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NEET 2027 · Physics · Class 11 · Ch. 1 Units and Measurement

Dimensional Analysis — Checking & Deriving

The two things you can do with a dimensional formula once you have it: test an equation for consistency, and build the form of a new relation from scratch. Plus the four things you cannot do.

PRIORITY 2TOPIC 8 + 11 OF 1520 QUESTIONS4 NCERT-BASEDNCERT §1.6.1 – 1.6.2

01Concept in plain language

Once you can write a dimensional formula, you can do two useful things with it: check whether an equation could be right, and work out the shape of a relation you have never seen.

Both rest on one idea, the principle of homogeneity: quantities can only be added or subtracted if they have the same dimensions. You can add a velocity to a velocity. You cannot add a velocity to a force, or a current to a temperature. It follows that every term in a valid physical equation must carry the identical dimensional formula.

Application one: checking

Reduce every term of a proposed equation. If any term disagrees, the equation is wrong — certainly, finally, with no appeal. If all terms agree, the equation has passed a test but has not been proved. NCERT states the asymmetry precisely: if an equation fails this consistency test, it is proved wrong, but if it passes, it is not proved right.

The reason is that dimensions are blind to pure numbers. T = 2π√(l/g) and T = 5√(l/g) are equally consistent, and only one describes a real pendulum.

Application two: deriving

Suppose you believe a quantity depends on three others but you do not know how. Assume a product of powers, write the dimensional equation, and match the exponents of M, L and T one base quantity at a time. Three equations, three unknowns, and the form of the relation drops out.

This is the method NCERT uses to obtain T = k√(l/g) for the pendulum. It delivers the shape of the answer and never the constant.

Two rules that unlock most exam questions

Rule of addition: anything added or subtracted must match. Whenever you see a sum containing an unknown constant, that constant's dimensions are already determined.

Rule of special functions: whatever sits inside sin, cos, tan, log or ex is dimensionless, and so is what comes out. Angles, ratios and exponents are always pure numbers.

02The two applications

CHECKING CONSISTENCY · EVERY TERM MUST REDUCE TO THE SAME DIMENSIONx = x₀ + v₀t + ½ a t²xLx₀Lv₀tL T⁻¹ × T = L½ a t²L T⁻² × T² = LAll four reduce to L, so the equation is dimensionally consistent.Note what the ½ did: nothing. Pure numbers are invisible to this test.
Fig. 1 — Checking a kinematic equation term by term. Note that the ½ contributes nothing: pure numbers are invisible to the test.
METHOD OF DIMENSIONS · ASSUME A PRODUCT, THEN SOLVE FOR THE EXPONENTS T = k lẅ gʸ mẔ [M⁰ L⁰ T¹] = [L]ẅ [L T⁻²]ʸ [M]Ẕ = MẔ Lẅ⁺ʸ T⁻²ʸ match M z = 0 match T −2y = 1 → y = −½ match L x + y = 0 → x = ½ T = k √(l/g) z = 0 says the period does not depend on the mass at all. The method cannot supply k = 2π — only experiment or theory can.
Fig. 2 — The method of dimensions applied to the pendulum. Matching M first gives z = 0 immediately, which is the statement that the period does not depend on the mass of the bob.
WHAT THE METHOD OF DIMENSIONS CANNOT DOcannot find pure numbersT = k√(l/g) — k = 2π must come from elsewherecannot separate like quantitieswork and torque are both M L² T⁻²fails on sums of several termsx = x₀ + v₀t + ½at² cannot be derived, only checkedlimited to three unknownsonly M, L and T give equations in mechanicsWhat it can do is decisive in one direction: an equation that fails the test is certainly wrong.
Fig. 3 — The four things the method cannot do. Each has appeared as a NEET statement or assertion–reason item.

03Checking consistency

The procedure

  1. Take the equation term by term, treating each side separately.
  2. Replace every symbol by its dimensional formula and collect exponents.
  3. Discard every pure number — ½, 2π, 6, 4/3 all vanish.
  4. Compare. If all terms match, the equation is consistent. If one does not, it is wrong.

NCERT's worked check

For x = x₀ + v₀t + ½at², all four terms reduce to L: x is L, x₀ is L, v₀t is (L T⁻¹)(T) = L, and ½at² is (L T⁻²)(T²) = L. The equation is dimensionally correct.

