SI units and dimensional analysis

Every equation in physics carries a hidden passport — its dimensions — and checking that passport takes just seconds but can catch errors that hours of algebra miss. Dimensional analysis is one of the oldest tools in the physicist's kit, used by Newton, Fourier, and Rayleigh to deduce laws before they were fully derived.

Definition — Dimension: the expression of a physical quantity in terms of the base quantities: mass M, length L, time T, electric current A, temperature K, amount of substance mol, and luminous intensity cd.

Definition — Dimensional formula: the algebraic expression showing the powers of base dimensions in a derived quantity. For example, velocity = L T⁻¹, force = M L T⁻², energy = M L² T⁻².

Definition — Dimensionless quantity: a quantity whose dimensional formula is M⁰ L⁰ T⁰ — it is a pure number. Examples: angle (radians), refractive index, strain, relative density, Reynolds number, π.

Dimensional Formulas of Common Quantities

Memorising these is mandatory for JEE, NEET, and RRB competitive exams.

Quantity	Formula	Dimensions
Velocity	distance/time	L T⁻¹
Acceleration	velocity/time	L T⁻²
Force	mass × acceleration	M L T⁻²
Work / Energy	force × distance	M L² T⁻²
Power	work/time	M L² T⁻³
Pressure	force/area	M L⁻¹ T⁻²
Momentum	mass × velocity	M L T⁻¹
Angular momentum	mass × velocity × radius	M L² T⁻¹
Torque	force × arm	M L² T⁻²
Gravitational constant G	from F = Gm₁m₂/r²	M⁻¹ L³ T⁻²
Planck's constant h	from E = hf	M L² T⁻¹
Coefficient of viscosity	from F = η A dv/dx	M L⁻¹ T⁻¹
Surface tension	force/length	M T⁻²

What Dimensional Analysis CAN Do

1. Check the Dimensional Consistency of an Equation

A physically correct equation must be dimensionally homogeneous: every additive term must have the same dimensions, and the two sides must match. If they don't, the equation is definitely wrong. If they do, it is possibly correct (you still need physics to confirm).

Example — check v = u + at:

• [v] = L T⁻¹
• [u] = L T⁻¹
• [at] = (L T⁻²)(T) = L T⁻¹ ✓

All terms have dimensions L T⁻¹. The equation passes the dimensional check.

Example — check KE = ½mv²:

• [½mv²] = M (L T⁻¹)² = M L² T⁻² = [Energy] ✓

2. Convert Units Between Systems

The value of any physical quantity in a new unit system equals its value in the old system multiplied by the ratio of the old unit to the new unit for each dimension.

Example — convert 1 joule to CGS (ergs):
1 J = 1 kg·m²·s⁻²

In CGS: 1 kg = 10³ g, 1 m = 10² cm, 1 s = 1 s.

So 1 J = (10³ g)(10² cm)²(s)⁻² = 10³ × 10⁴ g·cm²·s⁻² = 10⁷ ergs.

3. Derive Relationships (Up to a Constant)

This is the most powerful application and the one most tested. If you know which physical quantities govern a phenomenon, you can set up a power-law ansatz and solve for the exponents from dimensional consistency.

Classic example — time period of a simple pendulum:

Assume T depends on length L, mass m, and g:

T = k · Lᵃ · mᵇ · gᶜ

Writing dimensions on both sides:

[T¹] = [L]ᵃ [M]ᵇ [L T⁻²]ᶜ

T¹ M⁰ L⁰ = Mᵇ · L^(a+c) · T^(-2c)

Equating exponents:

• T: 1 = −2c → c = −½
• M: 0 = b → b = 0
• L: 0 = a + c → a = ½

Result: T = k √(L/g). The constant k = 2π emerges only from the full Newtonian analysis; dimensions cannot find it.

Why this matters: If an exam gives you four quantities and asks "which combination has dimensions of energy?" you can evaluate each using this method in under a minute.

What Dimensional Analysis CANNOT Do

1. Find Dimensionless Constants

The formula T = 2π√(L/g) contains 2π. Rayleigh's method finds the √(L/g) part but is blind to 2π. Similarly, the ½ in KE = ½mv² and the factor 4 in Coulomb's law are invisible to dimensions.

Implication: dimensional analysis gives proportionalities, not equations. Never claim a derived proportionality is a complete formula without experimental confirmation.

2. Distinguish Between Quantities of the Same Dimension

Work and torque both have dimensions M L² T⁻². They are physically different (work is a scalar, torque is a pseudo-vector), but their dimensional formulas are identical. A dimensional check cannot tell them apart.

