The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $\mathrm{T}_{1}$ to $\mathrm{T}_{2}$ resulting in increase of its length by $\Delta \mathrm{L}_{1}$. The temperature is further increased from $\mathrm{T}_{2}$ to $\mathrm{T}_{3}$ such that the increase in its length is $\Delta \mathrm{L}_{2}$. Given $\mathrm{T}_{3}+\mathrm{T}_{1}=2 \mathrm{~T}_{2}$ and $\mathrm{T}_{2}-\mathrm{T}_{1}=\Delta \mathrm{T}$, the value of $\Delta \mathrm{L}_{2}$ is $\_\_\_\_$ .

(D) $\Delta \mathrm{L}_{1}[1+\alpha \Delta \mathrm{~T}]$

(A) $\Delta \mathrm{L}_{1}\left[1+2 \alpha^{2}(\Delta \mathrm{~T})^{2}\right]$

(B) $\Delta \mathrm{L}_{1}\left[1+\alpha^{2}(\Delta \mathrm{~T})^{2}\right]$

(C) $\Delta \mathrm{L}_{1}[1+2 \alpha \Delta \mathrm{~T}]$

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Heat & Thermodynamics

The temperature of a metal strip having coefficient of linear expansion

α

is increased from

T_{1}

T_{2}

resulting in increase of its length by

Δ L_{1}

. The temperature is further increased from

T_{2}

T_{3}

such that the increase in its length is

Δ L_{2}

.
Given

T_{3} + T_{1} = 2 T_{2}

and

T_{2} - T_{1} = Δ T

, the value of

Δ L_{2}

____

Answer: (D)

The first elongation is

Δ L_{1} = L_{0} α (T_{2} - T_{1}) = L_{0} α Δ T

, leaving a length

L_{0} + Δ L_{1}

T_{2}

. From

T_{3} + T_{1} = 2 T_{2}

it follows that

T_{3} - T_{2} = T_{2} - T_{1} = Δ T

, so the second heating expands the already-lengthened strip:

Δ L_{2} = (L_{0} + Δ L_{1}) α Δ T = L_{0} α Δ T + Δ L_{1} α Δ T .

Since

L_{0} α Δ T = Δ L_{1}

, this gives

Δ L_{2} = Δ L_{1} (1 + α Δ T)

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option