🔥Thermal Expansion. ZERO to PYQs. JEE Main 2026.

Thermal Expansion

The lecture develops thermal expansion from basic definitions through JEE Main-style applications. It covers changes in length, area, volume, and density; constrained expansion and thermal stress; bimetallic strips; direction-dependent expansion; and the expansion of liquids.

1. Linear Expansion

A rod’s natural length is its length when it is free of stress. If its temperature changes by (\Delta T), its length becomes

[ L = L_0(1+\alpha\Delta T), ]

and its change in length is

[ \Delta L = \alpha L_0\Delta T. ]

Here, (\alpha) is the coefficient of linear expansion, which depends on the material. Its units are (\text{°C}^{-1}) or (\text{K}^{-1}). For many metals, it is roughly of the order of (10^{-5}\,\text{°C}^{-1}).

A positive temperature change causes expansion; a negative one causes contraction.

Example approach: When two rods have the same initial length and equal final lengths, write the expansion formula for each rod and equate the final lengths. In the cited 2019 problem, this gives a final temperature of (230^\circ\text{C}) for the second rod.

2. Equivalent Expansion Coefficient for Rods in Series

For rods joined end-to-end and heated by the same temperature change, their total expansion is the sum of their individual expansions. The equivalent coefficient is

[ \alpha_{\text{eq}} = \frac{\sum_i \alpha_i L_i}{\sum_i L_i}. ]

This is a length-weighted average. In a cited 2020 question, the effective coefficient for two joined wires is found using this relationship.

3. Expansion of Holes, Areas, and Volumes

Any distance within a uniformly expanding solid increases by the same linear expansion factor.

A hole in a sheet expands when heated. Its diameter changes as

[ d = d_0(1+\alpha\Delta T). ]

The material around the hole expands, increasing the hole’s circumference and therefore its diameter.

For small (\alpha\Delta T), use the binomial approximation ((1+x)^n \approx 1+nx):

• Area expansion: [ A = A_0(1+2\alpha\Delta T), \qquad \beta = 2\alpha. ] Here, (\beta) is the coefficient of areal or superficial expansion.

• Volume expansion: [ V = V_0(1+3\alpha\Delta T), \qquad \gamma = 3\alpha. ] Here, (\gamma) is the coefficient of volume or cubical expansion.

These first-order formulas apply to objects of any shape, assuming the material expands uniformly and (\alpha\Delta T) is small.

A cited 2023 hole problem uses the linear formula to find a diameter increase of (12\times10^{-3}\,\text{cm}). A cited 2022 cube problem uses the volume formula to calculate the change in volume from the cube’s surface area and temperature rise.

4. Density Change with Temperature

Heating does not change an object’s mass, but it changes its volume. Thus,

[ \rho = \frac{\rho_0}{1+\gamma\Delta T}. ]

For small (\gamma\Delta T),

[ \rho \approx \rho_0(1-\gamma\Delta T). ]

Therefore, heating generally lowers density, while cooling raises it. Since (\gamma=3\alpha), the percentage change in volume is about three times the percentage change in length. The density’s percentage change is approximately the negative of the volume’s percentage change.

Subtitle inconsistency: The cited 2020 density example gives conflicting length-change percentages in the subtitles (0.02% and 0.2%), and an answer that does not consistently follow either value. The reliable method is to use (\Delta\rho/\rho \approx -3\alpha\Delta T), or equivalently, the negative of the percentage volume change.

5. Thermal Stress When Expansion Is Constrained

A heated rod held between fixed supports cannot reach its new, freely expanded natural length. Its actual length stays fixed while its natural length increases, so it is compressed.

For small expansion, the magnitude of the thermal strain is approximately

[ \text{strain} \approx \alpha\Delta T. ]

Using Young’s modulus (Y),

[ \text{stress} = Y(\text{strain}), \qquad F = (\text{stress})A. ]

Thus, the magnitude of the compressive force is approximately

[ F = YA\alpha\Delta T. ]

The same stress–strain relationship applies when a load stretches a cooled rod back toward its original length. In that case, the rod’s thermal contraction determines the extension that must be restored.

Problem-solving sequence:

1. Find the rod’s free thermal length change.
2. Compare it with the constrained or restored length.
3. Calculate strain using the natural length, often approximated by (L_0).
4. Use (Y\epsilon) for stress, then multiply by the cross-sectional area to find force.

6. Bimetallic Strips

Two metals bonded together expand by different amounts if their (\alpha) values differ.

• When heated, the metal with the larger (\alpha) would become longer, so the strip bends with that metal on the outer, longer side.
• When cooled, the metal with the larger (\alpha) contracts more, becoming the shorter, inner side of the bend.

In the cited 2021 question, metal A has a larger expansion coefficient than metal B. Cooling makes A contract more, so the strip bends toward A’s side.

7. Anisotropic Expansion

In an anisotropic material, the linear expansion coefficient can differ by direction. For a cube with coefficients (\alpha_x), (\alpha_y), and (\alpha_z), the volume coefficient is

[ \gamma = \alpha_x + \alpha_y + \alpha_z. ]

This follows by expanding each dimension separately and neglecting products of small terms such as ((\alpha_x\Delta T)(\alpha_y\Delta T)).

The cited 2020 question applies this sum to obtain (\gamma=60\times10^{-6}\,\text{°C}^{-1}).

8. Expansion of Liquids and Overflow

A liquid has no fixed shape independent of its container, so the lecture treats its volume expansion rather than assigning it a linear expansion coefficient. Its volume follows

[ V = V_0(1+\gamma\Delta T). ]

If a container and its liquid are both heated, the overflow equals the liquid’s final volume minus the container’s final capacity. For an initially full container,

[ V_{\text{overflow}} = V_0(\gamma_{\text{liquid}}-\gamma_{\text{container}})\Delta T. ]

In a cited 2020 mercury-beaker problem, the condition that the unfilled volume remains constant is expressed by equating the changes in the beaker’s capacity and the mercury’s volume.

Speakers and Sources

• Vinay Uppal — lecturer and instructor.
• JEE Main previous-year questions — examples are attributed in the lecture to the 2019, 2020, 2021, 2022, and 2023 exams.