The radius of a metal sphere at room temperature $T$ is $R$, and the coefficient of linear expansion of the metal is $\alpha$. The sphere is heated a little by a temperature $\Delta T$ so that its new temperature is $T + \Delta T$. The increase in the volume of the sphere is approximately (a) $2\pi R \alpha \Delta T$ (b) $\pi R^2 \alpha \Delta T$ (c) $4\pi R^3 \alpha \Delta T / 3$ (d) $4\pi R^3 \alpha \Delta T$

The radius of a metal sphere at room temperature T is R, and the coefficient of linear expansion of the metal

NCERT Exemplar · Q7

T

(a)

2\pi R \alpha \Delta T

(b)

\pi R^2 \alpha \Delta T

(c)

4\pi R^3 \alpha \Delta T / 3

(d)

4\pi R^3 \alpha \Delta T

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Concept understanding — Thermal Expansion Coefficient

Thermal Expansion Coefficient: From Intuition to Precision

The Intuition: What Happens When Things Get Hot?

Think about a metal railway track on a summer day. The track is laid in sections with small gaps between them. On a hot afternoon, those gaps get smaller — sometimes the track even buckles. Why? Because the metal expands when heated.

Or consider a mercury thermometer. The liquid mercury sits in a bulb at the bottom. When your body warms the bulb, the mercury expands and rises up the narrow tube. The hotter you are, the higher it climbs.

This is thermal expansion: most materials get bigger when heated and smaller when cooled. The atoms inside vibrate more vigorously as temperature rises, pushing their neighbours slightly farther apart. The entire object grows in every direction.

But different materials expand by different amounts. A steel rod and an aluminium rod of the same length, heated by the same amount, will not end up the same length. Aluminium expands more. So we need a number that tells us how much a given material expands per degree of temperature change. That number is the thermal expansion coefficient.

The Precise Statement: Defining the Coefficient

There are actually three coefficients, depending on whether we care about length, area, or volume. For a first meeting, we focus on the most common one: the linear thermal expansion coefficient, denoted by the Greek letter $\alpha$ (alpha).

\alpha = \frac{1}{L_0} \cdot \frac{\Delta L}{\Delta T}

Where:

$L_0$ is the original length of the object (at some starting temperature)
$\Delta L$ is the change in length (final length minus original length)
$\Delta T$ is the change in temperature (final temperature minus initial temperature)

What this formula says in plain English: The coefficient $\alpha$ is the fractional change in length per degree of temperature change. If $\alpha = 2.5 \times 10^{-5} \, \text{per} \, ^\circ\text{C}$ , it means that for every $1^\circ\text{C}$ rise in temperature, the material expands by $0.0025\%$ of its original length.

How to Use It: The Working Formula

From the definition, we can rearrange to get the practical formula:

\Delta L = \alpha \, L_0 \, \Delta T

So the final length $L$ after a temperature change is:

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

Tip

For small temperature changes (say, less than $100^\circ\text{C}$ ), this linear approximation is excellent. For very large changes, the coefficient itself may change slightly with temperature, but at the introductory level we treat $\alpha$ as constant.

A Concrete Example

A steel bridge girder is $50.00 \, \text{m}$ long at $20^\circ\text{C}$ . The linear expansion coefficient of steel is $\alpha = 1.2 \times 10^{-5} \, /^\circ\text{C}$ . How much longer is it on a $40^\circ\text{C}$ day?

Step 1: Identify the quantities.

$L_0 = 50.00 \, \text{m}$
$\Delta T = 40 - 20 = 20^\circ\text{C}$
$\alpha = 1.2 \times 10^{-5} \, /^\circ\text{C}$

Step 2: Apply the formula.

\Delta L = \alpha \, L_0 \, \Delta T = (1.2 \times 10^{-5})(50.00)(20)

Step 3: Calculate.

\Delta L = 1.2 \times 10^{-5} \times 1000 = 0.012 \, \text{m} = 1.2 \, \text{cm}

So the girder expands by $1.2 \, \text{cm}$ . That is why bridges have expansion joints — without them, the structure would buckle.