Other pairs with shared dimensions:

• Pressure and bulk modulus (M L⁻¹ T⁻²)
• Angular momentum and Planck's constant (M L² T⁻¹)
• Frequency and decay constant (T⁻¹)

3. Handle Transcendental Functions

Arguments of sin, cos, ln, eˣ, and exponentials must be dimensionless. You cannot apply dimensional analysis to separate parts of an equation like x = A sin(ωt + φ), because the argument (ωt + φ) must itself be dimensionless. If you try to balance dimensions, the function-within-an-equation blocks you.

Rule: if any term in an equation involves a trigonometric, logarithmic, or exponential function, the argument must be dimensionless. This is a quick way to catch errors — e.g., writing sin(velocity) in an equation is dimensionally impossible.

4. Determine Which Physical Quantities Govern a Phenomenon

Dimensional analysis assumes you already know the relevant variables. If you leave out the correct variable or include an irrelevant one, the derived relationship will be wrong. This requires physical intuition, not algebra.

Worked Examples

Question: Using dimensional analysis, find the velocity of a wave on a string if it depends only on tension F (dimension M L T⁻²) and linear mass density μ (dimension M L⁻¹).

Step 1: Assume v = k Fᵃ μᵇ.

Step 2: [L T⁻¹] = [M L T⁻²]ᵃ [M L⁻¹]ᵇ = M^(a+b) · L^(a−b) · T^(−2a)

Step 3: Equating:

• T: −1 = −2a → a = ½
• L: 1 = a − b = ½ − b → b = −½
• M: 0 = a + b = ½ − ½ ✓

Conclusion: v = k √(F/μ). The true formula is v = √(F/μ) — dimensional analysis got the structure exactly right (here k = 1 happens to be confirmed by full analysis).

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Question: Check whether the equation P = ρgh is dimensionally correct (P = pressure, ρ = density, g = acceleration due to gravity, h = depth).

Step 1: [P] = M L⁻¹ T⁻²

Step 2: [ρgh] = (M L⁻³)(L T⁻²)(L) = M L⁻¹ T⁻² ✓

Conclusion: The equation is dimensionally consistent.

Common Misconceptions

Misconception 1: "If an equation is dimensionally correct, it must be physically correct."
Not true. The equation v = u − at is dimensionally correct (same dimensions as v = u + at) but physically wrong for uniform acceleration in the direction of motion. Dimensional analysis is a necessary but not sufficient condition.

Misconception 2: "Dimensionless means zero value."
No. Dimensionless means the dimension is M⁰L⁰T⁰ = 1 (the number 1, not zero). Angles, ratios, Reynolds numbers — all have definite non-zero values; they just don't carry units.

Misconception 3: "You can always use dimensional analysis for any formula."
Only when the relationship is a power law. Formulas involving sums/differences of different functional forms (like position x = ½at² + vt) can be individually checked term by term but cannot be derived as a whole by this method.

Cover the answer, recall, then check. 12 cards on SI base units and dimensional analysis for JEE Main.

Q1. Name the seven SI base quantities and their units.
A1. Length (metre, m), mass (kilogram, kg), time (second, s), electric current (ampere, A), temperature (kelvin, K), amount of substance (mole, mol), luminous intensity (candela, cd).

Q2. What is the dimensional formula of force?
A2. [F] = [MLT⁻²] (mass × acceleration).

Q3. Dimensional formula of work, energy and torque?
A3. All are [ML²T⁻²]. (Torque and energy share dimensions but are physically different quantities.)

Q4. Dimensional formula of pressure and stress?
A4. [ML⁻¹T⁻²] (force per unit area).

Q5. State the principle of homogeneity of dimensions.
A5. Every term added or equated in a physically correct equation must have the same dimensions. It is used to check equations and derive relations.

Q6. Dimensional formula of the universal gravitational constant G?
A6. From F = Gm₁m₂/r², [G] = [M⁻¹L³T⁻²].

Q7. Dimensional formula of Planck's constant h?
A7. From E = hν, [h] = [ML²T⁻¹] (same as angular momentum).

Q8. What are dimensionless quantities? Give two examples.
A8. Quantities with no dimensions: strain, refractive index, relative density, angle (radian), coefficient of friction, all pure numbers and trig/exponential arguments.

Q9. Dimensional formula of coefficient of viscosity η?
A9. [ML⁻¹T⁻¹] (from F = ηA·dv/dx).

