Suggested languages for you:

Americas

Europe

15P

Expert-verified
Found in: Page 541

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A steel rod is 3.00 cm in diameter at ${\mathbf{25}}{\mathbf{.}}{{\mathbf{00}}}^{{\mathbf{o}}}{\mathbf{C}}$. A brass ring has an interior diameter of 2.992 cm at ${\mathbf{25}}{\mathbf{.}}{{\mathbf{00}}}^{{\mathbf{o}}}{\mathbf{C}}$. At what common temperature will the ring just slide onto the rod?

The temperature at which ring just slides into the rod is $360.{0}^{\mathrm{o}}\mathrm{C}$

See the step by step solution

## Step 1: The given data

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

## Step 2: Understanding the concept of thermal expansion

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, $∆\mathrm{D}=\mathrm{D\alpha }∆\mathrm{T}$ …(i)

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

## Step 3: Calculation of the temperature

The total length after expansion is given as:

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

The total length after expansion is given as:

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

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:

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

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 $360.{0}^{\mathrm{o}}\mathrm{C}$