Dimensional analysis — checking equations, deriving relations: Question bank

True or false — justify

A dimensionally consistent equation is guaranteed to be physically correct.

False. Consistency is necessary but not sufficient — it cannot see the missing

\frac{1}{2}

s = u t + a t^{2}

, nor tell

E = m c^{2}

from

E = 3 m c^{2}

. Passing is a green light, not a proof.

If an equation fails the dimensional check, it might still be right.

False. A failed check is decisive: if two added terms carry different dimensions the equation is definitely wrong, no exceptions.

The quantities

km/h

and

m/s

have different dimensions.

False. They differ only in unit (chosen size); both are speeds with the same dimension

[L T^{- 1}]

. Dimension is the "kind," unit is the "size."

An angle in radians has dimension

[L]

because it is arc length over radius.

False. It is length ÷ length, so the

L

's cancel: an angle is dimensionless

[M^{0} L^{0} T^{0}]

. That's exactly why

sin θ

is allowed.

The exponent

c

1 0^{c}

must be dimensionless.

True. An exponent is a pure number; raising a base to "2 metres" is meaningless, so every exponent (like every angle and log argument) has dimension

[M^{0} L^{0} T^{0}]

Dimensional analysis can determine the numerical factor

2 π

in the pendulum period.

False. Dimensionless constants are invisible to the method — it gives

T \propto ℓ / g

but

k = 2 π

must come from full theory or experiment.

Both sides of a correct equation always share the same dimension.

True. Equating means "the same kind of thing on both sides," so

[LHS] = [RHS]

is forced by homogeneity.

The equation

s = u t + \frac{1}{2} a t^{2}

can be derived by dimensional analysis.

False. It is a sum of two terms, and the method only handles single products of powers — it cannot assemble multi-term formulas.

Two quantities with the same dimension must be the same physical quantity.

False. Torque and energy both have

[M L^{2} T^{- 2}]

, yet one is a turning effect and the other a capacity to do work — same dimension, different physics.

Spot the error

v = u + a t^{2}

— this is fine, both sides are velocities."

Wrong.

[a t^{2}] = (L T^{- 2}) (T^{2}) = L

, a length, not a velocity

L T^{- 1}

. The term clashes with

u

and

v

; the correct law is

v = u + a t

F = \frac{m v ^{2}}{r ^{2}}

for circular motion — dimensionally a force."

Wrong.

[m v^{2} / r^{2}] = M (L^{2} T^{- 2}) / L^{2} = M T^{- 2}

, which is not

M L T^{- 2}

. The right form is

m v^{2} / r

(one power of

r

), giving a genuine force.

θ = ω t + \frac{1}{2} α

— angle equals angular speed times time plus half angular acceleration."

Wrong.

[α] = T^{- 2}

is not dimensionless like an angle. The missing

t^{2}

(

\frac{1}{2} α t^{2}

) restores

T^{- 2} \cdot T^{2} =

dimensionless.

E = \frac{1}{2} m v

passed my quick glance, energy is mass times a speed."

Wrong.

[m v] = M L T^{- 1}

is a momentum, not energy

M L^{2} T^{- 2}

. Kinetic energy needs

v^{2}

\frac{1}{2} m v^{2}

"In

y = A sin (k x - ω t)

k

can have any dimension."

Wrong. The argument

k x - ω t

must be dimensionless, so each piece is:

[k x]

dimensionless

\Rightarrow [k] = L^{- 1}

, and

[ω t]

dimensionless

\Rightarrow [ω] = T^{- 1}

"Since

lo g

eats any positive number,

lo g (x)

with

x

in metres is fine."

Wrong. The argument of a log must be dimensionless —

lo g

expands into powers of

x

, which can't be added unless

x

is a pure number. Physicists write

lo g (x / x_{0})

to fix this.

Why questions

Why must the arguments of

sin

cos

e^{x}

, and

ln

be dimensionless?

These functions equal infinite sums like

x - \frac{x ^{3}}{6} + \dots

; adding

x

x^{3}

is only legal if

x

is a pure number, so the whole argument must be dimensionless.

Why does homogeneity forbid adding "5 metres + 3 seconds"?

Addition merges two amounts of the same kind; there is no number that is simultaneously a length and a time, so the sum has no meaning — nature never writes such a term.

Why does dimensional analysis reveal that a pendulum's period is independent of mass?

Matching the power of

M

gives

b = 0

, because neither

ℓ

(a length) nor

g

(an acceleration) carries any mass — so no combination of them can produce a mass in the answer.

Why can a passed dimensional check still hide a sign error like

v^{2} = u^{2} - 2 a s

+ 2 a s

Both terms

u^{2}

and

2 a s

share the dimension

L^{2} T^{- 2}

whichever sign joins them; dimensions describe kind, not the arithmetic

+

-

between same-kind terms.

Why does the power-matching trick give exactly three equations in mechanics?

Because

M

L

, and

T

are three independent "axes," and homogeneity demands the powers match on each axis separately — one equation per base dimension.

Why can dimensional analysis fail when a quantity depends on four variables?

Four unknown exponents but only three matching equations (

M, L, T

) leaves the system underdetermined — one exponent stays free, so the form isn't pinned down uniquely.

Why is the drop of the pure number

2

[2 a s] = L^{2} T^{- 2}

justified?

A pure number has dimension

[M^{0} L^{0} T^{0}]

, so multiplying by it changes nothing dimensionally — the "

2

" simply cannot be seen by the method.

Edge cases

Is a dimensionless quantity's "dimension" written as nothing, or as

M^{0} L^{0} T^{0}

They mean the same thing — a pure number carries every base to the power zero,

[M^{0} L^{0} T^{0}]

, which we often just call "dimensionless."

Can two dimensionally correct equations both be wrong at once?

Yes.

s = u t + a t^{2}

and

s = u t + 3 a t^{2}

are both consistent yet both wrong (only

\frac{1}{2}

is right) — consistency filters out garbage but keeps many impostors.

If every term in an equation is dimensionless, is the check automatic?

Effectively yes for homogeneity — all terms already share

[M^{0} L^{0} T^{0}]

— but you still gain nothing about the hidden numbers, signs, or functional form.

What does it mean if the exponents you solve for come out as fractions, like

c = - \frac{1}{2}

Nothing is wrong — fractional powers are legal (they signal a square root, as in

ℓ / g

); dimensions only require the powers balance, not that they be whole numbers.

Does dimensional analysis apply to an equation that is dimensionally correct but has an extra unit like an unmatched angle offset?

It cannot catch it — an angle offset (e.g.

sin (ω t + ϕ)

with wrong

ϕ

) is dimensionless, so it passes the check while the physics may still be off.

Can the method derive a purely numerical relation with no dimensions at all (like a ratio law)?

No. With every quantity dimensionless there are no

M, L, T

powers to match, so homogeneity gives zero constraints — the trick has nothing to bite on.

Connections

Dimensional analysis — checking equations, deriving relations — the parent this bank drills.
Equations of motion (kinematics) — source of many "spot the error" cases.
Units and the SI system — where the dimension-vs-unit trap comes from.
Buckingham Pi theorem — formalises the "unknowns vs equations" limit.
Vectors — components and addition — same-kind-adds-to-same-kind, geometrically.