Dimensional Analysis I: Checking Equations

Somewhere in every JEE-Main paper sits a question built on a formula you have never seen — the oscillation period of a liquid drop, the speed of a wave on a membrane. The setters know you haven't derived it. That is the point: three of the four options are impossible, and impossible in a way that takes thirty seconds to detect if you can check dimensions — and a panicked guess if you can't. The detector is the principle of homogeneity: every term that is added, subtracted, or equated in a physical equation must carry the same dimensions. Nature does not add three metres to five seconds. Any candidate formula that tries to is wrong before a single number is plugged in — no memory of the correct result required. Professional physicists run this audit reflexively on every line of a derivation; for you it converts a whole family of exam questions from "recall" into "routine". This module makes the audit a reflex: the term-by-term checking algorithm, forcing an equation to confess the dimensions of unknown constants hiding inside it (the $a$ and $b$ of the van der Waals equation are a board favourite), why the argument of every sine, exponential and logarithm must be a pure number — and the one thing the check can never do, which examiners test just as often. The timed version — killing MCQ options in under thirty seconds — is drilled in Dimensional Shortcuts on MCQs; this is where the method itself is built.

Key takeaways

Principle of homogeneity: in any valid physical equation $A = B + C$ , every term has identical dimensions — $[A] = [B] = [C]$ . Terms with different dimensions cannot be added, subtracted, or equated.
The audit is term-by-term: reduce each added term to powers of $\mathrm{M}$ , $\mathrm{L}$ , $\mathrm{T}$ using the dimensional formulae of the quantities in it, then compare.
Unknown constants inside an equation inherit whatever dimensions make the equation homogeneous — isolate the constant and read them off.
Arguments of $\sin$ , $\cos$ , $e^x$ , $\log$ must be dimensionless, and so are the values these functions return.
Passing the check is necessary but not sufficient: dimensionless factors like $\tfrac12$ or $2\pi$ are invisible to dimensions, so a consistent equation can still be wrong.
Key idea: dimensional homogeneity is a lie detector, not a truth machine — it can prove an equation wrong, never prove it right.

The core insight

Every term added or equated in a physical equation must have the same dimensional formula, so checking dimensions term by term can expose an impossible equation in seconds — while dimensionless factors remain invisible, which is why a passed check rules nothing in.

Where this lives today

The homogeneity audit is now written into software. Engineering and scientific codebases attach units to variables and let the compiler run the customs check on every line: Python's pint, C++ units types and F#'s built-in units-of-measure refuse to add newtons to newton-seconds — the beam balance executed a billion times a second. The motivation is expensive history: the Mars Climate Orbiter burned up in 1999 because one team's software emitted impulse in pound-force-seconds while another's consumed newton-seconds — a mismatch one dimensional audit at the interface would have flagged. Spacecraft, avionics and medical-dosage code now treat unit-checking as a standard defence layer. Theoretical physicists still open hard problems exactly as Rayleigh did: write the plausible variables, demand homogeneity, read off the only combination allowed — the famous 1950 estimate of the first atomic bomb's yield from fireball photographs, before the number was declassified, was precisely such an argument. When you kill three JEE options by dimensions, you are running the same engine at exam scale.

The physical picture

Picture a beam balance at a customs post, and think of each term of an equation as a parcel placed on it. The equation $v^2 = u^2 + 2as$ arrives as three parcels: $v^2$ on the left pan, $u^2$ and $2as$ on the right. The officer does not care what is inside each parcel — only what it weighs in the currency of dimensions. Weigh them: $[v^2] = \mathrm{L^2\,T^{-2}}$ , $[u^2] = \mathrm{L^2\,T^{-2}}$ , and $[2as] = (\mathrm{L\,T^{-2}})(\mathrm{L}) = \mathrm{L^2\,T^{-2}}$ . Three identical weights: the beam sits level, and the equation is allowed through.

