Thermal Expansion

Q: What is the anomalous expansion of water?

Between 0 and 4°C water contracts when heated instead of expanding, so its density is maximum at about 4°C. Because of this, lakes freeze from the top and the water at the bottom stays near 4°C, allowing fish to survive.

Q: How is thermal expansion asked in NEET?

NEET asks about the ratio α : β : γ, expansion of holes, the anomalous expansion of water, bimetallic strips and simple numericals on the change in length, area, volume or density. Thermal stress also appears with Young's modulus.

PhysicsThermal Properties Of MatterFor JEE aspirants

Thermal expansion is the increase in the length, area or volume of a body when its temperature rises: $Δ L = L_{0} α Δ T$ , $Δ A = A_{0} β Δ T$ , $Δ V = V_{0} γ Δ T$ , with $β = 2 α$ and $γ = 3 α$ . This page explains why matter expands, covers solids, liquids (real and apparent expansion), gases and the anomalous expansion of water, and then the applications behind most thermal expansion questions in NEET and JEE Main: bimetallic strips, pendulum clocks, measuring scales and thermal stress.

On this page1Why bodies expand2Linear, area, volume3β = 2α, γ = 3α4Density and holes5Liquids6Water anomaly7Gases8Applications9Thermal stress

Key Formulas - Quick Reference

★ Must learn $L = L_{0} (1 + α Δ T)$ , $A = A_{0} (1 + β Δ T)$ , $V = V_{0} (1 + γ Δ T)$
★ Must learnIsotropic solid: $α : β : γ = 1 : 2 : 3$ ; anisotropic: $γ = α_{x} + α_{y} + α_{z}$
Density: $ρ = \frac{ρ _{0}}{1 + γ Δ T} \approx ρ_{0} (1 - γ Δ T)$
★ Must learnLiquids: $γ_{real} = γ_{apparent} + γ_{vessel}$ ; overflow $= V_{0} (γ_{liquid} - γ_{vessel}) Δ T$
Ideal gas at constant pressure: $γ = \frac{1}{T}$ ( $3.66 \times 1 0^{- 3} K^{- 1}$ at $0^{\circ} C$ )
★ Must learnThermal stress (expansion prevented): $σ = Y α Δ T$ , force $F = Y A α Δ T$
Pendulum clock: $\frac{Δ T _{p}}{T _{p}} = \frac{1}{2} α Δ θ$ ; time lost per day $= \frac{1}{2} α Δ θ \times 86400 s$
Bimetallic strip: radius $R = \frac{d}{( α _{1} - α _{2} ) Δ T}$ ; scale reading: true value $=$ reading $\times (1 + α_{s} Δ θ)$

1. What Is Thermal Expansion and Why Does It Happen?

Most substances expand when heated and contract when cooled. A railway track, a bridge, a mercury column and a hot-air balloon all show it. The increase in dimensions of a body due to a rise in temperature is called thermal expansion.

Microscopic reason. Atoms in a solid vibrate about their mean positions in a potential energy well. The well is not symmetric: it rises steeply when atoms are pushed together (strong repulsion) and gently when they are pulled apart. At higher temperature the atoms vibrate with more energy between two turning points, and the midpoint of these points moves to a larger separation. So the average spacing, and the size of the body, grows.

Figure 1: The interatomic potential energy curve is not symmetric (drawn from a Morse potential). As temperature rises, atoms vibrate with more energy and the mid-point of their vibration (red dots) moves to larger separation: the solid expands. A symmetric (parabolic) well would give no expansion.

Key idea

Expansion comes from the asymmetry of the interatomic potential: more vibration energy means a larger average spacing.

2. Linear, Area and Volume Expansion

★ Must learn

For a small temperature change $Δ T$ , the fractional change in size is proportional to $Δ T$ :

\frac{Δ L}{L _{0}} = α Δ T, \frac{Δ A}{A _{0}} = β Δ T, \frac{Δ V}{V _{0}} = γ Δ T

$α$ = coefficient of linear expansion, $β$ = coefficient of area (superficial) expansion, $γ$ = coefficient of volume (cubical) expansion. Unit $K^{- 1}$ (same number per $^{\circ} C$ ). They are characteristic of the material and vary slightly with temperature.

Figure 2: The three kinds of thermal expansion (exaggerated). Dashed outlines are before heating.

α

β

and

γ

are the fractional changes in length, area and volume per kelvin.