Q10. Why must the argument of sin, log or eˣ be dimensionless?
A10. These functions expand as infinite series of the argument's powers; adding different powers requires each term (hence the argument) to be a pure number.

Q11. Dimensional formula of the gas constant R?
A11. From PV = nRT, [R] = [ML²T⁻²K⁻¹mol⁻¹] (energy per mole per kelvin).

Q12. Chief limitation of dimensional analysis.
A12. It cannot find dimensionless constants (like ½ or 2π), cannot handle equations with more than one term of the same dimension, and fails for trig/exponential relations.

Why it matters

Units and dimensions is the very first chapter of JEE Main and delivers 1–2 near-guaranteed marks every year. Questions are conceptual and fast: find a dimensional formula, check homogeneity, or spot which physical quantities share dimensions. Since Physics has only 20–25 questions, banking these is essential.

The seven SI base units

The SI system (Système International) is built on seven base quantities from which all others are derived.

Quantity	Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Current	ampere	A
Temperature	kelvin	K
Amount	mole	mol
Luminous intensity	candela	cd

Key dimensional formulae

• Velocity [LT⁻¹], acceleration [LT⁻²], force [MLT⁻²]
• Work/energy/torque [ML²T⁻²], power [ML²T⁻³]
• Pressure/stress [ML⁻¹T⁻²], momentum/impulse [MLT⁻¹]
• G = [M⁻¹L³T⁻²], h = [ML²T⁻¹], R = [ML²T⁻²K⁻¹mol⁻¹]

Exam Tricks & Tips

• 🎯 Quantities that share dimensions are favourite MCQs: work = energy = torque = [ML²T⁻²]; pressure = stress = energy density = [ML⁻¹T⁻²].
• 🎯 Angular momentum and Planck's constant both are [ML²T⁻¹] — a classic "match the pair" trap.
• 🎯 Arguments of sin, cos, log, eˣ are always dimensionless — use this to fix unknown constants inside them.
• 🎯 To find dimensions of a constant, isolate it in a known equation (e.g. G from F = Gm₁m₂/r²).
• 🎯 Strain, refractive index, relative density, angle, coefficient of friction are dimensionless — memorise this list.
• ❌ Common mistake: treating dimensional correctness as proof of physical correctness — a dimensionally right equation can still be wrong because pure numbers (½, 2π) are invisible to dimensions.

Expected exam pattern

1 question is almost certain: either "dimensional formula of X", "which pair has the same dimensions", or "check/derive by homogeneity". Occasionally combined with error analysis. Difficulty: easy to moderate; pure recall + reasoning.

Quick recap

Memorise the seven base units and the standard derived dimensional formulae. Use homogeneity to check equations and find constant's dimensions. Remember dimensionless quantities and that dimensional analysis cannot recover numerical constants or handle multi-term/trig equations.

Key formulas

• Seven SI base units: metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), candela (cd).
• Dimensional formula: mass [M], length [L], time [T]; e.g. force = [M L T⁻²], energy = [M L² T⁻²], pressure = [M L⁻¹ T⁻²], power = [M L² T⁻³].
• Principle of homogeneity: every term in a valid equation has the same dimensions.
• To convert a unit: n₂ = n₁ (M₁/M₂)^a (L₁/L₂)^b (T₁/T₂)^c.
• Uses: check equation correctness, derive relations (up to a constant), convert units.
• Limitation: cannot find dimensionless constants or distinguish quantities with equal dimensions (work vs torque).

Usage: use homogeneity to catch wrong options and to derive formulae to within a constant.

Worked Example

Problem: The velocity v of a wave on a stretched string may depend on the tension T (a force), the mass per unit length μ, and the length L. Use dimensional analysis to find how v depends on these quantities.

Solution:
Assume v = k · T^a · μ^b · L^c, where k is a dimensionless constant.
Write the dimensions (M, L, T for mass, length, time):

• [v] = L T⁻¹
• [T] = M L T⁻² (force)
• [μ] = M L⁻¹
• [L] = L

Substitute: L T⁻¹ = (M L T⁻²)^a (M L⁻¹)^b (L)^c
= M^(a+b) · L^(a−b+c) · T^(−2a)

Match powers on both sides:

• Mass: a + b = 0
• Time: −2a = −1 → a = 1/2
• Then b = −1/2
• Length: a − b + c = 1 → 1/2 − (−1/2) + c = 1 → c = 0

Answer: v = k · T^(1/2) · μ^(−1/2) = k √(T/μ). Dimensional analysis fixes the exponents but not the constant k; experiment gives k = 1, so v = √(T/μ).