Figure

The homogeneity balance — every term must weigh the same in dimensions

Notice what the officer did not certify. The balance cannot see pure numbers: $2as$ and $as$ weigh exactly the same, so $v^2 = u^2 + 7as$ would sit just as level. The check is one-directional — a tipped beam proves the equation false; a level beam proves only that it survives. This asymmetry is the most examined idea in the module, and it has a name to use in written answers: homogeneity is a necessary but not sufficient condition for correctness. Why does nature enforce homogeneity at all? Because adding quantities of different kinds has no meaning: three metres plus five seconds is not eight of anything. An equation claims two physical quantities are the same quantity — and sameness begins with being the same kind. Dimensions, built in Dimensions & Dimensional Formulae, are the bookkeeping of kind: every mechanical quantity reduces to powers of $\mathrm{M}$ , $\mathrm{L}$ , $\mathrm{T}$ , and two quantities are the same kind exactly when those powers match.

The homogeneity principle, stated precisely

The working statement is short. If a physical equation reads

A = B + C - D \qquad (1)

where: $A$ , $B$ , $C$ , $D$ — physical quantities (each possibly a product of several symbols, like $2as$ ); then validity requires $[A] = [B] = [C] = [D]$ . Read it physically: the equality sign and every plus or minus in physics joins quantities of one single kind. A corollary does most of the exam work: since multiplication may mix kinds ( $F = ma$ mixes mass and acceleration happily), the audit only ever compares complete terms — the chunks separated by $=$ , $+$ and $-$ — never individual symbols inside them. The audit algorithm, then:

Step 1. Slice the equation at every $=$ , $+$ , $-$ into complete terms. For $v^2 = u^2 + 2as$ the slices are $v^2$ , $u^2$ , $2as$ .

Step 2. Replace each quantity inside a term by its dimensional formula, drop all pure numbers, and simplify the powers of $\mathrm{M}$ , $\mathrm{L}$ , $\mathrm{T}$ .

Step 3. Compare the finished terms. All equal — the equation survives (say "dimensionally consistent", never "proved"). Any mismatch — the equation is dead, and you can say exactly why. The raw material for Step 2 is the bank of standard dimensional formulae built in the previous module — $[F] = \mathrm{M\,L\,T^{-2}}$ , $[E] = \mathrm{M\,L^2\,T^{-2}}$ , $[P] = \mathrm{M\,L^{-1}\,T^{-2}}$ , $[\rho] = \mathrm{M\,L^{-3}}$ and friends. This module assumes the bank is loaded.

Derivation AAuditing a correct equation and executing a wrong one

We want to run the full audit twice — once on the genuine kinematic equation $v^2 = u^2 + 2as$ , and once on a plausible-looking impostor for the pendulum period, $T = 2\pi\sqrt{g/L}$ , to watch the check actually kill something. The strategy is pure Step 1–2–3: slice into terms, reduce each to powers of $\mathrm{M}$ , $\mathrm{L}$ , $\mathrm{T}$ , compare. The thing to watch is how the square root handles dimensions — a root halves every exponent, which is where careless auditors slip.

Step 1. Slice the kinematic equation: terms $v^2$ , $u^2$ , $2as$ .

Step 2. Reduce each. Velocity is length per time, $[v] = [u] = \mathrm{L\,T^{-1}}$ , so $[v^2] = [u^2] = \mathrm{L^2\,T^{-2}}$ . Acceleration carries $[a] = \mathrm{L\,T^{-2}}$ and displacement $[s] = \mathrm{L}$ , so $[2as] = \mathrm{L\,T^{-2}} \times \mathrm{L} = \mathrm{L^2\,T^{-2}}$ — the 2 is dropped, it has no dimensions. Why drop the 2: dimensions track kind, not amount. Doubling a velocity-squared gives another velocity-squared.

Step 3. Compare: $\mathrm{L^2\,T^{-2}} = \mathrm{L^2\,T^{-2}} = \mathrm{L^2\,T^{-2}}$ . The equation is dimensionally consistent — it may be true (it is), but the audit alone cannot say so.

Step 4. Now the impostor $T = 2\pi\sqrt{g/L}$ . Left side: $[T] = \mathrm{T}$ . Right side: $[g] = \mathrm{L\,T^{-2}}$ and $[L] = \mathrm{L}$ , so $[g/L] = \mathrm{T^{-2}}$ , and the square root halves the exponent: $[\sqrt{g/L}] = \mathrm{T^{-1}}$ .

