Dimensions & Dimensional Formulae

A Falcon 9 leaving the pad delivers about $7.6 \times 10^6\ \mathrm{N}$ of thrust. A fast bowler's arm puts roughly $150\ \mathrm{N}$ on a cricket ball. Strip away the numbers and the units, and something remarkable remains: both quantities have exactly the same skeleton, $\mathrm{M\,L\,T^{-2}}$ — one factor of mass, one of length, two of time in the denominator. That skeleton is the dimensional formula, the deepest identity a physical quantity has. Change the country, the unit system, the century — the skeleton does not move. Exams know this. Every year, JEE-Main and NEET open their Unit-1 batting with dimensional-formula questions: find the dimensions of $b$ in this equation, which pair has identical dimensions. These are among the fastest marks on the paper — for the student who can derive a formula from its defining equation in three lines instead of recalling it from a memorised list that panic has just erased. This article gives you the machine: the grammar $[Q] = \mathrm{M}^a\mathrm{L}^b\mathrm{T}^c$ , the algorithm that generates any dimensional formula from a definition you already know, the short list actually worth memorising, and the famous coincidences — torque and work, Planck's constant and angular momentum — that examiners have turned into an industry. The next two modules weaponise the skill: checking equations, then deriving them. This is where the weapon is forged.

Key takeaways

Every physical quantity is a product of powers of the seven base dimensions $\mathrm{M, L, T, I, \Theta, N, J}$ — in mechanics, just $\mathrm{M, L, T}$ .
Never recall a dimensional formula cold: derive it from the defining equation — $F = ma$ gives $[F] = \mathrm{M\,L\,T^{-2}}$ in one line, and every harder case chains from it.
Dimensional arithmetic is exponent arithmetic: multiplying quantities adds exponents, dividing subtracts them, raising to a power multiplies them.
Arguments of $\sin$ , $\cos$ , $e^x$ and $\log$ are always dimensionless — which instantly fixes the dimensions of things like $\omega$ in $\sin(\omega t)$ and $k$ in $\sin(kx)$ .
Different quantities can share one skeleton — work and torque are both $\mathrm{M\,L^2\,T^{-2}}$ . Same dimensions never means same quantity.
Key idea: the dimensional formula records how a quantity is built from the base quantities — it is unit-system-proof, derivable in seconds from a defining equation, and the single most bankable skill in the measurement unit.

The core insight

The dimensional formula $[Q] = \mathrm{M}^a\mathrm{L}^b\mathrm{T}^c\ldots$ expresses any physical quantity as powers of the base quantities, and it is always generated — not remembered — by substituting known dimensions into the quantity's defining equation.

Where this lives today

Push the dimension game to its extreme and you reach the frontier of physics. Combine the three great constants — $G$ ( $\mathrm{M^{-1}L^3T^{-2}}$ ), $h$ ( $\mathrm{ML^2T^{-1}}$ ) and the speed of light $c$ ( $\mathrm{LT^{-1}}$ ) — and ask: is there a combination with the dimensions of pure length? Exponent algebra, exactly the kind you did in this article, gives one answer: $\sqrt{Gh/c^3} \approx 10^{-35}\ \mathrm{m}$ . That is the Planck length, the scale where physicists expect space itself to stop behaving classically — found not by experiment but by dimensional bookkeeping on three constants. Particle physicists go further and choose units where $c = 1$ and $h/2\pi = 1$ , collapsing the alphabet until mass, energy and frequency wear one skeleton; every "natural units" calculation at CERN is a descendant of equation (1). Closer to your own future work: modern engineering software carries dimensions inside its variables, so adding a force to a torque throws an error before anything gets built. Your three-mark skill is, at scale, how spacecraft avoid dying of unit slips.

The physical picture

Think of $\mathrm{M}$ , $\mathrm{L}$ and $\mathrm{T}$ as three kinds of atoms. Velocity is one length atom over one time atom. Force bonds a mass atom to a length atom and divides by time twice. Energy is force that has acted through another length. Nothing else is needed: the entire vocabulary of mechanics — momentum, pressure, power, viscosity, angular momentum — is molecules assembled from three atoms.

