JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

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 ______.

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

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

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Solution

Let the length of the metal strip at temperature $T_1$ be $L_1$ .

When the temperature is increased from $T_1$ to $T_2$ , the new length $L_2$ is given by:

$L_2 = L_1(1 + \alpha \Delta T)$

where $\Delta T = T_2 - T_1$ .

The increase in length is:

$\Delta L_1 = L_2 - L_1 = L_1 \alpha \Delta T$

When the temperature is further increased from $T_2$ to $T_3$ , the new length $L_3$ is:

$L_3 = L_2(1 + \alpha \Delta T')$

where $\Delta T' = T_3 - T_2$ .

Given $T_3 + T_1 = 2T_2$ , we have $T_3 - T_2 = T_2 - T_1 = \Delta T$ .

Thus, $\Delta T' = \Delta T$ .

The increase in length for this second temperature change is:

$\Delta L_2 = L_3 - L_2 = L_2 \alpha \Delta T$

Substituting $L_2 = L_1(1 + \alpha \Delta T)$ into the equation for $\Delta L_2$ :

$\Delta L_2 = L_1(1 + \alpha \Delta T) \alpha \Delta T$

$\Delta L_2 = (L_1 \alpha \Delta T)(1 + \alpha \Delta T)$

Since $\Delta L_1 = L_1 \alpha \Delta T$ , substituting this gives:

$\Delta L_2 = \Delta L_1(1 + \alpha \Delta T)$

Answer: $\Delta L_1[1+\alpha\Delta T]$

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Solution

Let the length of the metal strip at temperature $T_1$ be $L_1$ .

When the temperature is increased from $T_1$ to $T_2$ , the new length $L_2$ is given by:

$L_2 = L_1(1 + \alpha \Delta T)$

where $\Delta T = T_2 - T_1$ .

The increase in length is:

$\Delta L_1 = L_2 - L_1 = L_1 \alpha \Delta T$

When the temperature is further increased from $T_2$ to $T_3$ , the new length $L_3$ is:

$L_3 = L_2(1 + \alpha \Delta T')$

where $\Delta T' = T_3 - T_2$ .

Given $T_3 + T_1 = 2T_2$ , we have $T_3 - T_2 = T_2 - T_1 = \Delta T$ .

Thus, $\Delta T' = \Delta T$ .

The increase in length for this second temperature change is:

$\Delta L_2 = L_3 - L_2 = L_2 \alpha \Delta T$

Substituting $L_2 = L_1(1 + \alpha \Delta T)$ into the equation for $\Delta L_2$ :

$\Delta L_2 = L_1(1 + \alpha \Delta T) \alpha \Delta T$

$\Delta L_2 = (L_1 \alpha \Delta T)(1 + \alpha \Delta T)$

Since $\Delta L_1 = L_1 \alpha \Delta T$ , substituting this gives:

$\Delta L_2 = \Delta L_1(1 + \alpha \Delta T)$

Answer: $\Delta L_1[1+\alpha\Delta T]$

JEE Main 2026 — Physics Thermodynamics

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JEE Main 2026 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics