i cant find the answer

Question

A machinist wishes to insert a brass rod with a diameter of 7 mm into a hole with a diameter of 6.998 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

A web search gave me a thermal coefficient of linear expansion for brass ranging between

10.4 and 11.8×10^-6 C^-1.

I do not know which value you are supposed to use, so I will just denote it as α.

Then you can use the formula

ΔL = LαΔC

where

ΔL = 0.002 mm is the change in linear dimension,

L = 7 mm is the length of that dimension,

ΔC is the change in temperature.

From that

ΔC = ΔL/(Lα)

Depending on your chosen value of α you should get something under 28°C.