Thermal expansion is quietly one of the most conceptually clean topics in NEET physics — and the trap is that students treat it as too simple and then misapply the coefficients.
The core idea. When a solid is heated, its atoms vibrate with larger amplitude about their mean positions. The mean interatomic separation increases because the potential-energy curve is asymmetric (steeper on the repulsive side). This produces measurable changes in length, area, and volume (NCERT Class 11 Physics, Chapter 10, page 204).
Three coefficients, one relationship. For an isotropic solid:
The factors 2 and 3 come from the binomial approximation (higher-order α² terms are negligible for modest ΔT). This is where NEET questions test you: they give α and ask for β, or vice versa, and a common confusion is using the wrong multiplier.
Temperature vs heat. Temperature measures the average kinetic energy of molecules — it is a state variable. Heat is the energy transferred due to a temperature difference — it is a process quantity. The distinction matters: two bodies can be at the same temperature with vastly different heat contents. NCERT defines temperature operationally via the zeroth law of thermodynamics (Chapter 11, page 203): two systems each in thermal equilibrium with a third are in thermal equilibrium with each other.
Watch-out for NEET. Questions on thermal expansion typically test whether you can (a) correctly relate α, 2α, and 3α for the three types, (b) handle the expansion of a hole or cavity (it expands as if filled — the hole gets bigger, not smaller), and (c) convert between Celsius and Kelvin temperature changes (ΔT is numerically identical in both, but students sometimes second-guess themselves). These are direct-application problems; the arithmetic is light, but the conceptual step must be precise.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Temperature is a physical quantity that measures:
Answer: A. A is correct. Temperature is a measure of the average translational kinetic energy of the molecules of a substance (NCERT Class 11 Physics, Chapter 10, page 203).
Why B is wrong: B is wrong because total internal energy includes both kinetic and potential energy contributions and depends on the amount of substance, not just temperature.
Why C is wrong: C is wrong because heat is energy in transit due to a temperature difference — it is not a property stored in a body. A body has internal energy, not 'heat content' in the thermodynamic sense.
Why D is wrong: D is wrong because the rate of heat transfer depends on temperature difference, thermal conductivity, and geometry — it is not what temperature itself measures.
The zeroth law of thermodynamics provides the basis for the concept of:
Answer: B. B is correct. The zeroth law states that if two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other — this establishes temperature as a well-defined quantity (NCERT Class 11 Physics, Chapter 10, page 203).
Why A is wrong: A is wrong because internal energy is established through the first law of thermodynamics, not the zeroth law.
Why C is wrong: C is wrong because entropy is introduced via the second law of thermodynamics.
Why D is wrong: D is wrong because heat capacity is a material property defined through the relation Q = mcΔT, not through the zeroth law.
For an isotropic solid, if α is the coefficient of linear expansion, the coefficient of volume expansion β is:
Answer: C. C is correct. For isotropic solids, β = 3α. This follows from the binomial expansion of (1 + αΔT)³ ≈ 1 + 3αΔT when higher-order terms are negligible (NCERT Class 11 Physics, Chapter 10, page 204).
Why A is wrong: A is wrong because α is the linear coefficient — volume expansion involves three dimensions, giving a factor of 3.
Why B is wrong: B is wrong because the factor 2α corresponds to areal (superficial) expansion, not volume expansion.
Why D is wrong: D is wrong because α² is dimensionally incorrect for an expansion coefficient and does not arise from the expansion analysis.
A metal rod of length 1.00 m at 20 °C is heated to 120 °C. If the coefficient of linear expansion is α = 1.2 × 10⁻⁵ K⁻¹, the increase in length is:
Answer: D. D is correct. ΔL = L α ΔT = 1.00 × 1.2 × 10⁻⁵ × 100 = 1.2 × 10⁻³ m (NCERT Class 11 Physics, Chapter 10, page 204).
Why A is wrong: A is wrong — this results from using ΔT = 10 instead of 100, likely from a subtraction error.
Why B is wrong: B is wrong — this results from using β = 3α (volume coefficient) instead of α for a linear expansion problem: 1.00 × 3.6 × 10⁻⁵ × 100 = 3.6 × 10⁻³ m.
Why C is wrong: C is wrong — this results from using the final temperature, 120 °C, in place of the change ΔT = 100 K: 1.2 × 10⁻⁵ × 120 = 1.44 × 10⁻³ m.
A thin metal plate has a circular hole of diameter 2.000 cm at 25 °C. When heated to 125 °C, the diameter of the hole (α = 2.0 × 10⁻⁵ K⁻¹):
Answer: B. B is correct. A hole in a plate expands as if it were filled with the same material. ΔD = D α ΔT = 2.000 × 2.0 × 10⁻⁵ × 100 = 4.0 × 10⁻³ cm. New diameter = 2.000 + 0.004 = 2.004 cm.
Why A is wrong: A is wrong — this reflects the common misconception that a hole shrinks when the surrounding material expands. The hole expands exactly as a solid disc of the same material would.
Why C is wrong: C is wrong — the hole does not stay the same size; thermal expansion applies to all linear dimensions, including the diameter of a cavity.
Why D is wrong: D is wrong — this results from using 3α (volume coefficient) instead of α for this linear-dimension problem.
A metal sphere has a coefficient of linear expansion α = 1.5 × 10⁻⁵ K⁻¹. If its temperature increases by 200 K, the percentage increase in its volume is approximately:
Answer: D. D is correct. β = 3α = 4.5 × 10⁻⁵ K⁻¹. ΔV/V = β ΔT = 4.5 × 10⁻⁵ × 200 = 9.0 × 10⁻³ = 0.90% (NCERT Class 11 Physics, Chapter 10, page 204).