Step 5. Compare: $\mathrm{T} \neq \mathrm{T^{-1}}$ . The beam tips. $T = 2\pi\sqrt{g/L}$ is not a formula for a time — it is a formula for a frequency (in fact $\omega = \sqrt{g/L}$ is exactly the pendulum's angular frequency). The correct period is $T = 2\pi\sqrt{L/g}$ : then $[L/g] = \mathrm{T^2}$ and the root returns $\mathrm{T}$ , as required. Why this impostor matters: g-over-L versus L-over-g is precisely the flip your memory makes under exam pressure, and the audit catches it in one line — this is the single most valuable dimension-check in the syllabus. *Verified: $[2as] = (\mathrm{L\,T^{-2}})(\mathrm{L}) = \mathrm{L^2\,T^{-2}}$ recomputed; $[\sqrt{L/g}] = \sqrt{\mathrm{L} / \mathrm{L\,T^{-2}}} = \sqrt{\mathrm{T^2}} = \mathrm{T}$ recomputed; both against the NCERT Ch. 1 treatment of homogeneity.*

Hidden constants: make the equation confess

Boards love equations containing unlabelled constants: "the velocity of a particle varies as $v = At^2 + Bt + C$ ; find the dimensions of $A$ , $B$ and $C$ ." The cracking principle is homogeneity read in reverse: since each term must carry the dimensions of the left side, every constant inherits exactly the dimensions needed to make its term legal. Isolate and read off:

[X] = \frac{[\text{whole term's required dimensions}]}{[\text{everything else in the term}]} \qquad (2)

where: $X$ — the unknown constant; the numerator — the dimensions every term must share (here $[v] = \mathrm{L\,T^{-1}}$ ); the denominator — the dimensions of the known quantities multiplying $X$ . So $[A] = \mathrm{L\,T^{-1}}/\mathrm{T^2} = \mathrm{L\,T^{-3}}$ , $[B] = \mathrm{L\,T^{-1}}/\mathrm{T} = \mathrm{L\,T^{-2}}$ (an acceleration), and the lone $[C] = \mathrm{L\,T^{-1}}$ (a velocity). Constants are not necessarily dimensionless — they are whatever the equation forces them to be, as $G$ , $g$ , $h$ and $R$ all demonstrate.

Go deeper · Derivation B · 1 minExtracting a and b from the van der Waals equation

Derivation BExtracting a and b from the van der Waals equation

We want the dimensions of the constants $a$ and $b$ in the van der Waals equation of state $\left(P + \frac{a}{V^2}\right)(V - b) = nRT$ — the classic board and JEE question on hidden constants. The strategy is homogeneity applied inside each bracket: whatever is added to the pressure $P$ must itself be a pressure, and whatever is subtracted from the volume $V$ must itself be a volume. No knowledge of gas physics is needed — that is exactly the beauty being tested. Watch how each bracket becomes an independent one-line problem.

Step 1. First bracket: $P + \dfrac{a}{V^2}$ . Homogeneity demands $\left[\dfrac{a}{V^2}\right] = [P] = \mathrm{M\,L^{-1}\,T^{-2}}$ .

Step 2. Solve for $[a]$ using equation (2): $[a] = [P][V^2]$ . Volume has $[V] = \mathrm{L^3}$ , so $[V^2] = \mathrm{L^6}$ and $[a] = \mathrm{M\,L^{-1}\,T^{-2}} \times \mathrm{L^6} = \mathrm{M\,L^{5}\,T^{-2}}$ . Why multiply rather than divide: the constant sits in the numerator over $V^2$ , so clearing the fraction moves $V^2$ across as a product.

Step 3. Second bracket: $V - b$ . Homogeneity demands $[b] = [V] = \mathrm{L^3}$ directly — $b$ is a volume (physically, the volume excluded by the gas molecules themselves).