Material	$α$ ( $1 0^{- 5} K^{- 1}$ )	Material	$α$ ( $1 0^{- 5} K^{- 1}$ )
Aluminium	2.5	Gold	1.4
Brass	1.8	Iron	1.2
Copper	1.7	Glass (pyrex)	0.32
Silver	1.9	Invar	0.09

Metals have $α \sim 1 0^{- 5} K^{- 1}$ : a $1 m$ steel rod lengthens by only about $0.1 mm$ per $10 K$ . Invar (an iron-nickel alloy) barely expands, so it is used for pendulums, measuring tapes and precision instruments. Pyrex glass expands little, so it survives sudden heating.

2.1 Relation between α, β and γ

A square of side $L$ heated by $Δ T$ has side $L (1 + α Δ T)$ and area $L^{2} (1 + α Δ T)^{2} = L^{2} (1 + 2 α Δ T + α^{2} Δ T^{2})$ .
$α Δ T \sim 1 0^{- 3}$ , so the squared term is negligible: $Δ A = L^{2} (2 α) Δ T$ , i.e. $β = 2 α$ .
A cube: $V = L^{3} (1 + α Δ T)^{3} \approx L^{3} (1 + 3 α Δ T)$ , so $γ = 3 α$ .

Figure 3: The extra area is two strips

L Δ L

plus a corner

(Δ L)^{2}

that is negligible because

Δ L ≪ L

. Hence

β = 2 α

and, for volume,

γ = 3 α

★ Must learn

For an isotropic solid: $β = 2 α$ , $γ = 3 α$ , i.e. $α : β : γ = 1 : 2 : 3$ . For an anisotropic crystal with different expansivities along three axes: $β_{x y} = α_{x} + α_{y}$ and $γ = α_{x} + α_{y} + α_{z}$ .

3. Consequences: Holes, Cavities and Density

A hole expands like the material around it. A hole in a plate, the bore of a ring or the cavity of a hollow sphere grows on heating exactly as if it were filled with the same material. So a ring can be heated to slip over a slightly larger rod (shrink fitting), and a tight metal lid on a glass jar loosens in hot water.
Density falls on heating because the mass is unchanged while the volume grows: $ρ = \frac{ρ _{0}}{1 + γ Δ T} \approx ρ_{0} (1 - γ Δ T)$ .
Hollow and solid spheres of the same material and outer radius expand equally in outer size.

Figure 4: Heating is a uniform photographic enlargement (exaggerated here): every length, including the diameter of a hole, grows by the same factor

(1 + α Δ T)

. This is how a tight metal lid is loosened with hot water.

4. Expansion of Liquids: Real and Apparent

A liquid has no fixed shape, so only its volume expansion is defined. Liquids expand more than solids ( $γ \sim 1 0^{- 4}$ to $1 0^{- 3} K^{- 1}$ ). A liquid is always heated in a vessel that also expands, so what we observe is not the full story.

Figure 5: A heated liquid first seems to shrink (the vessel warms and expands first), then rises. What we see is the apparent expansion:

γ_{real} = γ_{apparent} + γ_{vessel}

★ Must learn

γ_{real} = γ_{apparent} + γ_{vessel}

Real expansion is the actual increase in volume of the liquid; apparent expansion is the increase seen against the vessel's markings. If the vessel is full, the volume that overflows is $V_{0} (γ_{liquid} - γ_{vessel}) Δ T$ . If $γ_{vessel} = γ_{liquid}$ nothing overflows; if $γ_{vessel}$ is larger, the level falls.

Substance	$γ$ ( $1 0^{- 5} K^{- 1}$ )	Substance	$γ$ ( $1 0^{- 5} K^{- 1}$ )
Alcohol (ethyl)	110	Glass (ordinary)	2.5
Paraffin	58.8	Glass (pyrex)	1.0
Water	20.7	Iron	3.55
Mercury	18.2	Aluminium	7.0

Exam Trick

Constant empty volume inside a vessel. If a glass vessel of volume $V$ holds mercury of volume $V_{m}$ and the space above the mercury must stay the same at every temperature, the two expansions must be equal: $V γ_{glass} = V_{m} γ_{Hg}$ , so $V_{m} = \frac{V γ _{glass}}{γ _{Hg}}$ . The same idea fixes a constant difference in length between two rods: $L_{1} α_{1} = L_{2} α_{2}$ .

