Chapter 13.4, Problem 43PP

To determine

Space between the plate and the rod.

Answer to Problem 43PP

There is a space of $1.7\times {10}^{-4}cm$ between the rod and the hole.

Explanation of Solution

Given:

Diameter of hole $d_h=0.85cm$

Initial temperature, $T_1=30.0°C$

Final temperature, $T_2=0.0°C$

Formula used:

  $\Delta d_x=\alpha \Delta T$

Where, $x=element,\alpha =25\times {10}^{-6}°C^{-1}$ for aluminum and $\alpha =12\times {10}^{-6}°C^{-1}$ for steel.

  $\Delta T$ is the change in temperature and $\alpha$ is the coefficient of thermal expansion.

Calculation:Aluminum shrinks more than steel.

  $\Delta d_{\text{aluminum}}=(25\times {10}^{-6}°C^{-1})\times (0.85cm)(0.0°C-30.0°C)$

  $\Delta d_{\text{aluminum}}=-6.38\times {10}^{-4}cm$

For steel, diameter has shrunk by:

  $\Delta d_{steel}=(12\times {10}^{-6}°C^{-1})\times (0.85cm)(0.0°C-30.0°C)$

  $\Delta d_{steel}=-3.06\times {10}^{-4}cm$

Hence, the spacing between hole and the rod will be:

  $\left(\frac{1}{2}\right)\left(6.38\times {10}^{-4}cm-3.06\times {10}^{-4}cm\right)=1.7\times {10}^{-4}cm$

Conclusion: Spacing between the rod and the hole is $1.7\times {10}^{-4}cm$ .