Step 4. Cross-check the whole equation: the left side is now pressure × volume, $[PV] = \mathrm{M\,L^2\,T^{-2}}$ — energy, matching $[nRT]$ (thermal energy) on the right. The audit closes consistently. Verified: $[a] = \mathrm{M\,L^5\,T^{-2}}$ and $[b] = \mathrm{L^3}$ recomputed against the standard NCERT exemplar answer; $[PV] = [nRT]$ cross-checked via work = force × distance.

The dimensionless-argument rule

There is a second homogeneity law, quieter but just as examinable: the argument of any trigonometric, exponential or logarithmic function must be a pure number, and the value the function returns is a pure number too. You may write $\sin(\omega t)$ but never $\sin(t)$ where $t$ is a raw time; $e^{-t/\tau}$ but never $e^{-t}$ . The reason appears the moment you remember what these functions are underneath — sums of powers. If the argument carried the dimension of time, the sine's sum would be adding a time to a time-cubed to a time-to-the-fifth: a triple violation of homogeneity inside one symbol. Only a dimensionless argument makes the sum legal:

\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \quad \Rightarrow \quad [x] = 1 \qquad (3)

where: $[x] = 1$ — the standard way of writing "dimensionless"; the displayed sum — the reason, not something to memorise: powers of $x$ are being added, so homogeneity forces all powers of $[x]$ to match, which only $[x]=1$ achieves. This rule is a free gift in constant-hunting questions: in $y = A\sin(kx)$ with $x$ a length, $[kx] = 1$ instantly gives $[k] = \mathrm{L^{-1}}$ — and $[y] = [A]$ , since the sine's output is a pure number. Radians ride along: an angle is arc over radius, $\mathrm{L}/\mathrm{L} = 1$ , so angles are dimensionless and $\omega t$ ( $\mathrm{T^{-1}} \times \mathrm{T} = 1$ ) is a legal argument.

Figure

The dimensionless window — what sin, exp and log will accept

Why a passed check proves nothing

Take the equation for distance fallen from rest, $s = \tfrac12 g t^2$ , and its evil twin $s = g t^2$ . Audit both: $[gt^2] = \mathrm{L\,T^{-2}} \times \mathrm{T^2} = \mathrm{L}$ . Both pass, identically. The factor $\tfrac12$ — the entire difference between right and wrong — is invisible to dimensions, and so are $2\pi$ , $\tfrac43$ , and every other pure number physics is decorated with. The check is blind in a second way: it cannot tell apart two different quantities that share a dimensional formula. Torque and energy both reduce to $\mathrm{M\,L^2\,T^{-2}}$ ; an audit would happily let you equate a torque to a kinetic energy — physical nonsense, one a turning effect, the other a capacity to do work. Both blindnesses point the same moral, the one to write in any "check the correctness" answer: homogeneity can only ever falsify. A tipped balance proves error; a level balance is an invitation to keep checking by other means — limiting cases, special values, or an actual derivation.

Limiting cases

Case	Result	Meaning
equation with a single term each side, e.g. $F = ma$	one comparison, $[F] = [ma]$	products may mix kinds freely — only $+$ , $-$ , $=$ are policed
pure-number factor changed, $2 \to 7$	audit unchanged, still passes	dimensionless factors are invisible: the check cannot fix coefficients
square root of a term	every exponent halves	roots and powers act on dimensions exactly as on symbols
trig, exp, log present	argument must reduce to $[x]=1$ ; output counts as 1	a second, independent audit inside every such function
two quantities share a formula (torque, energy)	audit cannot distinguish them	dimensional kind is coarser than physical kind
dimensionless equation, e.g. strain $= \Delta L / L$	all terms $[x] = 1$	homogeneity is satisfied trivially — the check has no grip

The last row matters later in the course: equations built from ratios (strain, refractive index, friction coefficients) pass any audit automatically, so for them the method offers zero protection.

Common trap“The equation checked out dimensionally, so it must be correct”

This is the most-marked misconception in the chapter, and examiners bait it deliberately: they present a dimensionally consistent equation with a wrong coefficient — $s = gt^2$ , $T = \pi\sqrt{L/g}$ , $E = mc^2/2$ — and the option "the equation is correct because it is dimensionally consistent". Consistency is a necessary condition, never a sufficient one. Dimensions cannot see pure numbers ( $\tfrac12$ , $2\pi$ ), cannot separate torque from energy, and cannot police relationships between dimensionless quantities at all. The correct exam language: "the equation is dimensionally consistent, hence it may be correct; dimensional analysis cannot confirm the numerical factor." Writing "proved correct" after a homogeneity check costs the reasoning mark even when the equation happens to be true.

