The temperature of a metal strip having coefficient of linear expansion α is increased from T 1 to T 2 resulting in increase of its length by Δ L 1 . The temperature is further increased from T 2 to T 3 such that the increase in its length is Δ L 2 . Given T 3 + T 1 = 2T 2 and T 2 - T 1 = Δ T , the value of Δ L 2 is

Physics · Thermal Properties of Matter

JEE Main 2026 — 4 April, Evening Shift — Question 9

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

Option A:
$\Delta L_1[1 + 2\alpha^2(\Delta T)^2]$
Option B:
$\Delta L_1[1 + \alpha^2(\Delta T)^2]$
Option C:
$\Delta L_1[1 + 2\alpha \Delta T]$
Option D:
$\Delta L_1[1 + \alpha \Delta T]$
Correct

Answer: D

Step-by-step solution

$\Delta L_1 = L_0 \alpha \Delta T$ . Final length at $T_2$ is $L_0 + \Delta L_1$ . Also $T_3 - T_2 = \Delta T$ . Then $\Delta L_2 = (L_0 + \Delta L_1) \alpha \Delta T = L_0 \alpha \Delta T + \Delta L_1 \alpha \Delta T = \Delta L_1(1 + \alpha \Delta T)$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Physics
Chapter: Thermal Properties of Matter
Topic: Thermal Expansion of Solids and its Applications