Show that the coefficient of area expansion,$(\Delta A / A) / \Delta T,$ of a rectangular sheet of a solid is twice its linear expansivity,$\alpha_{l}$.

(N/A) Consider a rectangular sheet of a solid material of length $a$ and breadth $b$. When the temperature increases by $\Delta T$,the length $a$ increases by $\Delta a = \alpha_{l} a \Delta T$ and the breadth $b$ increases by $\Delta b = \alpha_{l} b \Delta T$.
The increase in area $\Delta A$ is given by the sum of the three shaded regions in the diagram:
$\Delta A = \Delta A_{1} + \Delta A_{2} + \Delta A_{3}$
$\Delta A = a \Delta b + b \Delta a + (\Delta a)(\Delta b)$
Substituting the expressions for $\Delta a$ and $\Delta b$:
$\Delta A = a(\alpha_{l} b \Delta T) + b(\alpha_{l} a \Delta T) + (\alpha_{l} a \Delta T)(\alpha_{l} b \Delta T)$
$\Delta A = 2 \alpha_{l} ab \Delta T + \alpha_{l}^{2} ab (\Delta T)^{2}$
Since $A = ab$,we have:
$\Delta A = 2 \alpha_{l} A \Delta T + \alpha_{l}^{2} A (\Delta T)^{2}$
$\Delta A = \alpha_{l} A \Delta T (2 + \alpha_{l} \Delta T)$
Dividing by $A \Delta T$:
$\frac{\Delta A}{A \Delta T} = \alpha_{l} (2 + \alpha_{l} \Delta T)$
Since $\alpha_{l}$ is very small (typically $\approx 10^{-5} \text{ K}^{-1}$),the term $\alpha_{l} \Delta T$ is negligible compared to $2$. Therefore:
$\frac{\Delta A}{A \Delta T} \approx 2 \alpha_{l}$