Worked examples

Board level — audit and constant-hunt in one question (CBSE/ICSE pattern)

Check the dimensional consistency of $P = \rho g h$ for pressure at depth $h$ in a liquid of density $\rho$ . Then, for a particle whose velocity varies as $v = At^2 + Bt + C$ , find the dimensions and SI units of $A$ , $B$ and $C$ .

Step 1. Audit $P = \rho g h$ term by term. Left: $[P] = \mathrm{M\,L^{-1}\,T^{-2}}$ (force per area: $\mathrm{M\,L\,T^{-2}} / \mathrm{L^2}$ ).

Step 2. Right: $[\rho g h] = (\mathrm{M\,L^{-3}})(\mathrm{L\,T^{-2}})(\mathrm{L}) = \mathrm{M\,L^{-1}\,T^{-2}}$ . Left equals right — dimensionally consistent (state clearly: this shows the equation may be correct, and it is).

Step 3. Now the constants. Homogeneity forces every term of $v = At^2 + Bt + C$ to carry $[v] = \mathrm{L\,T^{-1}}$ .

Step 4. Read each constant off with equation (2). $[A] = \mathrm{L\,T^{-1}}/\mathrm{T^2} = \boxed{\mathrm{L\,T^{-3}}}$ , SI unit $\mathrm{m\,s^{-3}}$ . $[B] = \mathrm{L\,T^{-1}}/\mathrm{T} = \boxed{\mathrm{L\,T^{-2}}}$ , SI unit $\mathrm{m\,s^{-2}}$ — an acceleration. $[C] = \boxed{\mathrm{L\,T^{-1}}}$ , SI unit $\mathrm{m\,s^{-1}}$ — the velocity at $t = 0$ .

Step 5. The marks-scheme sentence: "By the principle of homogeneity, each added term must have the dimensions of velocity." Naming the principle is typically worth its own mark.

Competitive level — kill three options by weighing them (JEE-Main pattern)

The period $T$ of oscillation of a small liquid drop depends on its density $\rho$ , radius $r$ and surface tension $S$ . Which relation can be correct? *(A) $T = k\sqrt{\rho r^3 / S}$ (B) $T = k\sqrt{\rho r^2 / S}$ (C) $T = k\sqrt{S/\rho r^3}$ (D) $T = k\,\rho r^3 / S$ *

Step 1. Load the one non-standard formula: surface tension is force per length, $[S] = \mathrm{M\,L\,T^{-2}}/\mathrm{L} = \mathrm{M\,T^{-2}}$ . (This single recall is the whole difficulty of the question.)

Step 2. Weigh the core combination once: $[\rho r^3] = (\mathrm{M\,L^{-3}})(\mathrm{L^3}) = \mathrm{M}$ , so $\left[\dfrac{\rho r^3}{S}\right] = \dfrac{\mathrm{M}}{\mathrm{M\,T^{-2}}} = \mathrm{T^{2}}$ .

Step 3. Option (A): square root of $\mathrm{T^2}$ is $\mathrm{T}$ . A period. (A) survives.

Step 4. Execute the rest. (C) is the reciprocal root: $\mathrm{T^{-1}}$ — a frequency, dead. (D) skips the root: $\mathrm{T^2}$ — dead. (B): $[\rho r^2/S] = \mathrm{M\,L^{-1}}/\mathrm{M\,T^{-2}} = \mathrm{L^{-1}\,T^2}$ , whose root $\mathrm{L^{-1/2}\,T}$ is not even a clean quantity — dead. Answer: $\boxed{\text{(A)}}$ .

Step 5. Note what was not needed: any theory of drop oscillation. Homogeneity alone selected the survivor — though it cannot supply $k$ . Timing note: 45–75 seconds. Spend the first ten recalling $[S] = \mathrm{M\,T^{-2}}$ ; then weigh $\rho r^3/S$ once and reuse it across all four options. If two options survive, the question needs physics, not dimensions — that happens, and it is the setters' favourite twist on this classic.

