Consider the following statements: 1. The coefficient of linear expansion has dimension K⁻¹. 2. The coefficient of volume expansion has dimension K⁻¹. Choose the correct option:
- (a)Both 1 and 2 are correct
- (b)1 is correct but 2 is wrong
- (c)2 is correct but 1 is wrong
- (d)Both 1 and 2 are wrong
Correct — A, Both 1 and 2 are correct.
Both quantities are defined as a fractional change per degree of temperature rise. Linear expansion uses α = (ΔL/L) ÷ ΔT and volume expansion uses γ = (ΔV/V) ÷ ΔT.
The length cancels inside ΔL/L and the volume cancels inside ΔV/V, so each numerator is a pure number. A pure number divided by a temperature leaves K⁻¹, which is why both 1 and 2 are correct.
Carry this away: the dimension follows the definition, not the name. The word 'volume' in γ says what is changing, not what unit the coefficient itself carries.
- (b)1 is correct but 2 is wrong — This accepts α as K⁻¹ but denies the same to γ, and splitting them that way would require γ to carry a volume unit of its own.
It would be the right answer to a differently worded statement 2 that gave the coefficient of volume expansion a unit such as m³ K⁻¹. As statement 2 is printed, γ is a fraction of the existing volume per degree, so the m³ divides out.
- (c)2 is correct but 1 is wrong — This keeps γ at K⁻¹ but rejects α, which reverses the reasoning of option b without fixing it. Both definitions are built the same way, so a defect in one would have to appear in the other.
It would be the right answer to a version of statement 1 that described the increase in length of one particular rod per degree, ΔL ÷ ΔT. That quantity does carry a metre, so it is not the coefficient.
- (d)Both 1 and 2 are wrong — This treats the two coefficients as if the geometry they describe survived into their units, so α would be m K⁻¹ and γ would be m³ K⁻¹.
It would be the right answer to a pair of statements that attached a length unit to α and a volume unit to γ. The printed statements assign K⁻¹ to each, which is what the standard definitions give.
Thermal expansion is described by coefficients that measure change relative to the original size, not in absolute units. The linear coefficient α is the fractional increase in length per degree; the volume (cubical) coefficient γ is the fractional increase in volume per degree.
Because ΔL/L and ΔV/V are ratios of like quantities, both are dimensionless. Dividing either by a temperature interval therefore leaves the single dimension K⁻¹, written in full as [M⁰L⁰T⁰K⁻¹].
Defining them as fractions is what makes them material constants: the same figure works for a 1 cm wire and a 100 m span of the same metal.
This sits where heat meets dimensional analysis, so it can be settled without recalling a number for any substance. Reconstructing the defining equation and cancelling the like quantities is enough.
The same reasoning covers the area coefficient β, and it explains why α, β and γ can differ in magnitude while sharing one dimension.
It also underpins practical work — thermal stress, expansion joints, bimetallic strips, and the correction of a liquid's apparent expansion for the expansion of its container.
- The coefficient of linear expansion is α = (1/L)(dL/dT), a fractional change of length per degree, with SI unit K⁻¹.
- The coefficient of volume expansion is γ = (1/V)(dV/dT); the volume cancels inside the ratio, leaving the dimension K⁻¹.
- Each coefficient's dimensional formula is [M⁰L⁰T⁰Θ⁻¹] — no mass, length or time — often written [M⁰L⁰T⁰K⁻¹] since the SI temperature unit is the kelvin.
- For an isotropic solid over a small temperature range, the area coefficient β ≈ 2α and the volume coefficient γ ≈ 3α.
- A kelvin and a degree Celsius are equal in size, so a coefficient quoted as °C⁻¹ has the same numerical value in K⁻¹.
- For an ideal gas held at constant pressure, the volume expansion coefficient works out to 1/T, itself a quantity of dimension K⁻¹.
- Because these coefficients are fractional changes, their values do not depend on how large the sample is.
The first three rows divide by the original size before dividing by temperature, so each lands on K⁻¹. The last row skips that cancelling step and keeps a length unit.
- Reading a unit out of the name: 'volume expansion' describes what changes, while the coefficient is a fraction of the existing volume per degree, which leaves K⁻¹.
- Turning the factor in γ ≈ 3α into a difference of dimension; that factor compares magnitudes for an isotropic solid, and both quantities stay at K⁻¹.
- Answering from the length gained by one rod per degree, ΔL ÷ ΔT, which does carry a metre because it has not been divided by the original length.
- Assuming a figure quoted in °C⁻¹ must be rescaled before it can be called K⁻¹; the two degree sizes are identical, and only the zero point differs.
- Marking a two-statement item on the strength of the statement you checked first, instead of testing each statement on its own definition.
The shape backed by the item cited below is the numerical, as in the 2021 Geoscientist question: a coefficient of linear expansion is handed to you in per-degree form, and the change in length of a heated rod has to be worked out from it.
The UKPSC item in front of you shows a second shape — a pair of definitional statements, each to be judged against its own defining equation rather than against the name of the quantity.
GEO_GS_2021_Q622021Same quantity, opposite demand. The 2021 item supplies a coefficient of linear thermal expansion for copper, 17 × 10⁻⁶ per °C, and asks for the change in length of a 1 m rod heated by 100 °C — the definition used forwards, as arithmetic. The UKPSC item supplies no data and asks instead what dimension that coefficient carries. What links them is the one relation ΔL = L α ΔT: the numerical item works out because α is a per-degree fraction, and the UKPSC item asks you to say so.
- practice — not a real PYQ
What is the dimensional formula of the coefficient of volume expansion?
- (a)[M⁰L³T⁰K⁻¹]
- (b)[M⁰L⁰T⁰K⁻¹]
- (c)[M⁰L³T⁰K⁰]
- (d)[M⁰L⁰T⁰K¹]
Answerb — γ is defined as (ΔV/V) ÷ ΔT, and ΔV/V is a ratio of two volumes, hence dimensionless. Dividing it by a temperature therefore leaves the single dimension K⁻¹.Option a keeps a volume that has already cancelled. Option c is the dimension of volume itself, with no temperature in it. Option d is the reciprocal of the right answer, matching temperature rather than per-temperature.
- practice — not a real PYQ
For an isotropic solid heated through a small temperature rise, which relation connects the coefficients of linear (α), superficial (β) and cubical (γ) expansion?
- (a)β ≈ 2α and γ ≈ 3α
- (b)β ≈ 3α and γ ≈ 2α
- (c)β ≈ α/2 and γ ≈ α/3
- (d)β ≈ α² and γ ≈ α³
Answera — expanding (1 + αΔT)² and (1 + αΔT)³ and dropping the small higher-order terms gives an area gain of about 2αΔT and a volume gain of about 3αΔT.Option b swaps the two factors, giving area the growth that belongs to volume. Option c shrinks the coefficients instead of scaling them up, which is the wrong direction. Option d fails on dimension alone, since α² would be K⁻² and α³ would be K⁻³.
- practice — not a real PYQ
Which one of the following has a dimensional formula different from that of the other three?
- (a)Coefficient of linear expansion
- (b)Coefficient of superficial expansion
- (c)Coefficient of cubical expansion
- (d)Increase in the length of a given rod per degree rise in temperature
Answerd — this is ΔL ÷ ΔT for one particular rod. It has not been divided by the original length, so it retains a length and comes out in metre per kelvin, [M⁰L¹T⁰K⁻¹].Options a, b and c are each a fractional change per degree, so the original size cancels and each carries the dimension K⁻¹. That shared value is what leaves d as the odd one out.
- practice — not a real PYQ
A rod of length L, made of a material whose coefficient of linear expansion is α, is heated through a temperature rise ΔT. Which of the following combinations is dimensionless?
- (a)α ΔT
- (b)α L
- (c)α ÷ ΔT
- (d)L ΔT
Answera — α has the dimension K⁻¹ and ΔT has the dimension K, so their product cancels to a pure number. It is exactly the fractional change in length, ΔL/L.Option b is K⁻¹ multiplied by a length, giving metre per kelvin. Option c divides K⁻¹ by a temperature and gives K⁻². Option d multiplies a length by a temperature and gives metre kelvin.