Dimensions of Physical Quantities

Strip every unit of its accidentals — metres, feet, light years — and what remains is its dimension: the recipe of mass, length and time inside it. Dimensions are the grammar of physics equations: they tell you instantly whether an equation can be true, which quantities can be equated, and how a value transforms when you change unit systems. NTA tests dimensions in JEE Main and NEET more often than any other single skill from this chapter — as dimension-match MCQs, homogeneity Assertion–Reason items, and as the engine inside dimensional analysis questions. The complete catalogue of formulae is tabulated on the Dimensional Formulae and Equations page; here you learn to derive every entry yourself.

What are Dimensions of Physical Quantities? — Complete Theory

1. Definition: dimensions and the bracket notation

The dimension of a physical quantity states how many times, and with what powers, the base quantities enter its unit. We write the base quantities in square brackets — [M] for mass, [L] for length, [T] for time, [A] for current, [K] for temperature, [mol] for amount of substance — and assemble the dimensional formula. For example velocity = displacement/time gives [v] = [L T⁻¹]; force = mass × acceleration gives [F] = [M L T⁻²]. The bracket notation [Q] literally means "the dimensions of Q". Seven base quantities exist, but for the mechanics chapters three (M, L, T) generate nearly everything; electricity adds [A], heat adds [K].

2. Deriving dimensional formulae — the universal 3-step procedure

Step 1 — write a defining equation for the quantity in terms of quantities whose dimensions you already know.
Step 2 — replace every symbol by its dimensions.
Step 3 — simplify the powers of M, L, T by the algebra of exponents ([xᵃ][xᵇ] = [x]ᵃ⁺ᵇ, division subtracts powers).

Worked derivations (learn the procedure, not the answers):

Derivation A — Gravitational constant G. From Newton's law F = Gm₁m₂/r², isolate G = Fr²/m₁m₂. Substitute: [G] = [M L T⁻²][L²]/[M²]. Simplify: L: L¹⁺² = L³; M: M·M⁻² = M⁻¹; T: T⁻². So [G] = [M⁻¹ L³ T⁻²], unit N m² kg⁻². ✔ (check: 1 + 2 − 2 = 1 exponent on L? L¹ + L² = L³; M¹ − M² = M⁻¹; T⁻² — correct.)

Derivation B — Planck's constant h. From E = hν, h = E/ν = [M L² T⁻²]/[T⁻¹] = [M L² T⁻¹], unit J s. Note the T exponent: T⁻² − (−1) = T⁻¹. ✔

Derivation C — Coefficient of viscosity η. From the viscous-force law F = ηA(dv/dx), isolate η = F·x/(A·v). Substitute: [η] = [M L T⁻²][L]/([L²][L T⁻¹]). Numerator: M L² T⁻². Denominator: L³ T⁻¹. Ratio: [η] = [M L⁻¹ T⁻¹], unit Pa s. ✔

Derivation D — Magnetic field B. From F = qvB sinθ, B = F/(qv). Substitute: [qv] = [A T][L T⁻¹] = [A L] (the times cancel exactly). So [B] = [M L T⁻²]/[A L] = [M T⁻² A⁻¹], unit tesla. Notice B contains no L — a favourite trap. ✔

3. Dimensionless quantities

When every exponent is zero, the quantity is dimensionless: [M⁰ L⁰ T⁰]. Dimensionless quantities are either ratios of like quantities (strain ΔL/L, relative density ρ/ρ_water, refractive index c/v, Poisson's ratio, coefficient of friction, Mach number, efficiency) or pure counts/angles (plane angle in radian, solid angle in steradian, number of particles). "Dimensionless" does not mean "unitless" — radian is a unit without a dimension. Dimensionless quantities are exactly where dimensional analysis goes blind: 2, π, sin θ arguments and efficiency all vanish from its sight.

4. The principle of homogeneity of dimensions

Statement: a physically correct equation must be dimensionally homogeneous — every term connected by +, −, or = carries the same dimensions. Only like can be added to like: you cannot add a velocity to an acceleration any more than rupees to metres. The principle has two exam uses: checking an equation (the application developed on the dimensional analysis page) and finding unknown dimensions — e.g. if x = At² has [x] = [L], then [A] = [L T⁻²], i.e. A must be an acceleration. That "dimensional inference" trick — extracting [A] from a formula — is a standing JEE Main numerical pattern.