Exam traps

Auditing only one side. "LHS has dimensions of X" proves nothing until the RHS — every term of it — is weighed too. Write both columns explicitly.

Forgetting the argument rule. In $y = a\sin(2\pi t/T)$ questions, the marks are usually inside the sine: showing $[2\pi t/T] = 1$ and therefore $[y] = [a]$ . Students who audit only the outer equation miss the inner one.

Trying to check the number. "Confirmed because the 2 balances" is a red flag to any examiner — pure numbers have no dimensions and can never be confirmed or refuted dimensionally. If a question asks you to verify the coefficient, dimensional analysis is explicitly the wrong tool; say so.

Units in a dimensions answer (and vice versa). " $[B] = \mathrm{m\,s^{-2}}$ " loses the mark on a dimensions question; " $\mathrm{L\,T^{-2}}$ " loses it on a units question. Boards phrase it "dimensions and SI units" when they want the pair — read the question's noun.

Roots handled sloppily. $\sqrt{L/g}$ has dimensions $\mathrm{T}$ , not $\mathrm{T^2}$ — the root halves exponents. The $g/L$ -versus- $L/g$ flip in the pendulum formula is the canonical casualty, and exactly the error a ten-second audit exists to catch.

Claiming proof. "Hence proved correct" after a homogeneity check is a stated mark-loser in CBSE scheme notes and a wrong-option magnet in JEE/NEET. The licensed conclusion is "dimensionally consistent" — nothing stronger. A dimensional failure, though, is a complete proof of wrongness; claim it with confidence.

Radians treated as a dimension. $[\omega] = \mathrm{T^{-1}}$ , because the radian is $\mathrm{L}/\mathrm{L} = 1$ . An audit carrying a phantom "angle dimension" will wrongly execute correct equations like $v = r\omega$ .

Origins · 1 minLord Rayleigh and how this was discovered

Lord Rayleigh

1842–1919 · Cambridge and Terling Place

John William Strutt, third Baron Rayleigh, made the method of dimensions a working physicist's everyday instrument. Where Fourier had stated the homogeneity principle, Rayleigh used it relentlessly — his Theory of Sound (1877) opens problem after problem by writing the plausible variables and letting dimensions dictate the form of the answer; his dimensional argument for why sky-light scatters as $1/\lambda^4$ explained the blue sky essentially free of detailed theory. The same obsession with what tiny discrepancies must mean led him to notice that nitrogen from air was 0.5% denser than nitrogen from chemicals — error-chasing that uncovered argon and won the 1904 Nobel Prize. His 1915 Nature note, "The Principle of Similitude", remains the manifesto for maximum physics from minimum calculation.

History

1822

— Fourier's Analytical Theory of Heat states that every term of a physical equation must share the same dimensions.

1877

— Rayleigh's Theory of Sound turns the "method of dimensions" into a standard working tool.

1904

— Rayleigh wins the Nobel Prize for argon — discovered by refusing to ignore a 0.5% density discrepancy.

1914

— Buckingham publishes the pi theorem, the general framework behind every dimensional argument.

1915

— Rayleigh's Nature article "The Principle of Similitude" popularises dimensional reasoning across physics and engineering.

Going further

Texts. NCERT Physics Part 1, Ch. 1 — the CBSE/JEE canonical statement · H.C. Verma, Concepts of Physics Vol. 1, Ch. 1 (the best drill set on hidden constants) · CIE 9702 syllabus, "Physical quantities and units" · P.W. Bridgman, Dimensional Analysis (1922) for where the method leads.