Why checking is worth doing under exam pressure

A dimensional check takes about fifteen seconds and catches a whole class of algebra errors — a dropped square, an inverted fraction, a term multiplied where it should have been divided. It is worth running on any formula you have manipulated yourself in a long question.

It also has an advantage over checking units: you need not commit to a particular system, and you need not worry about multiples and sub-multiples.

04Deriving a relation

The procedure

  1. Assume the unknown quantity is a product of powers: X = k Aa Bb Cc, with k a dimensionless constant.
  2. Write the dimensional equation for both sides.
  3. Match the exponents of M, L and T separately. This gives three simultaneous equations.
  4. Solve, substitute back, and note that k must come from experiment or from a fuller theory.

Do M first

In most problems only one or two of the variables carry mass, so the M equation is the simplest and often fixes an exponent outright. In the pendulum problem it gives z = 0 in a single line — which is the physically interesting result that the period does not depend on the mass of the bob at all.

the method supplies column three and never column four
RelationAssumed dependenceDimensional resultConstant, from elsewhere
Simple pendulumT on l, g, mT ∝ √(l/g)k = 2π
Stokes' lawF on η, r, vF ∝ ηrvk = 6π
Speed of soundv on E, ρv ∝ √(E/ρ)k = 1
Wave on a stringv on F, μv ∝ √(F/μ)k = 1
Oscillating liquid dropT on ρ, r, ST ∝ √(ρr³/S)k found experimentally
Photon energyE on h, νE ∝ hνk = 1
Planck lengthfrom G, h, c√(Gh/c³)≈ 1.6 × 10⁻³⁵ m

05Formula sheet

The two principles
Homogeneity
every term in an equation has identical dimensions
the basis of both applications
Special functions
argument and value of sin, cos, tan, log, ex are dimensionless
so [ω] = T⁻¹ and [k] = L⁻¹ in a wave equation
Pure numbers
contribute nothing
½, 2π, 6π, 4/3 all vanish
The asymmetry
inconsistent ⇒ wrong ; consistent ⇒ not necessarily right
NCERT states this explicitly
The derivation method
Assumed form
X = k Aa Bb Cc
a product of powers, never a sum
Dimensional equation
[X] = [A]a [B]b [C]c
Solve
match M, then T, then L
three equations in three unknowns
Limit
at most three unknowns in mechanics
because only M, L and T supply equations
What is never obtained
the value of k
experiment or a fuller theory must supply it

06The four limitations

each of these has been examined directly
LimitationStandard illustration
1. Cannot find dimensionless constants. A pure number leaves no trace in a dimensional equation.T = k√(l/g) is as far as the method goes; k = 2π must come from experiment. Similarly the 6π in Stokes' law.
2. Cannot distinguish quantities with the same dimensions. The formula records exponents, not physical meaning or vector nature.Work and torque are both M L² T⁻², yet one is a scalar energy transfer and the other a vector turning effect.
3. Fails for sums, and for trigonometric, logarithmic or exponential relations. The method assumes a single product of powers.s = ut + ½at² can be checked but never derived. Nor can y = A sin(ωt − kx).
4. Limited to three unknowns. In mechanics only M, L and T supply equations, so at most three exponents can be found.A quantity depending on four others cannot be pinned down without extra physical input.

What the method does have going for it

Its one-directional power is real. A dimensionally inconsistent equation is wrong, and no amount of adjusting numerical factors can save it. That makes the check a genuinely reliable filter, and it costs almost nothing to run.

NCERT adds a second advantage: a dimensional test tells you no more and no less than a test of units, but it spares you from committing to any particular system of units and from worrying about multiples and sub-multiples.

07Exceptions and traps

Consistent does not mean correct

The most-tested idea in this section, usually as an assertion–reason item where both statements are true and the question is whether the reason explains the assertion.

An illegal addition is an instant disqualification

K = ½mv² + ma can be eliminated at a glance, because energy and force have been added. Scan every option for a sum of unlike terms before computing anything.

Check the numerator too

In the relativistic mass question, m and m₀ are both masses, so the prefactor must be exactly 1. That eliminates one option before you touch the bracket.

The method gives form, not formula

Deriving T ∝ √(l/g) is not the same as deriving T = 2π√(l/g). If an option includes a numerical constant, ask whether dimensional analysis could possibly have supplied it.

