Chapter 20, Problem 52P

(a)

To determine

To show that $\frac{L_A}{L_B}=\frac{{\alpha }_A}{{\alpha }_B}$ .

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length $L_A$ and $L_B$ is constant. The coefficient of linear expansions are ${\alpha }_A$ and ${\alpha }_B$ .

Write expression for change in length.

  $\Delta L=L_B-L_A$

Write expression for linear expansion.

  $\Delta L=\left(L_B+{\alpha }_B{L}_B\Delta T\right)-\left(L_A+{\alpha }_A{L}_A\Delta T\right)$

Solve above expression.

  $\Delta L=\left(L_B-L_A\right)+\left({\alpha }_B{L}_B-{\alpha }_A{L}_A\right)\Delta T$

Substitute $\Delta L$ for $L_B-L_A$ in above expression and solve.

  $\begin{array}{c}\Delta L=\Delta L+\left({\alpha }_B{L}_B-{\alpha }_A{L}_A\right)\Delta T\\ 0=\left({\alpha }_B{L}_B-{\alpha }_A{L}_A\right)\Delta T\\ 0=\left({\alpha }_B{L}_B-{\alpha }_A{L}_A\right)\\ \frac{L_A}{L_B}=\frac{{\alpha }_B}{{\alpha }_A}\end{array}$

Conclusion:

Thus, the equation is proved.

(b)

To determine

The value of $L_B$ .

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of $L_A$ is $250\text{cm}$ .

Formula used:

Write expression for change in length.

  $\Delta L=L_B-L_A$

Write expression for linear expansion.

  $\Delta L=\left(L_B+{\alpha }_B{L}_B\Delta T\right)-\left(L_A+{\alpha }_A{L}_A\Delta T\right)$

Solve above expression.

  $\Delta L=\left(L_B-L_A\right)+\left({\alpha }_B{L}_B-{\alpha }_A{L}_A\right)\Delta T$

Substitute $\Delta L$ for $L_B-L_A$ in above expression and solve.

  $\begin{array}{c}\Delta L=\Delta L+\left({\alpha }_B{L}_B-{\alpha }_A{L}_A\right)\Delta T\\ 0=\left({\alpha }_B{L}_B-{\alpha }_A{L}_A\right)\Delta T\\ 0=\left({\alpha }_B{L}_B-{\alpha }_A{L}_A\right)\end{array}$

Solve above expression.

  $\frac{L_A}{L_B}=\frac{{\alpha }_B}{{\alpha }_A}$   ..... (1)

Calculation:

Substitute $250\text{cm}$ for $L_A$ , $19\times {10}^{-6}{\text{K}}^{\text{-1}}$ for ${\alpha }_A$ and $11\times {10}^{-6}{\text{K}}^{\text{-1}}$ for ${\alpha }_B$ in equation (1).

  $\begin{array}{c}\frac{250\text{cm}}{L_B}=\frac{11\times {10}^{-6}{\text{K}}^{\text{-1}}}{19\times {10}^{-6}{\text{K}}^{\text{-1}}}\\ L_B=430\text{cm}\end{array}$

Conclusion:

Thus, the value of $L_B$ is $430\text{cm}$ .