Answer:
[tex]5.28\times 10^{-5}\ /^{\circ}C[/tex]
Explanation:
[tex]L_0[/tex] = Original length of rod
[tex]\alpha[/tex] = Coefficient of linear expansion = [tex]1.62\times 10^{-5}\ /^{\circ}C[/tex]
Initial temperature = 16°C
Final temperature = 260°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 23.83\times (260-16)\\\Rightarrow \Delta L=0.06395972\ cm[/tex]
Change in material rod length will be
[tex]23.83-23.59+0.0639572=0.3039572\ cm[/tex]
The coefficient of thermal expansion is given by
[tex]\alpha=\frac{\Delta L}{L_0\Delta T}\\\Rightarrow \alpha=\frac{0.3039572}{23.59\times (260-16)}\\\Rightarrow \alpha=5.28\times 10^{-5}\ /^{\circ}C[/tex]
The coefficient of thermal expansion for the material is [tex]5.28\times 10^{-5}\ /^{\circ}C[/tex]