Figure

Three atoms build all of mechanics — the dimensional molecules of velocity, force and energy

The full periodic table of dimensions has seven entries, one per SI base quantity: $\mathrm{M}$ , $\mathrm{L}$ , $\mathrm{T}$ , electric current $\mathrm{I}$ , temperature $\Theta$ , amount of substance $\mathrm{N}$ , luminous intensity $\mathrm{J}$ . In mechanics the first three do almost all the work; $\Theta$ appears the moment heat enters, and $\mathrm{I}$ waits for the electricity chapters. Some quantities are built from no atoms at all. An angle is arc over radius, strain is extension over original length, refractive index is a speed over a speed — the atoms cancel completely, leaving a dimensionless quantity, pure number, written $\mathrm{M^0L^0T^0}$ . These are not a footnote: a dimensionless combination is the only thing allowed inside a sine or an exponential, and exam setters hide answers exactly there. Notice what dimensions deliberately ignore: the size of the quantity, the unit it is quoted in, even its vector or scalar character. A dimensional formula is an X-ray — bones, not face. Units are dimensions dressed in a particular system, which is exactly why the dimension survives every unit conversion you did in the previous module.

The grammar of dimensions

Ask "what are the dimensions of $Q$ ?" and the answer is always a product of powers of the base symbols — nothing else is possible, because $Q$ was ultimately defined through multiplications and divisions of base quantities:

[Q] = \mathrm{M}^{a}\,\mathrm{L}^{b}\,\mathrm{T}^{c}\,\mathrm{I}^{d}\,\Theta^{e}\,\mathrm{N}^{f}\,\mathrm{J}^{g} \qquad (1)

where: $[Q]$ — read "the dimensions of $Q$ " (the square brackets are an operator, not decoration); $a, b, c, \ldots$ — the dimensional exponents, rational numbers counting how many of each atom the quantity contains, negative values meaning "in the denominator". Velocity has $(a,b,c) = (0,1,-1)$ ; force has $(1,1,-2)$ . In mechanics you may quietly drop the last four symbols and write $[Q] = \mathrm{M}^a\mathrm{L}^b\mathrm{T}^c$ . The arithmetic of these formulae is nothing but the law of exponents you have used since Class 8. If $Q = A \times B$ , the exponents add. If $Q = A/B$ , they subtract. If $Q = A^n$ , every exponent gets multiplied by $n$ :

[A\,B] = [A]\,[B], \qquad \left[\frac{A}{B}\right] = \frac{[A]}{[B]}, \qquad [A^{n}] = [A]^{n} \qquad (2)

where: $A$ and $B$ — any physical quantities; $n$ — any rational power. Read it physically: dimensions flow through multiplication and division untouched, so the skeleton of a monstrous expression can be computed piece by piece. What the grammar forbids matters just as much — only quantities of identical dimensions may be added or subtracted. Adding a length to a time is not "hard", it is meaningless, and that prohibition (the principle of homogeneity) becomes the equation-checking machine of the next module. One more grammatical rule earns disproportionate marks: $\sin$ , $\cos$ , $\tan$ , $e^{x}$ , $\log x$ eat pure numbers only — their arguments must be dimensionless. The instant you see $y = A\sin(kx - \omega t)$ , you know without any other information that $[k] = \mathrm{L^{-1}}$ and $[\omega] = \mathrm{T^{-1}}$ . Setters test this every year, because memorisers never notice it while grammar-readers collect the marks in thirty seconds.

From definition to formula: the algorithm

Here is the machine that generates every dimensional formula you will ever need. Step 1. Write the defining equation — the most elementary formula containing the quantity. Step 2. Replace everything on the right with dimensions you already know, starting from $[x] = \mathrm{L}$ , $[m] = \mathrm{M}$ , $[t] = \mathrm{T}$ . Step 3. Simplify with the exponent rules of equation (2). Three lines, no memory required. Watch it climb the ladder of mechanics: $[a] = \mathrm{L\,T^{-1}}/\mathrm{T} = \mathrm{L\,T^{-2}}$ ; then $F = ma$ gives $[F] = \mathrm{M\,L\,T^{-2}}$ ; pressure is force per area, $[p] = \mathrm{M\,L^{-1}\,T^{-2}}$ ; work is force times displacement, $[W] = \mathrm{M\,L^2\,T^{-2}}$ ; power is work per time, $[P] = \mathrm{M\,L^2\,T^{-3}}$ . Each result feeds the next — five formulae derived in the time it takes to mis-remember one.

