The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$ resulting in increase of its length by $\Delta L_1$. The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta \mathrm{L}_2$.
Given $T_3+T_1=2 T_2$ and $T_2-T_1=\Delta T$, the value of $\Delta L_2$ is
Sol.
1. Initial state and first expansion Let the initial length at temperature $T_1$ be $L_1$. The increase in length $\Delta L_1$ when temperature increases from $T_1$ to $T_2$ is:
$$
\begin{aligned}
& \Delta L_1=L_1 \alpha\left(T_2-T_1\right) \\
& =L_1 \alpha \Delta T
\end{aligned}
$$
2. Second expansion The length at temperature $T_2$ is:
$$
L_2=L_1+\Delta L_1=L_1(1+\alpha \Delta T)
$$
The increase in length $\Delta L_2$ when temperature increases from $T_2$ to $T_3$ is:
$$
\Delta L_2=L_2 \alpha\left(T_3-T_2\right)
$$
Given $T_3+T_1=2 T_2$, we have:
$$
T_3-T_2=T_2-T_1=\Delta T
$$
3. Calculation of ${ }^{\Delta L_2}$ Substituting $L_2$ and $\left(T_3-T_2\right)$ into the expression for $\Delta L_2$ :
$$
\begin{aligned}
& \Delta L_2=\left[L_1(1+\alpha \Delta T)\right] \alpha \Delta T \\
& =\left(L_1 \alpha \Delta T\right)(1+\alpha \Delta T) \\
& =\Delta L_1(1+\alpha \Delta T)
\end{aligned}
$$