Question

A steel rod is 3.418 cm in diameter at 26.00°C. A brass ring has an interior diameter of 3.410 cm at 26.00°C. At what common temperature will the ring just slide onto the rod? The linear expansion coefficient of steel is 11.00 × 10^-6 1/C°. The linear expansion coefficient of brass is 19.00 × 10^-6 1/C°.

Answer 1

The ring will just slide onto the rod when diameters of both become same.

Let at temperature T diameters of both are same.

let D_s and D_B are the initial diameters of steel rod and brass ring respectively.While D_s^' and D_B^' are the diameters of Steel rod and brass ring at temperature T respectively.

we can write

$D_{S}^{'}=D_{S}(1+\alpha_{S}(T-26))$ --------(1)

$D_{B}^{'}=D_{B}(1+\alpha_{B}(T-26))$   ---------(2)

where $\alpha_{S}$ and $\alpha_{B}$ are the linear expansion coefficient for steel and brass respectively.

According to question-

$D_{S}^{'}=D_{B}^{'}$

$\Rightarrow$$D_{S}(1+\alpha_{S}(T-26))=D_{B}(1+\alpha_{b}(T-26))$

$\Rightarrow$$3.418\times(1+11\times 10^{-6}(T-26))=3.410(1+19\times 10^{-6}(T-26))$

$\Rightarrow$$3.418\times11\times 10^{-6}(T-26)-3.410\times 19\times 10^{-6}(T-26))=3.410-3.418$

$\Rightarrow$$-27.192\times 10^{-6}(T-26)=-0.008$

$\Rightarrow T-26=294.20$$T=268.20^{0}C$

$\Rightarrow T=320.20^{0}C$