Answer
a.
we should heat the ring.
b.
$430^{\circ}\mathrm{C}$
Work Step by Step
Heating the ring, the diameter of the hole expands with the same linear coefficient as the material of which the ring is made.
Making the diameter greater means that the steel bar (which does not expand, since its $\Delta T=0$ ) could fit inside it.
So,
a.
We should heat the ring.
b.
Table 16-1 gives $\alpha_{\mathrm{A}\mathrm{l}}=2.4\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1}$
Use equation 16-4, $\Delta L=\alpha L_{0}\Delta T, $
with $\mathrm{L}$ being the diameter of the hole in the ring.
$\displaystyle \Delta T=\frac{\Delta L}{\alpha L_{0}}$
$T-T_{0}=\displaystyle \frac{\Delta L}{\alpha L_{0}}$
$T=T_{0}+\displaystyle \frac{\mathrm{L}-\mathrm{L}_{0}}{\alpha L_{0}}$
$=10.00^{\circ}\displaystyle \mathrm{C}+\frac{4.040\mathrm{c}\mathrm{m}-4.000\mathrm{c}\mathrm{m}}{(2.4\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1})(4.000\mathrm{c}\mathrm{m} )}=430^{\circ}\mathrm{C}$