Sanity-check the direction of a derived dependence

After solving for exponents, ask whether the physics agrees. A larger surface tension should make a drop oscillate faster, so S must sit in the denominator of the period. This catches inverted answers, which are the commonest distractor.

Dimensional constants are not pure numbers

G, h, k, σ and R all carry dimensions. Only things like 2π, ½ and counted integers are dimensionless. Treating G as a pure number breaks every gravitational check.

08Standard results to remember

recognising the form lets you eliminate three options without any algebra
RelationForm from dimensional analysisConstant
Simple pendulumT = k√(l/g)k = 2π
Stokes' lawF = kηrvk = 6π
Terminal velocityv = k r²(ρ−σ)g/ηk = 2/9
Speed of sound in a solidv = √(Y/ρ)—
Speed of sound in a gasv = √(γP/ρ)—
Transverse wave on a stringv = √(F/μ)—
Frequency of a stretched stringn = (1/2l)√(F/μ)—
Oscillating liquid dropT ∝ √(ρr³/S)found experimentally
Escape velocityv = √(2GM/R)the 2 from energy conservation
Photon energyE = hνk = 1
Planck length√(Gh/c³) ≈ 1.6 × 10⁻³⁵ m—
Planck time√(Gh/c⁵) ≈ 5.4 × 10⁻⁴⁴ s—
Planck mass√(hc/G) ≈ 2.2 × 10⁻⁸ kg—
Speed of light from μ₀, ε₀c = 1/√(μ₀ε₀)—

09Scientists to remember

high-frequency names in statement and matching items
NameWhat to attach to the name
Joseph FourierStated the principle of dimensional homogeneity in his 1822 treatise on heat. The founding figure of the subject.
Lord RayleighTurned homogeneity into a working derivation technique. The 'method of dimensions' used in NCERT Example 1.5 is Rayleigh's method.
Edgar BuckinghamThe π-theorem (1914), which says how many independent dimensionless groups a problem has. Named in passing, not examined numerically.
George Gabriel StokesStokes' law F = 6πηrv. Dimensional analysis gives ηrv; Stokes supplied the 6π from hydrodynamics.
Max PlanckConstructed the Planck length, time and mass from G, h and c alone — the purest illustration of the method's reach.
Albert EinsteinThe relativistic mass relation of NCERT Exercise 1.13, where homogeneity alone tells you where c belongs.
Daniel BernoulliBernoulli's equation, in which every term must be a pressure — a standard homogeneity illustration.

10Twenty worked questions

About the labels

Items tagged PYQ here are NCERT worked examples and chapter-end exercises, which NTA has drawn on repeatedly. Every dimensional claim on this page was checked by symbolic exponent algebra before printing.

Q01PYQ · NCERT Example 1.4Ruling out formulae

The SI unit of energy is J = kg m² s⁻². Of the formulae for kinetic energy below, which one cannot be ruled out on the basis of dimensional arguments? (a) K = m²v³ · (b) K = ½mv² · (c) K = ma · (d) K = ½mv² + ma

  1. K = ½mv²
  2. K = m²v³
  3. K = ma
  4. K = ½mv² + ma
Show full solution
Given
[K] = M L² T⁻²; four candidate formulae.
Asked
Which survives the dimensional test.
Concept
Test each candidate against M L² T⁻². Also check that no candidate adds two quantities of different dimensions, which is illegal regardless of the total.
Formula
[K] = M L² T⁻²
Baby steps
  1. m²v³ = M²(L T⁻¹)³ = M² L³ T⁻³. Ruled out.
  2. ½mv² = M(L T⁻¹)² = M L² T⁻². Survives. ✓
  3. ma = M(L T⁻²) = M L T⁻². Ruled out.
  4. ½mv² + ma adds M L² T⁻² to M L T⁻² — illegal, so ruled out.
  5. Only K = ½mv² survives among the options offered.
Answer
(a) K = ½mv²
Why not others
(a)Correct.
(b)M² L³ T⁻³ — wrong in all three exponents.
(c)M L T⁻² is force, not energy.
(d)Two different dimensions have been added, which is never permitted.
Shortcut
Look for illegal additions first. A sum of unlike terms can be eliminated at a glance, without computing anything.
Where it goes wrong
NCERT's original list also contains K = (3/16)mv², which is equally consistent dimensionally. Dimensions cannot choose between ½ and 3/16 — only the actual definition of kinetic energy can. That is the point of the example.
Q02PYQ · NCERT Example 1.3Checking