Why A is wrong: A is wrong — this uses α directly instead of β = 3α, giving only the linear fractional change applied as if it were volume change.
Why B is wrong: B is wrong — this uses 2α (the areal coefficient) instead of 3α for a volume problem.
Why C is wrong: C is wrong — this likely results from using β = 3α but with ΔT = 100 K instead of the given 200 K.
A bimetallic strip is made of brass (higher α) bonded to iron (lower α). When heated uniformly, the strip bends such that:
Answer: C. C is correct. Brass has a higher coefficient of linear expansion, so it elongates more than iron for the same temperature rise. The longer side forms the outer (convex) curve. This is the working principle of bimetallic thermostats (NCERT Class 11 Physics, Chapter 10, page 204).
Why A is wrong: A is wrong — both metals expand, but by different amounts, which is precisely what causes bending. Equal expansion would keep it straight only if both had the same α.
Why B is wrong: B is wrong — iron expands less, so it becomes the shorter (inner, concave) side of the curved strip.
Why D is wrong: D is wrong — the bending direction is deterministic: the metal with higher α always ends up on the convex side.
A steel rail is 10.0 m long at 15 °C. The rail is laid without expansion gaps on a day when the temperature is 15 °C. If the temperature rises to 45 °C (α_steel = 1.2 × 10⁻⁵ K⁻¹, Y_steel = 2.0 × 10¹¹ Pa, cross-sectional area = 4.0 × 10⁻³ m²), the compressive force developed in the rail if it is prevented from expanding is:
Answer: A. A is correct. Thermal strain = αΔT = 1.2 × 10⁻⁵ × 30 = 3.6 × 10⁻⁴. Thermal stress = Y × αΔT = 2.0 × 10¹¹ × 3.6 × 10⁻⁴ = 7.2 × 10⁷ Pa. Force = stress × area = 7.2 × 10⁷ × 4.0 × 10⁻³ = 2.88 × 10⁵ N.
Why B is wrong: B is wrong — this results from dropping the 4.0 from the area: 7.2 × 10⁷ Pa × 1.0 × 10⁻³ m² = 7.2 × 10⁴ N.
Why C is wrong: C is wrong — this is ten times too small: the right stress, 7.2 × 10⁷ Pa, multiplied by 4.0 × 10⁻⁴ m², a slip of one power of ten in the area.
Why D is wrong: D is wrong — this comes from using the final temperature, 45 °C, in place of the rise ΔT = 30 K: 2.0 × 10¹¹ × 1.2 × 10⁻⁵ × 45 × 4.0 × 10⁻³ = 4.32 × 10⁵ N.
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Given
A brass rod has length L = 1.000 m at temperature T₁ = 20 °C. It is heated to T₂ = 220 °C. The coefficient of linear expansion of brass is α = 1.80 × 10⁻⁵ K⁻¹.
Required
Find: (a) the increase in length ΔL, and (b) the percentage increase in volume.
Concept
Thermal expansion: for a linear dimension, ΔL = LαΔT. For volume, ΔV/V = 3αΔT (isotropic solid, binomial approximation).
Formula
ΔL = LαΔT
ΔV/V = 3αΔT
Substitution
ΔT = 220 − 20 = 200 K (note: ΔT is numerically the same in °C and K).
(a) ΔL = 1.000 × 1.80 × 10⁻⁵ × 200
(b) ΔV/V = 3 × 1.80 × 10⁻⁵ × 200
Calculation
(a) ΔL = 1.000 × 1.80 × 10⁻⁵ × 200 = 3.60 × 10⁻³ m = 3.60 mm
(b) ΔV/V = 5.40 × 10⁻⁵ × 200 = 1.08 × 10⁻² = 1.08%
Note on exact values: The temperature values 20, 220, and the factor 3 are exact (counting/defined). They do not limit significant figures. The result is governed by the 3 significant figures in α.
Final answer
(a) ΔL = 3.60 × 10⁻³ m (3 significant figures, matching α).
(b) Percentage volume increase = 1.08%.
Common trap
Using α directly for volume expansion (getting 0.36% instead of 1.08%). Always check which type of expansion the question asks — linear, areal, or volume — and apply the correct multiplier (1, 2, or 3).
Similar NEET-style question
An iron sphere of radius 10.00 cm at 0 °C is heated to 100 °C. If α_iron = 1.2 × 10⁻⁵ K⁻¹, find the percentage increase in the surface area of the sphere. (Answer: ΔA/A = 2αΔT = 2 × 1.2 × 10⁻⁵ × 100 = 2.4 × 10⁻³ = 0.24%.)
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Heat (Q) is energy in transfer due to a temperature difference. Temperature is a measure of the average kinetic energy of molecules. Heat flows spontaneously from higher T to lower T.
-- NCERT Class 11 Physics, Ch. 10, p. 203Linear: ΔL/L = α ΔT. Area: ΔA/A = 2α ΔT. Volume: ΔV/V = β ΔT, where β = 3α (isotropic). α is the coefficient of linear expansion (units: 1/K).
-- NCERT Class 11 Physics, Ch. 10, p. 205Fractional change in length, area, volume per degree temperature change.
| Symbol | Quantity | SI Unit |
|---|---|---|
| alpha | linear coefficient | 1/K |
| beta | volume coefficient | 1/K |
| Delta_T | temperature change | K |
More in Properties of Bulk Matter: 5 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
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