Answer:
[tex]\alpha = 4.12 \times 10^{-5} per Degree C[/tex]
Explanation:
As we know that when we increase the temperature of the rod and the ruler then in that case there will be change in the length of the both.
Due to this the ruler measurement is different from actual length of the rod
So by thermal expansion we know that
[tex]L = L_o(1 +\alpha \Delta T)[/tex]
[tex]L = 19.11 (1 + \alpha(280 - 19))[/tex]
[tex]L = 19.11(1 + 261\alpha)[/tex]
Now for length of unit division of steel scale
[tex]1 unit = 1( 1 + \alpha_s \Delta T)[/tex]
[tex]1 unit = (1 + (11 \times 10^{-6})(280 - 19))[/tex]
[tex]1 unit = 1.0029[/tex]
now the measured length from the scale is given as
[tex]L_{measured} = \frac{L}{1.0029}[/tex]
[tex]L_{measured} = \frac{19.11(1 + 261\alpha)}{1.0029}[/tex]
[tex]19.26 = \frac{19.11(1 + 261\alpha)}{1.0029}[/tex]
[tex]\alpha = 4.12 \times 10^{-5} per Degree C[/tex]