Provenance. All checks recomputable by hand. A verifier should confirm: $[2as] = \mathrm{L^2\,T^{-2}}$ ; $[\sqrt{g/L}] = \mathrm{T^{-1}}$ vs $[\sqrt{L/g}] = \mathrm{T}$ ; $[A] = \mathrm{L\,T^{-3}}$ , $[B] = \mathrm{L\,T^{-2}}$ , $[C] = \mathrm{L\,T^{-1}}$ in $v = At^2 + Bt + C$ ; van der Waals $[a] = \mathrm{M\,L^5\,T^{-2}}$ , $[b] = \mathrm{L^3}$ ; $[S] = \mathrm{M\,T^{-2}}$ and $[\sqrt{\rho r^3/S}] = \mathrm{T}$ in the drop-oscillation MCQ; $[k] = \mathrm{L^{-1}}$ in $y = A\sin(kx)$ ; $[b] = \mathrm{M\,T^{-1}}$ in $v = v_0 e^{-bt/m}$ .

Prerequisites. Dimensions & Dimensional Formulae (the formula bank this audit runs on). Leads to. Dimensional Analysis II: Deriving Relations & Limits (the same principle run forwards) · Dimensional Shortcuts on MCQs (the timed drill built on this module).

Why exactly must the argument of sin or log be dimensionless?

Because those functions are secretly sums of different powers of their argument: $\sin x = x - x^3/6 + \cdots$ , $e^x = 1 + x + x^2/2 + \cdots$ . If $x$ carried the dimension of time, the sum would add a pure number to a time to a time-squared — homogeneity violated many times inside one symbol. Only a dimensionless argument makes every power the same kind; the output is then a pure number too, which is why in $y = A\sin(kx)$ all of $y$ 's dimensions come from $A$ . In a board answer, the accepted sentence is simply: "arguments of trigonometric, exponential and logarithmic functions must be dimensionless by the principle of homogeneity."

Can I check with units instead of dimensions?

Yes — checking that both sides carry the same SI units is logically the same audit, and A-Level papers phrase it exactly that way. Dimensions are units with the arbitrary choices stripped off: $\mathrm{m\,s^{-2}}$ and $\mathrm{km\,h^{-2}}$ are different units but one dimension, $\mathrm{L\,T^{-2}}$ . For Indian boards, answer in dimensional notation when the question says "dimensions" — mixing the vocabularies is a known mark-loser.

Torque and energy both have ML²T⁻² — so is a torque a kind of energy?

No. Sharing a dimensional formula means the audit cannot distinguish them, not that they are the same quantity. Torque is a turning effect with a direction; energy is a scalar capacity to do work — the newton-metre and the joule stay separate names precisely to preserve the distinction that dimensions erase. It is a stock exam question ("name two different physical quantities with the same dimensional formula" — torque/energy, and at a subtler level angular momentum/Planck's constant) and a genuine limitation of the method.

If g is called a constant, why does it have dimensions?

"Constant" means the value does not change in context — it says nothing about dimensions. $g = 9.8\ \mathrm{m\,s^{-2}}$ carries $\mathrm{L\,T^{-2}}$ ; $G$ carries $\mathrm{M^{-1}\,L^3\,T^{-2}}$ ; $R$ carries energy per mole per kelvin. Only pure numbers — $2\pi$ , $\tfrac12$ — are dimensionless. When a question asks for "the dimensions of the constant", the phrasing itself is the hint: they are non-trivial, and homogeneity will hand them to you.

The exam asked me to "check the correctness" of an equation. Is a dimensional check a full answer?

It is the expected answer, with one crucial caveat. Run the term-by-term audit, show every term's reduction to M-L-T powers, conclude "dimensionally consistent" — then add the licensed limitation: dimensional analysis cannot verify dimensionless factors, so consistency shows the equation can be correct, not that it is. CBSE schemes allot a mark for that limitation line. If the audit fails, conclude flatly that the equation is incorrect — falsification is the one direction in which the method is absolute.

How do I audit an equation like v = v₀e^(−bt/m)?

Two nested audits. Inner: the exponent must be dimensionless, so $[bt/m] = 1$ , giving $[b] = \mathrm{M\,T^{-1}}$ — often the actual question. Outer: $e^{-bt/m}$ is then a pure number, so $[v] = [v_0]$ , trivially satisfied. Work inside-out on anything containing $\sin$ , $\cos$ , $e^x$ or $\log$ : force each argument dimensionless first (this usually determines the unknown constant), then treat the function's value as 1 in the outer audit.