91Ó°ÊÓ

1. Diameter of the steel rod at 25^oC is ${\mathrm{D}}_{\mathrm{s}}=3.000\mathrm{cm}$
2. Diameter of the brass ring at 25^oC is${\mathrm{D}}_{\mathrm{b}}=2.992\mathrm{cm}$
3. Expansion coefficient for steel is${\mathrm{Î\pm }}_{\mathrm{s}}=11×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}$
4. Expansion coefficient for brass is${\mathrm{Î\pm }}_{\mathrm{b}}=19×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}$

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. Thermal expansion can be occurring due to an increase in temperature. For the given problem, we have to use the formula for linear expansion to calculate the final temperature.

Formula:

The linear expansion of diameter of a body,$∆\mathrm{D}=\mathrm{\P ÙÎ\pm }∆\mathrm{T}$ …(¾±)

Where $\mathrm{Î\pm }$is the coefficient of linear expansion of body, $∆\mathrm{T}$is the temperature difference and D is the diameter

The total length after expansion is given as:

$\begin{array}{rcl}\mathrm{D}& =& {\mathrm{D}}_{\mathrm{s}}+∆{\mathrm{D}}_{\mathrm{s}}\\ & =& {\mathrm{D}}_{\mathrm{s}}+{\mathrm{D}}_{\mathrm{s}}{\mathrm{Î\pm }}_{\mathrm{s}}∆\mathrm{T}..............\left(\mathrm{a}\right)\left(\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{the}\mathrm{change}\mathrm{of}\mathrm{expansion}\right)\end{array}$

The total length after expansion is given as:

$\begin{array}{rcl}\mathrm{D}& =& {\mathrm{D}}_{\mathrm{b}}+∆{\mathrm{D}}_{\mathrm{b}}\\ & =& {\mathrm{D}}_{\mathrm{b}}+{\mathrm{D}}_{\mathrm{b}}{\mathrm{Î\pm }}_{\mathrm{b}}∆\mathrm{T}..............\left(\mathrm{b}\right)\left(\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{the}\mathrm{change}\mathrm{of}\mathrm{expansion}\right)\end{array}$

The rod will fit through the ring only if the diameter of the ring and the rod is the same after the expansion so that, from equations (a) and (b), it is given as:

$\begin{array}{rcl}{\mathrm{D}}_{\mathrm{s}}+{\mathrm{D}}_{\mathrm{s}}{\mathrm{Î\pm }}_{\mathrm{s}}∆\mathrm{T}& =& {\mathrm{D}}_{\mathrm{b}}+{\mathrm{D}}_{\mathrm{b}}{\mathrm{Î\pm }}_{\mathrm{b}}∆\mathrm{T}\\ {\mathrm{D}}_{\mathrm{s}}-{\mathrm{D}}_{\mathrm{b}}& =& {\mathrm{D}}_{\mathrm{b}}{\mathrm{Î\pm }}_{\mathrm{b}}∆\mathrm{T}-{\mathrm{D}}_{\mathrm{s}}{\mathrm{Î\pm }}_{\mathrm{s}}∆\mathrm{T}\\ {\mathrm{D}}_{\mathrm{s}}-{\mathrm{D}}_{\mathrm{b}}& =& \left({\mathrm{D}}_{\mathrm{b}}{\mathrm{Î\pm }}_{\mathrm{b}}-{\mathrm{D}}_{\mathrm{s}}{\mathrm{Î\pm }}_{\mathrm{s}}\right)∆\mathrm{T}\\ ∆\mathrm{T}& =& \frac{{\mathrm{D}}_{\mathrm{s}}-{\mathrm{D}}_{\mathrm{b}}}{{\mathrm{D}}_{\mathrm{b}}{\mathrm{Î\pm }}_{\mathrm{b}}-{\mathrm{D}}_{\mathrm{s}}{\mathrm{Î\pm }}_{\mathrm{s}}}\\ & & \end{array}$

By substituting the given values in the above equation, we get the temperature change as:

$\begin{array}{rcl}∆\mathrm{T}& =& \frac{3.000\mathrm{cm}-2.992\mathrm{cm}}{\left(2.992\mathrm{cm}×19×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}-3.000\mathrm{cm}×11×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}\right)}\\ & =& \frac{0.008\mathrm{cm}}{\left(56.848-33\right)×{10}^{-6}\mathrm{cm}/{}^{\mathrm{o}}\mathrm{C}}\\ & =& \frac{0.{008}^{\mathrm{o}}\mathrm{C}}{\left(23.848\right)×{10}^{-6}}\\ & =& 335.0^{\mathrm{o}}\mathrm{C}\end{array}$

Using this value, we can calculate the total temperature is given as:

$\begin{array}{rcl}\mathrm{T}& =& {\mathrm{T}}_{\mathrm{i}}+∆\mathrm{T}\\ & =& 25.{00}^{\mathrm{o}}\mathrm{C}+335.0^{\mathrm{o}}\mathrm{C}\\ & =& 360.0^{\mathrm{o}}\mathrm{C}\end{array}$

Hence, the value of the temperature is$360.0^{\mathrm{o}}\mathrm{C}$