Question

A machinist wishes to insert a brass rod with a diameter of 3 mm into a hole with a diameter of 2.996 mm. By how much would the machinist have to lower the temperature (in °C) of the rod to make it fit the hole?

Answer 1

$$d_r(1-\alpha \Delta T)=d_h\\ 3(1-19\cdot10^{-6} \Delta T)=2.996\\ \Delta T=70.18\degree C$$