Chapter 13, Problem 88A

(a)

To determine

Change in length of sides of the square.

Answer to Problem 88A

Change in length of the sides of square will be $3.8\times {10}^{-4}\text{m}$ .

Explanation of Solution

Given:

Linear coefficient of expansion of iron, $\alpha \text{ = 12}\times {10}^{-6}\text{per}°\text{C}$

Length of the plate, $L\text{ = 0}\text{.33}\text{m}$

Initial temperature, $T_1\text{= 0}°\text{C}$

Final temperature, $T_2\text{= 95}°\text{C}$

Formula used:

Formula of Linear thermal expansion is $\text{Δ}L\text{ = α }{L}_1\text{Δ}T$ .

Where, $\Delta L$ is the change in length  $L_1$ , $\alpha$ is a coefficient of linear expansion, $\Delta T$ is the change in temperature and $L_1$ is the length at temperature $T$ .

Calculation:

Substitute all the given values in formula

$\text{Δ}L\text{ = α }{L}_1\text{Δ}T$ or $\text{Δ}L\text{ = α }{L}_1\left(T_2-T_1\right)$

  $\text{=12}\times {10}^{-6}\text{per}°\text{C}\times \text{0}\text{.33}\text{m}\left(\text{95}°\text{C}-\text{0}°\text{C}\right)$

  $\text{=3}\text{.8}\times {\text{10}}^{-4}\text{m}$

Conclusion:

Change in length of the square if heated from $\text{0}°\text{C}$ to $\text{95}°\text{C}$ will be $3.8\times {10}^{-4}\text{m}$ .

(b)

To determine

Change in area of the square.

Answer to Problem 88A

Change in area will be $2.3\times {10}^{-3}\text{m}$ .

Explanation of Solution

Given:

Length of the plate $\text{= 0}\text{.33}\text{m}$

Change in length, $\Delta L\text{= }3.8\times {10}^{-4}\text{m}$

Formula used:

Area of square is $\left(A\right)=L^2$ , where, $\text{L}$ is length.

Calculation:

Area of square is $\left(A\right)=L^2$ .

  $\begin{array}{c}\text{Change = }\frac{\Delta A}{A}=\frac{A_2-A_1}{A_1}\\ =\frac{L_2{}^2-L_1{}^2}{L_1{}^2}\\ =\frac{{\left(L_1+\Delta L\right)}^2-L_1{}^2}{L_1{}^2}\end{array}$

Substitute the given values is formula, $\frac{{\left(L_1+\Delta L\right)}^2-L_1{}^2}{L_1{}^2}$

  $\text{=}\frac{{\left(\text{0}\text{.33}\text{m}+3.8\times {10}^{-4}\text{m}\right)}^2-{\left(\text{0}\text{.33}\text{m}\right)}^2}{{\left(\text{0}\text{.33}\text{m}\right)}^2}=\text{2}\text{.3}\times {10}^{-3}\text{m}$

Conclusion:

Change in area of a square will be $2.3\times {10}^{-3}\text{m}$ .