5. Anomalous Expansion of Water

Water behaves unusually between $0^{\circ} C$ and $4^{\circ} C$ : on heating, it contracts. Its density is maximum (about $1000 kg m^{- 3}$ ) at about $4^{\circ} C$ ; above $4^{\circ} C$ it expands like other liquids. On freezing, water expands by about 9%, which is why ice floats and why water pipes burst in winter.

Figure 6: Water is densest at about

4^{\circ} C

(curve from the Kell equation). Between

0

and

4^{\circ} C

it contracts on heating (

γ < 0

); above

4^{\circ} C

it expands normally. Volume of a fixed mass is least at

4^{\circ} C

Why lakes freeze from the top. As a lake cools, the cooler surface water is denser and sinks until the whole lake reaches $4^{\circ} C$ . Below $4^{\circ} C$ the surface water becomes lighter and stays on top, cools to $0^{\circ} C$ and freezes. Ice is a poor conductor, so the water underneath stays at about $4^{\circ} C$ and aquatic life survives.

Figure 7: Water at

4^{\circ} C

is densest and sinks; colder water and ice float. Ice, a poor conductor, insulates the water below, so lakes freeze from the top down and aquatic life survives.

Quick Recall: tap to check

At what temperature is the volume of a given mass of water least?

About

4^{\circ} C

What is

γ

for water between

0

and

4^{\circ} C

Negative: water contracts on heating in this range.

Why does a glass bottle full of water crack in a freezer?

Water expands by about 9% on freezing.

6. Expansion of Gases

Gases expand far more than solids and liquids, and their expansion depends strongly on pressure, so it is quoted at constant pressure. For an ideal gas $p V = μ R T$ ; at constant $p$ , $V \propto T$ (Charles's law), so

γ = \frac{1}{V} (\frac{\partial V}{\partial T})_{p} = \frac{1}{T}

At $0^{\circ} C$ this is $\frac{1}{273.15} = 3.66 \times 1 0^{- 3} K^{- 1}$ , hundreds of times larger than for solids, and it decreases as temperature rises.

Figure 8: At constant pressure

V \propto T

(Charles's law), so

γ = \frac{1}{V} \frac{d V}{d T} = \frac{1}{T}

: about

3.66 \times 1 0^{- 3} K^{- 1}

0^{\circ} C

, much larger than for solids or liquids, and it depends on temperature.

7. Applications of Thermal Expansion

7.1 Bimetallic strip

Two metals with different $α$ (say brass and iron) are riveted or welded along their length. On heating, brass expands more, so the strip bends with brass on the convex side. On cooling it bends the other way. For strips of thickness $d$ each, the radius of curvature is $R = \frac{d}{( α _{1} - α _{2} ) Δ T}$ . Used in thermostats (electric irons, ovens), fire alarms, car indicator flashers and dial thermometers.

Figure 9: A bimetallic strip bends because the two metals expand by different amounts (curvature exaggerated). The metal with the larger

α

is on the outside of the curve. Used in thermostats, fire alarms and flashers.

7.2 Pendulum clocks

Period $T_{p} = 2 π \frac{L}{g}$ , so $\frac{Δ T _{p}}{T _{p}} = \frac{1}{2} \frac{Δ L}{L} = \frac{1}{2} α Δ θ$ .
A longer period means fewer oscillations per day: in summer the clock loses time; in winter it gains.
Time lost or gained in a time $t$ : $Δ t = \frac{1}{2} α Δ θ \times t$ ; per day, $t = 86400 s$ .

Figure 10: A metal pendulum lengthens on heating, so its period grows by the fraction

\frac{1}{2} α Δ θ

and the clock loses time (drawn exaggerated). Invar pendulums (

α

tiny) keep good time.

7.3 Measuring scales

A metal scale is correct at the temperature at which it was calibrated. At a higher temperature each division is longer, so the scale reads less than the true length: true length $=$ reading $\times (1 + α_{s} Δ θ)$ , where $α_{s}$ is for the scale and $Δ θ$ is measured from the calibration temperature. If the object also expands, use the difference of expansivities.

7.4 Everyday examples

Gaps are left between rails and in bridges (expansion joints); telephone and power lines sag more in summer.
Thick glass tumblers crack when hot water is poured in: the inside expands before the outside. Pyrex, with small $α$ , does not.
Platinum is sealed into glass because their $α$ values are nearly equal.
Rivets are put in red-hot: on cooling they contract and grip the plates tightly. Iron tyres are heated before fitting on wooden wheels.