Consider the equation ½mv² = mgh, where m is mass, v velocity, g the acceleration due to gravity and h a height. Dimensionally, this equation is:

  1. incorrect, because the ½ is dimensionless
  2. incorrect, because the left side has M and the right has M²
  3. incorrect, because heights and velocities cannot be equated
  4. correct, because both sides reduce to M L² T⁻²
Show full solution
Given
½mv² = mgh.
Asked
Whether the equation is dimensionally correct, and why.
Concept
Reduce each side separately and compare. Numerical factors are discarded.
Formula
LHS = M(L T⁻¹)² ; RHS = M(L T⁻²)(L)
Baby steps
  1. LHS: [M][L T⁻¹]² = M L² T⁻².
  2. RHS: [M][L T⁻²][L] = M L² T⁻².
  3. Both sides agree.
  4. So the equation is dimensionally correct.
Answer
(d) correct, because both sides reduce to M L² T⁻²
Why not others
(a)The ½ being dimensionless is exactly why it does not matter.
(b)Both sides carry M to the first power.
(c)It is not h and v that are equated but the two full expressions.
(d)Correct — the NCERT worked example.
Shortcut
Reduce the two sides independently and only then compare. Trying to cancel symbols across the equals sign before reducing invites mistakes.
Where it goes wrong
Dimensional correctness does not prove the equation is right. NCERT is explicit: a dimensionally consistent equation need not be exact, but a dimensionally wrong one must be wrong.
Q03PYQ · NCERT Example 1.5Deriving

The period T of a simple pendulum is assumed to depend on its length l, the mass m of the bob and g, as T = k lẅ gʸ mẔ. Solving the dimensional equations gives:

  1. x = ½, y = ½, z = 0
  2. x = 1, y = −1, z = 0
  3. x = ½, y = −½, z = 0
  4. x = ½, y = −½, z = ½
Show full solution
Given
T = k lẅ gʸ mẔ, with k dimensionless.
Asked
The values of the exponents x, y and z.
Concept
Write the dimensional equation, then match the exponents of M, L and T one base quantity at a time. Three equations, three unknowns.
Formula
[M⁰ L⁰ T¹] = [L]ẅ [L T⁻²]ʸ [M]Ẕ
Baby steps
  1. RHS = Lẅ⁺ʸ T⁻²ʸ MẔ.
  2. Match M: z = 0 — the period does not depend on the mass at all.
  3. Match T: −2y = 1 → y = −½.
  4. Match L: x + y = 0 → x = +½.
  5. So T = k l½ g−½ = k√(l/g), and experiment supplies k = 2π.
Answer
(c) x = ½, y = −½, z = 0
Why not others
(a)y has the wrong sign; g must appear in the denominator.
(b)These exponents would give T ∝ l/g, whose dimensions are T², not T.
(c)Correct.
(d)z must be zero — nothing in the dimensional equation allows a mass dependence.
Shortcut
Always match M first when one of the variables carries mass alone. It usually collapses to zero immediately and removes one unknown for free.
Where it goes wrong
The method delivers the form but never the constant. k = 2π cannot be obtained dimensionally, and NEET has tested that limitation directly.
Q04Checking

Which of the following equations is dimensionally incorrect?

  1. v² = u² + 2as
  2. s = ut + ½at
  3. v = u + at
  4. s = ut + ½at²
Show full solution
Given
Four kinematic relations.
Asked
The dimensionally inconsistent one.
Concept
Reduce every term of every equation. The moment one term disagrees with the others, that equation is wrong.
Formula
check each term separately against the dimension of the left side
Baby steps
  1. v² = u² + 2as: L²T⁻² = L²T⁻² + (LT⁻²)(L) = L²T⁻². Consistent.
  2. s = ut + ½at: L = (LT⁻¹)(T) + (LT⁻²)(T) = L + LT⁻¹. The second term is a velocity, not a length. Inconsistent.
  3. v = u + at: LT⁻¹ = LT⁻¹ + (LT⁻²)(T) = LT⁻¹. Consistent.
  4. s = ut + ½at²: L = L + (LT⁻²)(T²) = L. Consistent.
Answer
(b) s = ut + ½at
Why not others
(a)All three terms reduce to L²T⁻².
(b)Correct — the last term should carry t², not t.
(c)Both terms reduce to a velocity.
(d)This is the correct form of option (b).
Shortcut
Scan for the term that looks like a familiar formula with a power altered. Papers generate wrong options by changing exactly one exponent in a correct equation.
Where it goes wrong
Getting the right answer for the wrong reason. The ½ is not the problem in option (b); the missing square on t is.
Q05Deriving