Visualising Dimensions of Physical Quantities

Use the pipeline map as a mental subway: M, L, T are the three terminus stations; every physical quantity is a route with a specified number of stops on each line. [M L² T⁻²] means "one stop on M, two on L, two back on T". When NTA asks "which has the same dimensions as X?", you are really asked "which route has the same stop-pattern?" — energy and torque share the route (M¹L²T⁻²), momentum and impulse share (M¹L¹T⁻¹), h and angular momentum share (M¹L²T⁻¹). Same route, different passengers.

Solved Examples on Dimensions of Physical Quantities (Step-by-Step)

Solved Example 1 — Extract an unknown dimension (NTA pattern)

Q. The position of a particle is x = At + Bt², where x is in metres and t in seconds. Find the dimensions of A and B.

Step 1 (homogeneity). [x] = [L]; each right-hand term must equal [L].

Step 2. [At] = [A][T] = [L] → [A] = [L T⁻¹] (a velocity). [Bt²] = [B][T²] = [L] → [B] = [L T⁻²] (an acceleration).

Step 3 (sense check). At is the initial-velocity term, Bt² the acceleration term of the kinematic equation — the physics agrees. ✔

Solved Example 2 — Derive and verify [thermal conductivity]

Q. Heat Q flows in time t through a slab of thickness d and area A with temperature difference ΔT: Q = k·A·ΔT·t/d. Find [k].

Step 1. Isolate k = Q·d/(A·ΔT·t).

Step 2. [Q] = [M L² T⁻²]; so numerator = [M L² T⁻²][L] = [M L³ T⁻²]. Denominator = [L²][K][T] = [L² K T].

Step 3. [k] = [M L³ T⁻²]/[L² K T] = [M L T⁻³ K⁻¹], unit W m⁻¹ K⁻¹. ✔

Practice Questions on Dimensions of Physical Quantities (With Solutions)

Practice Q1 (Numerical-value type)

Q. Among strain, stress, plane angle and refractive index, how many are dimensionless?

Solution. Strain = ΔL/L ✔ dimensionless; plane angle = arc/r ✔ dimensionless; refractive index = c/v ✔ dimensionless. Stress = F/A has [M L⁻¹ T⁻²] ✘. Count = 3.

Practice Q2 (Single-correct MCQ)

Q. Energy density (energy per unit volume) has the same dimensions as: (a) force (b) pressure (c) power (d) surface tension

Solution. [Energy/volume] = [M L² T⁻²]/[L³] = [M L⁻¹ T⁻²] = dimensions of pressure (and stress, and Young's modulus). Force is [M L T⁻²], power [M L² T⁻³], surface tension [M T⁻²].

Practice Q3 (Assertion–Reason)

Q. Assertion (A): If an equation is dimensionally homogeneous, it is certainly physically correct. Reason (R): All physically correct equations must be dimensionally homogeneous.

Solution. R is true (homogeneity is necessary). A is false — v² = u² + as is homogeneous but wrong (the coefficient of "as" must be 2). So A false, R true.

Key Formulas & Takeaways

FAQs on Dimensions of Physical Quantities

What is the dimension of a physical quantity?

It shows how, and with what powers, the base quantities (M, L, T, A, K, mol, cd) enter the unit of that quantity — e.g. force has dimensions [M L T⁻²] because F = ma.

What is the principle of homogeneity of dimensions?

In any correct physical equation, every term added or equated must have identical dimensions; only quantities of the same nature can be added, subtracted or equated.

Do dimensionless quantities have units?

Some do — the plane angle (radian) and solid angle (steradian) have special units yet zero exponents of M, L, T; strain, refractive index and relative density are pure numbers with no unit at all.

Which pairs of quantities have the same dimensions?

Work, energy, heat and torque all reduce to [M L² T⁻²]; Planck's constant and angular momentum both have [M L² T⁻¹]. Same dimensions do not make quantities identical — one is a scalar, the other a vector.