8. Thermal Stress

If a rod is fixed between rigid walls and heated, it cannot expand. The walls compress it by exactly the length it wanted to gain, $Lα Δ T$ , so its strain is $α Δ T$ and

thermal stress = Y α Δ T, F = Y A α Δ T

The stress does not depend on the length of the rod. On cooling a clamped rod, the stress is tensile. Thermal stress is why rails buckle in heat waves if expansion gaps are too small.

Figure 11: If expansion is prevented, the body is strained by

α Δ T

and a thermal stress

Y α Δ T

appears, whatever its length. Gaps in rails and bridges, and loops in pipelines, avoid this.

JEE Advanced

Temperature-dependent $α$ : $\frac{d L}{L} = α (T) d T$ gives $L = L_{0} exp (\int_{T_{0}}^{T} α d T) \approx L_{0} (1 + \int_{T_{0}}^{T} α d T)$ . Two rods in series between walls (lengths $L_{1}$ , $L_{2}$ , same area): total free expansion $L_{1} α_{1} Δ T + L_{2} α_{2} Δ T$ must be cancelled, and the same force acts in both: $F (\frac{L _{1}}{A Y _{1}} + \frac{L _{2}}{A Y _{2}}) = (L_{1} α_{1} + L_{2} α_{2}) Δ T$ . Moment of inertia of a heated body: $I \propto L^{2}$ , so $\frac{Δ I}{I} = 2 α Δ T$ , and a freely spinning disc slows down ( $I ω$ constant): $\frac{Δ ω}{ω} = - 2 α Δ T$ .

Figure 12: Four problem types cover almost every thermal expansion question. Identify which one before choosing a formula.

Figure 13: Mind map of thermal expansion for quick revision.

9. Solved Examples

Solved Example 1

An aluminium rod is

1.0 m

long at

2 0^{\circ} C

. Find its length at

7 0^{\circ} C

. (

α = 2.5 \times 1 0^{- 5} K^{- 1}

)

Solution:

$Δ L = L_{0} α Δ T = (1.0) (2.5 \times 1 0^{- 5}) (50) = 1.25 \times 1 0^{- 3} m$ .

Answer: $L = 1.00125 m$ (it grows by $1.25 mm$ ).

Solved Example 2

A blacksmith fixes an iron ring on the rim of a wooden wheel. At

2 7^{\circ} C

the diameters of the rim and the ring are

5.243 m

and

5.231 m

. To what temperature must the ring be heated to fit the rim? (

α_{iron} = 1.20 \times 1 0^{- 5} K^{- 1}

)

Solution:

The ring's diameter must grow by $0.012 m$ : $0.012 = 5.231 \times 1.20 \times 1 0^{- 5} \times Δ T$ , so $Δ T = 191 K$ .

Answer: $T \approx 27 + 191 = 21 8^{\circ} C$ .

Solved Example 3

A hole is drilled in a copper sheet. Its diameter is

4.24 cm

2 7^{\circ} C

. What is its diameter at

22 7^{\circ} C

? (

α_{Cu} = 1.70 \times 1 0^{- 5} K^{- 1}

)

Solution:

The hole expands like copper: $Δ d = d α Δ T = 4.24 \times 1.70 \times 1 0^{- 5} \times 200 = 1.44 \times 1 0^{- 2} cm$ .

Answer: $d \approx 4.2544 cm$ (it increases).

Solved Example 4

A brass wire

1.8 m

long at

2 7^{\circ} C

is held taut with little tension between two rigid supports. If it is cooled to

- 3 9^{\circ} C

, what is the tension? (Diameter

2.0 mm

α = 2.0 \times 1 0^{- 5} K^{- 1}

Y = 0.91 \times 1 0^{11} Pa

)

Solution:

The wire wants to shorten by $Lα Δ T$ but cannot: strain $= α Δ T$ , so $F = Y A α Δ T$ .

$F = (0.91 \times 1 0^{11}) π (1.0 \times 1 0^{- 3})^{2} (2.0 \times 1 0^{- 5}) (66)$ .

Answer: $F \approx 3.8 \times 1 0^{2} N$ (tension; independent of length).

Solved Example 5

A steel rod is clamped between two rigid walls at room temperature and heated by

5 0^{\circ} C

. Find the thermal stress. (

Y = 2 \times 1 0^{11} Pa

α = 1.2 \times 1 0^{- 5} K^{- 1}

)

Solution:

Stress $= Y α Δ T = (2 \times 1 0^{11}) (1.2 \times 1 0^{- 5}) (50)$ .

