Q 11-10-048JEE MainJEE Main 2026 (4 Apr, Shift 2)Medium
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) $\Delta L_1[1 + \alpha\Delta T]$
Let the length at $T_1$ be $L$. Then $\Delta L_1 = L\alpha\Delta T$ and the length at $T_2$ is $L(1 + \alpha\Delta T)$.
$T_3 - T_2 = T_2 - T_1 = \Delta T$, so the second expansion starts from the longer length:
$$\Delta L_2 = L(1 + \alpha\Delta T)\alpha\Delta T = \Delta L_1(1 + \alpha\Delta T)$$
Solution by Sreeraj P, M.Sc Physics