The difference in length between the two rods = 0.0025m
Explanations:Linear expansivity of a material is given by the formula:
[tex]\alpha\text{ = }\frac{l_2-l_1}{l_1(\theta_2-\theta_1)}[/tex]For the Aluminium rod:
[tex]\begin{gathered} l_{A1}\text{ = 5.0m} \\ \theta_{A1}=20^0C \\ \theta_{A2}=70^0C \\ \alpha_A\text{ = }23\times10^{-6}C^{-1} \\ \alpha_A\text{ = }\frac{l_{A2}-l_{A1}}{l_{A1}(\theta_{A2}-\theta_{A1})} \\ \text{ }23\times10^{-6}\text{ = }\frac{l_{A2}-5}{5(70-20)} \\ 5\times50\times\text{ }23\times10^{-6}=\text{ }l_{A2}-5 \\ l_{A2}=\text{ (}5750\text{ }\times10^{-6})\text{ + 5} \\ l_{A2}=\text{ 0.00575+5} \\ l_{A2}=\text{ 5.00575m} \end{gathered}[/tex]For the Nickel rod:
[tex]\begin{gathered} l_{N1}\text{ = 5.0m} \\ \theta_{N1}=20^0C \\ \theta_{N2}=70^0C \\ \alpha_N=\text{ 13}\times10^{-6}C^{-1} \\ \alpha_N\text{ = }\frac{l_{N2}-l_{N1}}{l_{N1}(\theta_{N2}-\theta_{N1})} \\ \text{ 1}3\times10^{-6}\text{ = }\frac{l_{N2}-5}{5(70-20)} \\ 5\times50\times\text{ 1}3\times10^{-6}=\text{ }l_{A2}-5 \\ l_{N2}=\text{ (32}50\text{ }\times10^{-6})\text{ + 5} \\ l_{N2}=\text{ }0.00325+5 \\ l_{N2}=\text{ 5.00325m} \end{gathered}[/tex]The difference in length between the two rods will be given as:
[tex]\begin{gathered} l_{A2}-l_{N2}=\text{ 5.00575-5.00325} \\ l_{A2}-l_{N2}=0.0025m \end{gathered}[/tex]The difference in length between the two rods = 0.0025m
