The temperature of a metal strip having coefficient of linear expansion alpha is increased from T1 to T2 resulting in increase of its length by Delta L1. The temperature is further increased from T2 to T3 such that the increase in its length is Delta L2. Given T3 + T1 = 2T2 and T2 - T1 = Delta T, the value of Delta L2 is _______.

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q34 – JEE-MAIN 2026

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Home Physics Thermal PropertiesQuestion #490

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}

is _______.

Clue 1 of 3: Conceptual Principle

Recall the fundamental formula for linear thermal expansion, which describes how the length of a material changes with temperature.

🧠 Full Solution Path

Step 1: Express

Δ L_{1}

and the length at

T_{2}

✦ Active

Let the initial length of the metal strip at temperature $T_{1}$ be $L_{0}$ . The coefficient of linear expansion is $α$ . The temperature change from $T_{1}$ to $T_{2}$ is $Δ T_{1} = T_{2} - T_{1}$ . We are given $T_{2} - T_{1} = Δ T$ . Thus, the increase in length $Δ L_{1}$ is:

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

The length of the strip at temperature $T_{2}$ is $L_{2} = L_{0} + Δ L_{1}$ .

Step 2: Express

Δ L_{2}

using the length at

T_{2}

○ Expand

The temperature is further increased from $T_{2}$ to $T_{3}$ . The change in temperature for this phase is $Δ T_{2} = T_{3} - T_{2}$ . We are given $T_{3} + T_{1} = 2 T_{2}$ . Rearranging this gives $T_{3} - T_{2} = T_{2} - T_{1}$ . Since $T_{2} - T_{1} = Δ T$ , we have $T_{3} - T_{2} = Δ T$ . The increase in length $Δ L_{2}$ is based on the length at $T_{2}$ ( $L_{2}$ ):

Δ L_{2} = L_{2} α (T_{3} - T_{2})

Δ L_{2} = (L_{0} + Δ L_{1}) α Δ T

💡 Teacher's Secret Hint

Remember to use the expanded length at $T_{2}$ as the initial length for the second expansion step.

Step 3: Substitute

Δ L_{1}

and simplify○ Expand

Expand the expression for $Δ L_{2}$ :

Δ L_{2} = L_{0} α Δ T + Δ L_{1} α Δ T

From Step 1, we know that $L_{0} α Δ T = Δ L_{1}$ . Substitute this into the equation for $Δ L_{2}$ :

Δ L_{2} = Δ L_{1} + Δ L_{1} α Δ T

Factor out $Δ L_{1}$ :

Δ L_{2} = Δ L_{1} (1 + α Δ T)

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Physics Question 34 – JEE-MAIN 2026

Clue 1 of 3: Conceptual Principle

🧠 Full Solution Path