Answer: $1.2 \times 1 0^{8} Pa$ , compressive.

Solved Example 6

A pendulum clock with a steel pendulum (

α = 1.2 \times 1 0^{- 5} K^{- 1}

) keeps correct time at

2 0^{\circ} C

. How many seconds does it lose or gain per day at

4 0^{\circ} C

Solution:

$\frac{Δ T _{p}}{T _{p}} = \frac{1}{2} α Δ θ = \frac{1}{2} (1.2 \times 1 0^{- 5}) (20) = 1.2 \times 1 0^{- 4}$ .

Per day: $1.2 \times 1 0^{- 4} \times 86400$ .

Answer: it loses about $10.4 s$ per day (period longer, clock slow).

Solved Example 7

A glass flask of volume

1000 cm^{3}

is completely filled with mercury at

0^{\circ} C

. How much mercury overflows when both are heated to

10 0^{\circ} C

? (

γ_{Hg} = 18.2 \times 1 0^{- 5} K^{- 1}

γ_{glass} = 2.5 \times 1 0^{- 5} K^{- 1}

)

Solution:

Overflow $= V_{0} (γ_{Hg} - γ_{glass}) Δ T = 1000 (15.7 \times 1 0^{- 5}) (100)$ .

Answer: $15.7 cm^{3}$ .

Solved Example 8

The density of mercury is

13.6 g cm^{- 3}

0^{\circ} C

. Find its density at

10 0^{\circ} C

. (

γ = 18.2 \times 1 0^{- 5} K^{- 1}

)

Solution:

$ρ = \frac{ρ _{0}}{1 + γ Δ T} = \frac{13.6}{1 + 0.0182}$ .

Answer: $ρ \approx 13.36 g cm^{- 3}$ .

Solved Example 9

A steel scale is correct at

2 0^{\circ} C

. A length measured with it at

4 0^{\circ} C

reads

50.00 cm

. What is the true length? (

α_{steel} = 1.2 \times 1 0^{- 5} K^{- 1}

)

Solution:

At $4 0^{\circ} C$ each division is longer by the factor $(1 + α Δ θ)$ , so the scale reads low: true $= 50.00 (1 + 1.2 \times 1 0^{- 5} \times 20)$ .

Answer: $50.012 cm$ .

Solved Example 10

A bimetallic strip is made of brass (

α = 1.8 \times 1 0^{- 5} K^{- 1}

) and iron (

α = 1.2 \times 1 0^{- 5} K^{- 1}

) strips, each

0.5 mm

thick. Find its radius of curvature when heated by

100 K

Solution:

$R = \frac{d}{( α _{1} - α _{2} ) Δ T} = \frac{0.5 \times 1 0 ^{- 3}}{( 0.6 \times 1 0 ^{- 5} ) ( 100 )}$ .

Answer: $R \approx 0.83 m$ , with brass on the outer side.

Solved Example 11

The coefficient of linear expansion of a solid is

1.0 \times 1 0^{- 5} K^{- 1}

. Its coefficient of volume expansion is
(A)

1.0 \times 1 0^{- 5}

(B)

2.0 \times 1 0^{- 5}

(C)

3.0 \times 1 0^{- 5}

(D)

3.3 \times 1 0^{- 6} K^{- 1}

Solution:

Answer: (C). For an isotropic solid $γ = 3 α$ .

Solved Example 12

Water at

0^{\circ} C

is heated to

1 0^{\circ} C

. Its volume
(A) increases steadily
(B) decreases steadily
(C) first decreases, then increases
(D) first increases, then decreases

Solution:

Answer: (C). Water contracts from $0$ to $4^{\circ} C$ (volume least at $4^{\circ} C$ ) and then expands.

Solved Example 13

Find the coefficient of volume expansion of an ideal gas at constant pressure at

2 7^{\circ} C

Solution:

$γ = \frac{1}{T} = \frac{1}{300 K}$ .

Answer: $γ \approx 3.3 \times 1 0^{- 3} K^{- 1}$ .