Two Important Cousins: Area and Volume Expansion

For a thin sheet (like a metal plate), we care about area expansion. The area expansion coefficient is approximately $2\alpha$ . For a solid object, the volume expansion coefficient is approximately $3\alpha$ . These come from the same idea: if every linear dimension grows by a factor $(1 + \alpha \Delta T)$ , then area grows by $(1 + \alpha \Delta T)^2 \approx 1 + 2\alpha \Delta T$ , and volume by $(1 + \alpha \Delta T)^3 \approx 1 + 3\alpha \Delta T$ .

Watch out

These approximations ( $2\alpha$ and $3\alpha$ ) are valid only when $\alpha \Delta T$ is small compared to 1. For most solids and modest temperature changes, this is true. For gases, the expansion is much larger and a different treatment is needed.

What the Coefficient Tells Us About Materials

| Material | $\alpha$ (per $^\circ\text{C}$ ) | Behaviour |

|----------|--------------------------------|-----------| …

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When a metal sphere is heated, its radius expands linearly. This linear expansion leads to a cubic expansion in its volume. Using the binomial approximation for small temperature changes, the increase in volume is approximately $4\pi R^3 \alpha \Delta T$ .

When a material is heated, its constituent atoms or molecules vibrate with greater amplitude. This increased vibration leads to an increase in the average distance between them, causing the material to expand. This phenomenon is known as thermal expansion.

For a solid, this expansion can be observed in its length, area, or volume. We define the coefficient of linear expansion, $\alpha$ , as the fractional change in length per unit change in temperature. That is, if an object of initial length $L$ undergoes a temperature change $\Delta T$ , its new length $L'$ is given by $L' = L(1 + \alpha \Delta T)$ .

Similarly, the coefficient of volume expansion, $\gamma$ , is the fractional change in volume per unit change in temperature. If an object of initial volume $V$ undergoes a temperature change $\Delta T$ , its new volume $V'$ is given by $V' = V(1 + \gamma \Delta T)$ . The increase in volume is then $\Delta V = V' - V = V \gamma \Delta T$ .

For isotropic materials (materials that expand equally in all directions), there is a direct relationship between the coefficient of linear expansion ( $\alpha$ ) and the coefficient of volume expansion ( $\gamma$ ). If each dimension of an object expands linearly, then its volume expands cubically.

Consider a cube of side length $L$ . Its initial volume is $V = L^3$ . If its temperature increases by $\Delta T$ , its new side length becomes $L' = L(1 + \alpha \Delta T)$ . The new volume $V'$ is then:

$V' = (L')^3 = [L(1 + \alpha \Delta T)]^3 = L^3 (1 + \alpha \Delta T)^3$ .

Since $V = L^3$ , we have $V' = V (1 + \alpha \Delta T)^3$ .

Since $\alpha \Delta T$ is typically a very small quantity (e.g., $10^{-5} \times 100 = 10^{-3}$ ), we can use the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$ . In this case, $x = \alpha \Delta T$ and $n=3$ .

So, $(1 + \alpha \Delta T)^3 \approx 1 + 3\alpha \Delta T$ .

Substituting this back into the expression for $V'$ :

$V' \approx V (1 + 3\alpha \Delta T)$ .

Comparing this with the general volume expansion formula $V' = V(1 + \gamma \Delta T)$ , we can see that for isotropic materials, the coefficient of volume expansion is approximately three times the coefficient of linear expansion:

$\gamma \approx 3\alpha$ .

Now we can apply this concept to the metal sphere.

Initial Volume of the Sphere: The initial volume of the sphere at room temperature $T$ with radius $R$ is given by the formula:

V = \frac{4}{3}\pi R^3

Volume Expansion Formula: The increase in volume $\Delta V$ due to a temperature change $\Delta T$ is given by:

\Delta V = V \gamma \Delta T

where $V$ is the initial volume and $\gamma$ is the coefficient of volume expansion.

