A brass ring with inner diameter 2.00 cm and outer diameter 3.00 cm needs to fit over a 2.00-cm-diameter steel rod, but at 20$°$C the hole through the brass ring is 50 mm too small in diameter. To what temperature, in$°$C, must the rod and ring be heated so that the ring just barely slips over the rod?

A brass ring with inner diameter 2.00 cm and outer diameter 3.00 cm needs to fit over a 2.00-cm-diameter steel rod, but at 20$°$C the hole through the brass ring is 50 mm too small in diameter.

In increasing the length of the circumference, the requirement for ${\delta }_d$is

$\delta =\pi {\delta }_d$

There is a matter of change in length $∆L_x$of part with linear coefficient ${\alpha }_x$would be

localid="1649077598252" $\frac{∆L_z}{L}={\alpha }_x∆T\Rightarrow ∆L_x={\alpha }_{x}L∆T$

Thus, the necessary temperature difference should be

localid="1649077590486" $∆T=\frac{\delta }{({\alpha }_b-{\alpha }_8)L}$

Thus, there can be a replacement of the circumference in both the numerator and denominator and find

$∆T=\frac{\pi {\delta }_n}{({\alpha }_b-{\alpha }_8)\pi D}=\frac{{\delta }_n}{({\alpha }_b-{\alpha }_8)D}$

Thus, the substitution of numerics in order to find $∆T$is:

$∆T=\frac{5.{10}^{-5}}{(1.9-1.1).{10}^{-5}.0.02}=313°C$

Therefore, in $313°C$the rod and ring be heated so that the ring just barely slips over the rod.