A steel rod is 3.000 cm at `25^@C`. A brass ring has an interior diameter of 2.992 cm at `25^@C`. At what common temperature will the ring just slide on to the rod?

Explanation

`alpha_("steel")=12xx10^-6K^-1` and `alpha_("brass")=18xx10^-6K^-1`
Let t be the required common temperature Then `:. DeltaT=t-25`. At the common temperature, both must have the same diameter.
`D=3.000(1+12xx10^-6DeltaT)`
`=2.992(1+18xx10^-6DeltaT)`
`DeltaT=448^@Cimpliest=448+25`
`=473^@C`