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 $\Delta L_2$.
Given $T_3 + T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, the value of $\Delta L_2$ is ______.
Answer: (D) $\Delta L_1[1 + \alpha\Delta T]$
Concept. $\Delta L=L\alpha\Delta T$ uses the length at the start of each rise. The second rise starts from the longer length $L_1(1+\alpha\Delta T)$.
Step 1. Equal rises. $T_3-T_2=T_2-T_1=\Delta T$ (from $T_3+T_1=2T_2$).
Step 2. Ratio. $\dfrac{\Delta L_2}{\Delta L_1}=\dfrac{L_1(1+\alpha\Delta T)}{L_1}=1+\alpha\Delta T$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
