## Physics: Principles with Applications (7th Edition)

Increase in length for Super Invar is $2.2\times10^{-6}m$. Increase in length for steel is $1.3\times10^{-4}m$.
Calculate the increase in length of the table, first for Super Invar, then for steel. $$\Delta \mathcal{l}=\alpha \mathcal{l}_o \Delta T=(0.20\times 10^{-6}/C^{\circ})(1.8m)(6 C^{\circ})=2.2\times10^{-6}m$$ $$\Delta \mathcal{l}=\alpha \mathcal{l}_o \Delta T=(12\times 10^{-6}/C^{\circ})(1.8m)(6 C^{\circ})=1.3\times10^{-4}m$$ The change in length for Super Invar is $\frac{1}{60}$ of the change for steel.