The viscous force on a small sphere of radius r moving with velocity v through a fluid of viscosity η is assumed to be F = k ηa rb vc. Dimensional analysis gives:

  1. a = 1, b = 1, c = 1
  2. a = 1, b = 2, c = 1
  3. a = 2, b = 1, c = 1
  4. a = 1, b = 1, c = 2
Show full solution
Given
F = k ηa rb vc; [η] = M L⁻¹ T⁻¹.
Asked
The exponents a, b and c.
Concept
Only η carries mass, so matching M fixes a immediately. The rest follows in two more lines.
Formula
[F] = M L T⁻² = (M L⁻¹ T⁻¹)a Lb (L T⁻¹)c
Baby steps
  1. RHS = Ma L−a+b+c T−a−c.
  2. Match M: a = 1.
  3. Match T: −a − c = −2 → c = 1.
  4. Match L: −a + b + c = 1 → −1 + b + 1 = 1 → b = 1.
  5. So F = k η r v — and experiment gives k = 6π, which is Stokes' law.
Answer
(a) a = 1, b = 1, c = 1
Why not others
(a)Correct.
(b)b = 2 would make the force depend on the cross-sectional area, which is not what the exponents give.
(c)a = 2 contradicts the M equation, since only η carries mass.
(d)c = 2 contradicts the T equation.
Shortcut
Whenever exactly one variable carries M, that exponent is fixed in one step. Do M first, then T, then L — in that order the system usually solves without simultaneous equations.
Where it goes wrong
Dimensional analysis gives F ∝ ηrv but never the 6π. Stokes derived that factor from the full hydrodynamic equations.
Q06Checking

The speed of sound in a medium of bulk modulus E and density ρ is proposed as v = √(E/ρ). Dimensionally this expression is:

  1. inconsistent — it gives L² T⁻²
  2. inconsistent — it gives T L⁻¹
  3. consistent — it gives L T⁻¹
  4. inconsistent — it gives M L T⁻¹
Show full solution
Given
[E] = M L⁻¹ T⁻² (a modulus, i.e. pressure dimensions); [ρ] = M L⁻³.
Asked
Whether √(E/ρ) has the dimensions of a speed.
Concept
Divide first, then take the square root. Expect the mass to cancel — a speed cannot contain M.
Formula
v = √(E/ρ)
Baby steps
  1. E/ρ = (M L⁻¹ T⁻²) / (M L⁻³).
  2. M cancels; L: −1 − (−3) = +2; T: −2.
  3. E/ρ = L² T⁻², which is (speed)².
  4. √(E/ρ) = L T⁻¹. Consistent with a speed.
Answer
(c) consistent — it gives L T⁻¹
Why not others
(a)That is E/ρ before the square root is taken.
(b)That is the reciprocal of a speed.
(c)Correct — the same structure appears in v = √(γP/ρ) for a gas and v = √(T/μ) for a string.
(d)M cancels completely; no mass can survive in a speed.
Shortcut
A whole family shares this shape: √(elastic property ÷ inertial property) is always a speed. Bulk modulus over density, tension over linear density, γP over density — all the same idea.
Where it goes wrong
Forgetting the square root and answering L² T⁻². Option (a) is there for exactly that.
Q07Checking

The terminal velocity of a sphere falling through a viscous fluid is written as v = 2r²(ρ − σ)g / 9η. This expression is:

  1. dimensionally incorrect
  2. dimensionally correct only if σ = 0
  3. dimensionally correct
  4. dimensionally incorrect because ρ and σ are subtracted
Show full solution
Given
r radius, ρ and σ densities, g acceleration due to gravity, η viscosity.
Asked
Whether the expression is dimensionally consistent with a velocity.
Concept
Substitute and reduce. