The temperature of a metal strip having coefficient of linear expansion α is increased from T₁ to T₂ resulting in increase of its length by Δ L₁. The temperature is further increased from T₂ to T₃ such that the increase in its length is Δ L₂. Given T₃ + T₁ = 2T₂ and T₂ - T₁ = Δ T, the value of Δ L₂ is ______. (a) Δ L₁ [1 + 2α² (Δ T)²] (b) Δ L₁ [1 + α² (Δ T)²] (c) Δ L₁ [1 + 2α Δ T] (d) Δ L₁ [1 + α Δ T]

Answer: (d). Δ L₁ = L₁αΔ T and the length at T₂ is L₂ = L₁(1+αΔ T). From T₃ + T₁ = 2T₂: T₃ - T₂ = T₂ - T₁ = Δ T. Δ L₂ = L₂αΔ T = L₁αΔ T(1+αΔ T) So the answer is (d) Δ L₁[1+αΔ T].

The temperature of a metal strip having coefficient of linear… | JEE Main 2026 (4 Apr, Shift 2)

Q8JEE Main 2026 · 4 APR · SHIFT 2Medium

Thermometry & Thermal Expansion

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 ______.

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Answer: (d)

$\Delta L_1 = L_1\alpha\Delta T$ and the length at $T_2$ is $L_2 = L_1(1+\alpha\Delta T)$ .
From $T_3 + T_1 = 2T_2$ : $T_3 - T_2 = T_2 - T_1 = \Delta T$ .
$\Delta L_2 = L_2\alpha\Delta T = L_1\alpha\Delta T(1+\alpha\Delta T)$
So the answer is (d) $\Delta L_1[1+\alpha\Delta T]$ .

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