Practice Questions

A $10 m$ steel rail ( $α = 1.2 \times 1 0^{- 5} K^{- 1}$ ) is laid at $1 5^{\circ} C$ . What gap is needed for $4 5^{\circ} C$ ?Answer: $3.6 mm$
A metal plate's area grows by $0.2%$ when heated by $50 K$ . Find $α$ .Answer: $2 \times 1 0^{- 5} K^{- 1}$
By what percentage does the density of a solid ( $α = 2 \times 1 0^{- 5} K^{- 1}$ ) fall when heated by $100 K$ ?Answer: about $0.6%$
An iron rod and a copper rod must differ in length by $10 cm$ at all temperatures. Find their lengths. ( $α_{Fe} = 1.2$ , $α_{Cu} = 1.7$ , both $\times 1 0^{- 5} K^{- 1}$ )Answer: iron $34 cm$ , copper $24 cm$
A clock with a brass pendulum ( $α = 1.9 \times 1 0^{- 5} K^{- 1}$ ) is correct at $2 5^{\circ} C$ . Does it gain or lose at $5^{\circ} C$ , and by how much per day?Answer: gains about $16.4 s$
Why does a thick glass tumbler crack when boiling water is poured into it?Answer: The inner surface expands before the outer; the uneven expansion sets up stresses
A liquid has $γ = 5 \times 1 0^{- 4} K^{- 1}$ and is in a vessel with $γ = 5 \times 1 0^{- 5} K^{- 1}$ . Find its apparent expansivity.Answer: $4.5 \times 1 0^{- 4} K^{- 1}$

Common Mistakes to Avoid

Watch out

Thinking a hole in a plate shrinks on heating. It expands, like the material around it.
Using $γ = 3 α$ for a liquid. Liquids have only $γ$ ; $α$ is not defined for them.
Forgetting the vessel: the observed (apparent) expansion is $γ_{real} - γ_{vessel}$ .
Saying a pendulum clock gains time in summer. The pendulum lengthens, so the clock loses time.
Writing thermal stress as $Y α L Δ T$ : the stress $Y α Δ T$ is independent of length.
Assuming water always expands on heating; between $0$ and $4^{\circ} C$ it contracts.
Putting the metal with the larger $α$ on the concave side of a heated bimetallic strip; it is on the convex side.
Taking $γ$ of a gas as a constant; at constant pressure it is $\frac{1}{T}$ and falls as $T$ rises.

Frequently Asked Questions

What is thermal expansion?

Thermal expansion is the increase in length, area or volume of a body when its temperature rises. For a small temperature change the fractional change is proportional to the temperature change, with coefficients $α$ (length), $β$ (area) and $γ$ (volume).

What is the relation between alpha, beta and gamma?

For an isotropic solid the area coefficient is twice and the volume coefficient three times the linear coefficient, so $α : β : γ = 1 : 2 : 3$ . It follows from squaring or cubing $(1 + α Δ T)$ and dropping small terms.

Why do solids expand on heating?

The potential energy curve between two atoms is steeper on the compression side than on the stretching side. When atoms vibrate with more energy at higher temperature, the midpoint of their vibration shifts to a larger separation, so the average spacing increases.

What is the anomalous expansion of water?

Between $0$ and $4^{\circ} C$ water contracts when heated instead of expanding, so its density is maximum at about $4^{\circ} C$ . Because of this, lakes freeze from the top and the water at the bottom stays near $4^{\circ} C$ , allowing fish to survive.

What is the difference between real and apparent expansion of a liquid?

Real expansion is the actual increase in volume of the liquid. Apparent expansion is what is observed against the markings of the vessel, which also expands. $γ_{real} = γ_{app} + γ_{vessel}$ .

Why does a pendulum clock lose time in summer?

Heat lengthens the metal pendulum, and the time period is proportional to the square root of the length. A longer period means fewer ticks per day, so the clock runs slow. The fractional change in period is $\frac{1}{2} α Δ θ$ for a temperature rise $Δ θ$ .

How is thermal expansion asked in NEET?

NEET asks about the ratio $α : β : γ$ , expansion of holes, the anomalous expansion of water, bimetallic strips and simple numericals on the change in length, area, volume or density. Thermal stress also appears with Young's modulus.

Which thermal expansion problems are common in JEE Main?

JEE Main uses pendulum clocks losing or gaining time, apparent expansion and overflow of liquids, thermal stress and force in clamped rods, rods with a constant length difference, scale corrections and the change in moment of inertia or angular speed on heating.

Previous year questions on Thermal Expansion

5 questions from past papers, each with a step-by-step solution.

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