An iron ring is to fit snugly on a cylindrical iron rod. At `20°C`, the diameter of the rod is `6.445\ cm` and the inside diameter of the ring is `6.420\ cm`. To slip over the rod, the ring must be slightly larger than the rod diameter by about `0.008\ cm`. To what temperature must the ring be brought if its hole is to be large enough so it will slip over the rod? Given: `d_0 = 6.420\ times 10^(-2)\ m` `d = 6.445 +0.008 =6.453 times 10^(-2)\ m` Known: `α = 12 times 10^(-6)\ (°C)^(-1)` Equation: `∆L = αL_0∆T` Solution: `∆L = αL_0∆T` `=>` `∆T=(∆L) /(αL_0)` `∆T=((6.453 times 10^(-2)\ m)-(6.420\ times 10^(-2)\ m)) /((12 times 10^(-6)\ (°C)^(-1))(6.420\ times 10^(-2)\ m))` `∆T=430°C` `T = 20°C + ∆T`