JEE Main 2026 — Thermal Properties of Matter Question with 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:

$L_3=L_2(1+\alpha \Delta T^{\prime })$

where $\Delta T^{\prime }=T_3-T_2$.

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

Thus, $\Delta T^{\prime }=\Delta T$.

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