91影视

The linear expansion relation is,

$鈭�L=L_0伪鈭�T$

...(i)

$鈭�T=\frac{鈭�L}{L_0伪}$

...(ii)

$$\begin{gathered} 鈭�L=2.5020鈥�2.5000 \\ =0.0020in \end{gathered}$$

Substitute all the values in equation (ii),

$$\begin{gathered} 鈭�T=\frac{0.0020in}{\left(2.5000in\right)\left(1.2脳{10}^{-5}{\left(掳C\right)}^{-1}\right)} \\ =66.7掳C \end{gathered}$$

Therefore, the temperature to which the ring should be warmed is,

$$\begin{gathered} T=T_0+鈭�T \\ =20.0掳C+66.7掳C \\ =87掳C \end{gathered}$$

The linear expansion equation for the diameter of the brass shaft is,

$L_b=L_{0b}\left(1+{伪}_b鈭�T\right)$

The linear expansion equation for the diameter of the hole in the steel ring,

$L_s=L_{0s}\left(1+{伪}_s鈭�T\right)$

For same rise in temperature $鈭�T$,

$$\begin{gathered} L_{0b}\left(1+{伪}_b鈭�T\right)=L_{0s}\left(1+{伪}_s鈭�T\right) \\ {L}_{0b}+L_{0b}{伪}_b鈭�T=L_{0s}+L_{0s}{伪}_s鈭�T \end{gathered}$$

Rearranging for ,

$$\begin{gathered} 鈭�T=\frac{L_{0b}-L_{0s}}{L_{0s}{伪}_s-L_{0b}{伪}_b} \\ =\frac{2.5020in.-2.5000in.}{\left(2.5000in.\right)\left(1.2脳{10}^{-5}{\left(掳C\right)}^{-1}\right)-\left(2.5020in.\right)\left(2.0脳{10}^{-5}{\left(掳C\right)}^{-1}\right)} \\ =-100掳C \end{gathered}$$

Thus, the temperature at which the ring will just slip off the shaft is,

$$\begin{gathered} T=T_0+鈭�T \\ =20.0掳C-100掳C \\ =-80掳C \end{gathered}$$