By the definition of linear thermal expansion:
[tex]\begin{gathered} \Delta L=\alpha L_0\Delta T \\ L-L_0=\alpha L_0(T-T_0) \end{gathered}[/tex]Here, L_0=4.997 cm=0.04997 m is the initial diameter of the steel plug at some temperature T_0, L=5 cm=0.05 m is the diameter of steel plug at T=30°C, α=11*10^(-6) °C^(-1) is the linear expansion of steel.
Rearranging the above equation to get an expression for T_0,
[tex]\begin{gathered} T-T_0=\frac{L-L_0}{\alpha L_0} \\ T_0=T-\frac{L-L_0}{\alpha L_0} \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} T_0=(30^{\circ}C)-\frac{(0.05\text{ m})-(0.04997\text{ m})}{(11\times10^{-6\circ}C^{-1})\times(0.04997\text{ m})} \\ =-24.58\degree C \end{gathered}[/tex]Therefore, at -24.58°C the stell plug will fit exactly into a hole of the constant diameter of 4.997 cm.
