91Ó°ÊÓ

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 ${\mathrm{δ}}_d$is

$\mathrm{δ}=\mathrm{π}{\mathrm{δ}}_d$

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

localid="1649077598252" $\frac{∆L_z}{L}={\mathrm{Î\pm }}_x∆T⇒∆L_x={\mathrm{Î\pm }}_{x}L∆T$

Thus, the necessary temperature difference should be

localid="1649077590486" $∆T=\frac{\mathrm{δ}}{({\mathrm{Î\pm }}_b-{\mathrm{Î\pm }}_8)L}$

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

$∆T=\frac{\mathrm{Ï€}{\mathrm{δ}}_n}{({\mathrm{Î\pm }}_b-{\mathrm{Î\pm }}_8)\mathrm{Ï€}D}=\frac{{\mathrm{δ}}_n}{({\mathrm{Î\pm }}_b-{\mathrm{Î\pm }}_8)D}$

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.