
Initial temperature = 24°C
Final temperature = 280°C
Change in length of a Steel is given by
[tex]\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=11\times 10^{-6}\times 20.53\times (280-24)\\\Rightarrow \Delta L=0.05781248\ cm[/tex]
Change in material rod length will be
[tex]20.53-20.37+0.05781248=0.21781248\ cm[/tex]
The coefficient of thermal expansion is given by
[tex]\alpha=\frac{\Delta L}{L_0\Delta T}\\\Rightarrow \alpha=\frac{0.21781248}{20.37\times (280-24)}\\\Rightarrow \alpha=4.17\times 10^{-5}\ /^{\circ}C[/tex]
The coefficient of thermal expansion for the material is [tex]4.17\times 10^{-5}\ /^{\circ}C[/tex]