3. Relating Volume and Linear Expansion Coefficients: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2024Set 2024-E1 markMCQ
$49 \times 10^{-5} \mathrm{~K}^{-1}$
›Reveal solutionSolution
The fractional change in density equals the negative of the fractional change in volume. Using $\gamma = 49 \times 10^{-5}\,\mathrm{K^{-1}}$ and $\Delta T = 50^\circ\mathrm{C}$ , the percentage change in density is $-\gamma \Delta T \times 100\% = -2.45\%$ , so the magnitude is 2.45%, matching option (D).

Concept & Intuition

When a substance is heated, its volume expands. Since mass stays constant, density ( $\rho = m/V$ ) must decrease. The volume expansion coefficient $\gamma$ tells us the fractional change in volume per degree: $\Delta V / V = \gamma \,\Delta T$ . Because density is inversely proportional to volume, the fractional change in density is the negative of the fractional change in volume (for small changes). So we can directly compute the percentage change in density from $\gamma$ and $\Delta T$ without needing initial values.

Step-by-step solution

Recall the relation between density and volume Density $\rho = m/V$ . For a fixed mass $m$ , a small change in volume $V$ gives

$\frac{\Delta \rho}{\rho} \approx -\frac{\Delta V}{V}.$
This is exact for infinitesimal changes; for finite but small changes (like here), it is an excellent approximation.

Express the volume change The coefficient of volume expansion $\gamma$ is defined by

$\frac{\Delta V}{V} = \gamma \, \Delta T.$
Given $\gamma = 49 \times 10^{-5}\,\mathrm{K^{-1}}$ and $\Delta T = 50^\circ\mathrm{C}$ (a change of 50 K, since the size of a Celsius degree equals a kelvin),
$\frac{\Delta V}{V} = (49 \times 10^{-5}) \times 50 = 49 \times 50 \times 10^{-5} = 2450 \times 10^{-5} = 0.0245.$

Find the fractional change in density Using step 1:
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KCET 2021Set B-21 markMCQ
$Q. Which of the following curves represent the variation of coefficient of volume expansion of an ideal gas at constant pressure? (A) (B) (C) (D)$
›Reveal solutionSolution
For an ideal gas at constant pressure, the coefficient of volume expansion $\gamma$ is inversely proportional to the absolute temperature $T$ ( $\gamma = 1/T$ ), so the correct graph is a rectangular hyperbola — option (B).

The coefficient of volume expansion, $\gamma$ , tells you how much the volume of a substance changes per unit change in temperature, relative to its original volume. For solids and liquids, $\gamma$ is often nearly constant over small temperature ranges. But for an ideal gas, the behaviour is different — and beautifully simple — because the gas obeys a precise equation of state.

At constant pressure, the ideal gas law is $PV = nRT$ . Since $P$ , $n$ , and $R$ are fixed, volume $V$ is directly proportional to absolute temperature $T$ :
$V = \left( \frac{nR}{P} \right) T$
This is a straight line through the origin when you plot $V$ vs $T$ (in kelvin). Now, the definition of the coefficient of volume expansion is:
$\gamma = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P$
That is, the fractional change in volume per degree temperature change, at constant pressure.

Find $\frac{\partial V}{\partial T}$ at constant $P$ . From $V = \frac{nR}{P} T$ , differentiate with respect to $T$ :

$\left( \frac{\partial V}{\partial T} \right)_P = \frac{nR}{P}$

Substitute into the definition of $\gamma$ .

$\gamma = \frac{1}{V} \cdot \frac{nR}{P}$

Replace $V$ using the ideal gas law. Since $V = \frac{nR}{P} T$ , we get:

$\gamma = \frac{1}{\frac{nR}{P} T} \cdot \frac{nR}{P} = \frac{1}{T}$
The $nR/P$ cancels perfectly. So for an ideal gas at constant pressure:
$\gamma = \frac{1}{T}$
This is a key result: the coefficient of volume expansion is not constant — it varies inversely with the absolute temperature.
$\gamma_{\text{ideal gas, const. }P} = \frac{1}{T}$

Interpret the graph. The relation $\gamma = 1/T$ is a rectangular hyperbola. As $T$ increases, $\gamma$ decreases; as $T$ approaches zero, $\gamma$ blows up (but remember, the gas liquefies before absolute zero, so the ideal gas law fails there). The curve is smooth and decreasing, never touching either axis. …
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