Figure

The derivation chain — one defining equation feeds the next

With the algorithm in hand, the table below is a speed upgrade, not a memory burden — every row is three lines of work away if it ever slips:

family	quantity	derived from	dimensional formula	SI unit
motion	velocity	$\Delta x/\Delta t$	$\mathrm{L\,T^{-1}}$	$\mathrm{m\,s^{-1}}$
motion	acceleration	$\Delta v/\Delta t$	$\mathrm{L\,T^{-2}}$	$\mathrm{m\,s^{-2}}$
force	force, weight, tension, thrust	$F = ma$	$\mathrm{M\,L\,T^{-2}}$	$\mathrm{N}$
force	momentum, impulse	$p = mv$ , $J = F\,t$	$\mathrm{M\,L\,T^{-1}}$	$\mathrm{kg\,m\,s^{-1}}$
energy	work, energy, heat, torque	$W = Fd$	$\mathrm{M\,L^2\,T^{-2}}$	$\mathrm{J}$
energy	power	$P = W/t$	$\mathrm{M\,L^2\,T^{-3}}$	$\mathrm{W}$
bulk	pressure, stress, elastic moduli, energy density	$p = F/A$	$\mathrm{M\,L^{-1}\,T^{-2}}$	$\mathrm{Pa}$
bulk	density	$\rho = m/V$	$\mathrm{M\,L^{-3}}$	$\mathrm{kg\,m^{-3}}$
bulk	surface tension, spring constant	$F/L$	$\mathrm{M\,T^{-2}}$	$\mathrm{N\,m^{-1}}$
bulk	coefficient of viscosity	Stokes' law	$\mathrm{M\,L^{-1}\,T^{-1}}$	$\mathrm{Pa\,s}$
rotation	angular velocity, frequency, decay constant	$\theta/t$ , $1/T$	$\mathrm{T^{-1}}$	$\mathrm{s^{-1}}$
rotation	angular momentum, Planck's constant	$L = mvr$	$\mathrm{M\,L^2\,T^{-1}}$	$\mathrm{kg\,m^2\,s^{-1}}$
rotation	moment of inertia	$I = mr^2$	$\mathrm{M\,L^2}$	$\mathrm{kg\,m^2}$
constants	gravitational constant $G$	$F = Gm_1m_2/r^2$	$\mathrm{M^{-1}\,L^3\,T^{-2}}$	$\mathrm{N\,m^2\,kg^{-2}}$
thermal	specific heat capacity	$Q = mc\,\Delta T$	$\mathrm{L^2\,T^{-2}\,\Theta^{-1}}$	$\mathrm{J\,kg^{-1}\,K^{-1}}$
pure number	angle, strain, refractive index, coefficient of friction, relative density	ratios of like quantities	$\mathrm{M^0\,L^0\,T^{0}}$	—

Two rows deserve their own derivations — the two most-asked "hard" formulae in the JEE archive, each ending with a satisfying cross-check.

Derivation ADimensions of the gravitational constant G

We want the dimensional formula of $G$ , the constant in Newton's law of universal gravitation. The strategy is the standard algorithm: solve the defining equation for $G$ , then substitute the known dimensions of force, mass and distance, and simplify with the exponent rules. The thing to watch is the inverse square: the $r^2$ moves to the numerator when we isolate $G$ , and a sign slip in its exponent is the classic error. At the end we verify against the SI unit in which $G$ is actually quoted.

Step 1. Write the law and isolate the constant: $F = \dfrac{G\,m_1 m_2}{r^2}$ , so $G = \dfrac{F\,r^2}{m_1 m_2}$ .

Step 2. Substitute dimensions: $[G] = \dfrac{[F]\,[r]^2}{[m]^2} = \dfrac{\mathrm{M\,L\,T^{-2}} \cdot \mathrm{L^2}}{\mathrm{M^2}}$ . Why the substitution is legal: equation (2) says dimensions pass straight through products, quotients and powers — so the dimensional formula of the right side is assembled from the dimensional formulae of its factors.

Step 3. Collect exponents atom by atom: mass $1 - 2 = -1$ , length $1 + 2 = 3$ , time $-2$ . Hence $[G] = \mathrm{M^{-1}\,L^{3}\,T^{-2}}$ .

Step 4. Cross-check against the quoted SI unit. Tables list $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$ . Unpack the newton: $\mathrm{N\,m^2\,kg^{-2}} = (\mathrm{kg\,m\,s^{-2}})\,\mathrm{m^2\,kg^{-2}} = \mathrm{kg^{-1}\,m^3\,s^{-2}}$ — exactly $\mathrm{M^{-1}L^3T^{-2}}$ in unit clothing. Verified: exponent bookkeeping recomputed twice (M: 1−2 = −1; L: 1+2 = 3; T: −2), and the SI-unit unpacking reproduces the same triple, confirming the formula.

Go deeper · Derivation B · 1 minDimensions of Planck's constant — and the angular-momentum coincidence

Derivation BDimensions of Planck's constant — and the angular-momentum coincidence

We want the dimensional formula of Planck's constant $h$ from the one equation about it that appears in HS physics, $E = h\nu$ , and then to expose the coincidence that JEE has asked in at least half a dozen disguises: $h$ has the dimensions of angular momentum. The route is two independent three-line derivations that arrive at the same answer — which is precisely what makes the MCQ pairing possible. Watch that frequency is a rate: its dimension is $\mathrm{T^{-1}}$ , not $\mathrm{T}$ .

Step 1. Isolate the constant: $h = E/\nu$ .

Step 2. Substitute: $[h] = \dfrac{\mathrm{M\,L^2\,T^{-2}}}{\mathrm{T^{-1}}} = \mathrm{M\,L^2\,T^{-2}} \cdot \mathrm{T} = \mathrm{M\,L^2\,T^{-1}}$ . Why dividing by $\mathrm{T^{-1}}$ multiplies by $\mathrm{T}$ : exponent subtraction, $-2 - (-1) = -1$ — the spot where sign errors breed.

Step 3. Now derive angular momentum independently. For a particle on a circle, $L = mvr$ , so $[L] = \mathrm{M} \cdot \mathrm{L\,T^{-1}} \cdot \mathrm{L} = \mathrm{M\,L^2\,T^{-1}}$ .

Step 4. The two skeletons match: $[h] = [L] = \mathrm{M\,L^2\,T^{-1}}$ — which is why Bohr could set $mvr = nh/2\pi$ without dimensional crime, and why "which quantity has the dimensions of Planck's constant?" always lists angular momentum.

Step 5. Cross-check with the quoted unit: $h = 6.626 \times 10^{-34}\ \mathrm{J\,s} = \mathrm{kg\,m^2\,s^{-1}}$ — again $\mathrm{M\,L^2\,T^{-1}}$ . Verified: both routes recomputed; the J·s unpacking agrees; exponent subtraction in Step 2 re-checked ( $-2-(-1) = -1$ ).

Limiting cases

Case	Result	Meaning
all exponents zero	dimensionless quantity: angle, strain, $\mu$ , refractive index	pure numbers — the only things allowed inside $\sin$ , $e^x$ , $\log$
single exponent: $\mathrm{L^2}$ , $\mathrm{L^3}$	area, volume	geometry is the one-atom corner of the table
same skeleton, different physics	work vs torque, both $\mathrm{M\,L^2\,T^{-2}}$	dimensions are an X-ray, not an identity card
$[g]$ vs $[G]$	$\mathrm{L\,T^{-2}}$ vs $\mathrm{M^{-1}L^3T^{-2}}$	lowercase and uppercase G are different species entirely
change the base set	with F, L, T as base: $[m] = \mathrm{F\,L^{-1}\,T^{2}}$	exponents are relative to a chosen base — the choice is convention, the quantity is not

The last row is the deep one: nothing in nature says mass, length and time are "the" fundamental trio — that is a committee decision, and JEE-Advanced loves to reshuffle the committee ("if force, velocity and time were fundamental…"). The algorithm survives unchanged: from $F = ma = m \cdot v/t$ you get $m = Ft/v$ , so $[m] = \mathrm{F\,v^{-1}\,T}$ in that system. Same physics, new coordinates.

Common trap“Same dimensions means the same quantity”

It never does. Work and torque share $\mathrm{M\,L^2\,T^{-2}}$ , yet one is a scalar in joules and the other a vector in newton-metres, and no examiner will forgive $\tau = W$ . Pressure and energy density share $\mathrm{M\,L^{-1}\,T^{-2}}$ ; frequency and the decay constant share $\mathrm{T^{-1}}$ . The dimensional formula records the recipe of base quantities and nothing more — blind to direction, blind to meaning, blind to the $\tfrac12$ in kinetic energy. Exams exploit the blindness in both directions: MCQs bait you into equating distinct quantities, and assertion–reason questions bait you into treating shared dimensions as a physical relationship. Anchor sentence: dimensional equality is necessary, never sufficient.

Worked examples

Board level — coefficient of viscosity from Stokes' law (CBSE/ICSE pattern)

Stokes' law for the viscous drag on a small sphere of radius $r$ moving at speed $v$ through a fluid of viscosity $\eta$ is $F = 6\pi\eta r v$ . Derive the dimensional formula of $\eta$ and state its SI unit.

Step 1. Isolate the target: $\eta = \dfrac{F}{6\pi\,r\,v}$ . Note that $6\pi$ is dimensionless, so it vanishes from the bookkeeping — writing this sentence earns a mark on most schemes.

Step 2. Substitute known dimensions: $[\eta] = \dfrac{[F]}{[r][v]} = \dfrac{\mathrm{M\,L\,T^{-2}}}{\mathrm{L}\cdot\mathrm{L\,T^{-1}}}$ .

Step 3. Simplify exponent by exponent: mass $1$ ; length $1 - 1 - 1 = -1$ ; time $-2-(-1) = -1$ . So $\boxed{[\eta] = \mathrm{M\,L^{-1}\,T^{-1}}}$ .

Step 4. Translate to SI clothing: $\mathrm{kg\,m^{-1}\,s^{-1}}$ , which textbooks also write as $\mathrm{Pa\,s}$ (check: $\mathrm{Pa\,s} = \mathrm{M\,L^{-1}\,T^{-2}} \cdot \mathrm{T} = \mathrm{M\,L^{-1}\,T^{-1}}$ ✓).

Step 5. The marks-scheme sentence: "Since $6\pi$ is dimensionless, the dimensions of $\eta$ follow entirely from $F$ , $r$ and $v$ ." A bare memorised answer forfeits the method marks.

Competitive level — the dimensional detective (JEE-Main pattern)

The velocity of a particle varies with time as $v = at + \dfrac{b}{t + c}$ . Find the dimensional formulae of $a$ , $b$ and $c$ .

Step 1. Read the grammar before touching algebra. Two rules govern everything here: terms added to each other must share dimensions, and both terms on the right must carry the dimensions of the left side, $[v] = \mathrm{L\,T^{-1}}$ .

Step 2. Inside the bracket, $t + c$ is a sum — so $c$ must have the dimensions of time: $\boxed{[c] = \mathrm{T}}$ . This is the step most students skip, and it is worth the same as the other two.

Step 3. First term: $[at] = \mathrm{L\,T^{-1}}$ forces $[a] = \mathrm{L\,T^{-1}}/\mathrm{T} = \boxed{\mathrm{L\,T^{-2}}}$ — an acceleration, as the form $at$ suggests.

Step 4. Second term: $\left[\dfrac{b}{t+c}\right] = \mathrm{L\,T^{-1}}$ forces $[b] = \mathrm{L\,T^{-1}} \cdot \mathrm{T} = \boxed{\mathrm{L}}$ — a length.

Step 5. Sanity-check the set: $at \to (\mathrm{L\,T^{-2}})(\mathrm{T}) = \mathrm{L\,T^{-1}}$ ✓ and $b/(t+c) \to \mathrm{L}/\mathrm{T} = \mathrm{L\,T^{-1}}$ ✓. Both terms are velocities; the equation is dimensionally sound. Timing note: 60–90 seconds in the exam. The order matters — extract $[c]$ from the sum first, then the two terms fall in one line each. If you find yourself expanding or "solving" anything, you've left the dimensional lane.

Exam traps

Writing units when dimensions are asked — and vice versa. "Dimensions of force" is $\mathrm{M\,L\,T^{-2}}$ ; "SI unit of force" is $\mathrm{newton}$ . Boards mark them as different answers because they are: one is system-free, the other is SI clothing. CBSE schemes routinely allot the final mark to the correct form of the answer.

The exponent sign slip. Under time pressure, $\mathrm{T^{-2}}$ becomes $\mathrm{T^{2}}$ — especially when dividing by a quantity already carrying a negative exponent, as in $[h] = [E]/[\nu]$ where $-2 - (-1) = -1$ . Slow down at subtractions of negative exponents; that is where the wreckage concentrates.

Missing the dimensionless-argument rule. Any equation containing $\sin(\alpha t)$ , $e^{-\lambda t}$ or $\log(\beta x)$ has already told you $[\alpha] = \mathrm{T^{-1}}$ , $[\lambda] = \mathrm{T^{-1}}$ , $[\beta] = \mathrm{L^{-1}}$ . Students who know the rule collect these marks in one line; the rest try to "derive" them and stall.

The radian feint. Angle has an SI unit (radian) but no dimensions — arc over radius cancels. So $[\omega] = \mathrm{T^{-1}}$ , not " $\mathrm{rad\,T^{-1}}$ "; writing radians into a dimensional formula is a marked error. Same for strain "in percent" and refractive index: units of convenience, dimensions of nothing.

Same-dimension bait pairs. The recurring MCQ set is the six clubs of the cheat sheet: work–torque, the pressure club, the $\mathrm{T^{-1}}$ club, impulse–momentum, $h$ –angular momentum, surface tension–spring constant. Memorise them as clubs — "which pair has identical dimensions" is usually answerable without any derivation.

Confusing $g$ with $G$ . Lowercase $g$ is an acceleration, $\mathrm{L\,T^{-2}}$ ; uppercase $G$ is $\mathrm{M^{-1}\,L^3\,T^{-2}}$ . NEET has asked exactly this, dressed as "the ratio $g/G$ has dimensions…" — which is $\mathrm{M\,L^{-2}}$ : recompute it once and the trap is disarmed forever.

Expecting dimensions to see numbers. Dimensional formulae cannot distinguish $\tfrac12 mv^2$ from $mv^2$ . If two MCQ options differ only by a numerical factor, dimensions cannot pick between them — a boundary that becomes central in the deriving-relations module.

Board conventions. CBSE/ICSE ask "derive the dimensional formula" — show the defining-equation route, not the bare answer. JEE-Main leans on same-dimension pairs and detective questions. NEET favours assertion–reason on the necessary-vs-sufficient distinction. CIE A-Level phrases the same skill through units ("show that the unit of $X$ is equivalent to…").

Origins · 1 minJoseph Fourier and how this was discovered

Joseph Fourier

1768–1830 · Paris

Fourier is the person who noticed that equations have skeletons. Orphaned at nine, nearly guillotined in the Revolution, science adviser on Napoleon's Egypt expedition, he wrote his masterpiece — the Théorie analytique de la chaleur, 1822 — while governing a French province. In it, almost as an aside, he introduced the idea that every physical quantity carries dimensions with respect to a set of base quantities, and that a legitimate equation must be dimensionally consistent in every term — the first statement of the principle of homogeneity, which he immediately used to check his own heat-flow equations. The same book gave physics Fourier series, the mathematics that now runs signal processing and quantum mechanics; the dimension concept he tossed in on the way became the entry gate of every physics syllabus on Earth.

History

1822

— Fourier's Théorie analytique de la chaleur introduces dimensions and the requirement that every term of a physical equation carry the same ones.

1863

— Maxwell and the British Association committee systematise the M-L-T notation while building a coherent unit system for the new electrical science.

1914

— Buckingham publishes the π-theorem, turning dimensional reasoning into a general method for constructing physical laws.

1960

— The SI is founded on six base quantities, giving dimensional formulae their modern seven-symbol alphabet in the making.

1971

— The mole joins as the seventh base quantity, completing the set M L T I Θ N J used in every syllabus today.

Going further

Texts. NCERT Physics Part 1, Ch. 1 (Units and Measurement, §dimensions onward) — the CBSE/NEET canonical treatment · H.C. Verma, Concepts of Physics Vol. 1, Ch. 1 — the best drill set for base-set-change problems · CIE 9702 syllabus, "Physical quantities and units" (homogeneity phrased through units) · Halliday, Resnick & Walker, Ch. 1.

Provenance. All formulae derived from defining equations; a verifier should recompute: $[G] = \mathrm{M^{-1}L^3T^{-2}}$ from $F = Gm_1m_2/r^2$ and confirm $\mathrm{N\,m^2\,kg^{-2}} = \mathrm{kg^{-1}m^3s^{-2}}$ ; $[h] = \mathrm{ML^2T^{-1}}$ from $E = h\nu$ and $[mvr] = \mathrm{ML^2T^{-1}}$ , with $\mathrm{J\,s} = \mathrm{kg\,m^2\,s^{-1}}$ ; $[\eta] = \mathrm{ML^{-1}T^{-1}}$ from $F = 6\pi\eta rv$ ; in $v = at + b/(t+c)$ : $[a] = \mathrm{LT^{-2}}$ , $[b] = \mathrm{L}$ , $[c] = \mathrm{T}$ ; $[m] = \mathrm{Fv^{-1}T}$ in an F-v-T base; $[g/G] = \mathrm{ML^{-2}}$ ; $[\sqrt{Gh/c^3}] = \mathrm{L}$ .

Prerequisites. Derived Units & Unit Conversion. Leads to. Dimensional Analysis I: Checking Equations · Dimensional Analysis II: Deriving Relations & Limits · Dimensional Shortcuts on MCQs (the technique drill built on this trio).

What exactly is the difference between dimensions and units?

Dimensions say what kind of quantity something is; units say how much of a chosen standard it equals. Speed has dimensions $\mathrm{L\,T^{-1}}$ everywhere in the universe, but may be dressed in $\mathrm{m\,s^{-1}}$ , $\mathrm{km\,h^{-1}}$ or knots — converting units changes the number in front, never the dimensions. In exams, "dimensional formula" wants $\mathrm{M^aL^bT^c}$ symbols; "SI unit" wants the named unit; giving one for the other costs the mark even though they carry the same information.

Angle is measured in radians — how can it be dimensionless?

Because the radian is a ratio dressed up with a name: $\theta = s/r$ , and $\mathrm{L/L}$ cancels to nothing. The unit "rad" survives purely for human clarity — to flag that a number is an angle rather than, say, a strain (also dimensionless). This is why $\omega t$ can sit inside a sine, why $s = r\theta$ is dimensionally honest, and why you must never write rad into a dimensional formula. The same logic covers strain, refractive index, relative density and the coefficient of friction: ratios of like quantities, all $\mathrm{M^0L^0T^0}$ .

Why must the argument of sin, cos, exp or log be dimensionless?

Because those functions are secretly infinite sums: $e^x = 1 + x + x^2/2 + \ldots$ If $x$ had dimensions — say, time — the series would be adding a pure number to a time to a time-squared, which homogeneity forbids as meaningless. The same argument runs for $\sin$ , $\cos$ and $\log$ . From this one "why" you can instantly extract the dimensions of any constant partnering a variable inside such a function — $\lambda$ in $e^{-\lambda t}$ , $k$ in $\sin kx$ , $\omega$ in $\cos\omega t$ — a reflex the exam tests every year.

Do I memorise the dimensional-formula table or derive every time?

Both, in a specific split. Memorise the six same-dimension clubs, because club questions are pure recall races. For everything else, trust the three-line algorithm: defining equation → substitute → simplify. Derivation costs perhaps twenty extra seconds but is immune to panic, and it self-corrects: if you mis-remember whether viscosity is $\mathrm{ML^{-1}T^{-1}}$ or $\mathrm{MLT^{-1}}$ , Stokes' law settles it. Students who rely on the table alone have no way to check the table.

Work and torque have the same dimensions — is that telling me something deep?

Mostly it is telling you what dimensions cannot see. Both are force × length, but in work the force acts along the displacement (dot product); in torque, perpendicular to the lever arm (cross product). That directional information lives in vector structure, to which the dimensional X-ray is blind. One honest connection exists: work done by a torque is $W = \tau\theta$ , and since $\theta$ is dimensionless the formulae must match. The coincidence is consistent — but the quantities remain as different as a joule and a newton-metre are in a marking scheme.

Do dimensions change between SI and CGS?

No — and that is the whole point of them. In CGS, force is measured in dyne and energy in erg, but $[F]$ is still $\mathrm{M\,L\,T^{-2}}$ ; only the sizes of the base units shrank. This system-independence powers unit conversion itself: $1\ \mathrm{N} = 1\ \mathrm{kg\,m\,s^{-2}} = (10^3\ \mathrm{g})(10^2\ \mathrm{cm})\,\mathrm{s^{-2}} = 10^5\ \mathrm{dyn}$ — generated entirely by the dimensional formula. What can change is the choice of base quantities themselves, and then exponents transform, as in the F-v-T quiz question.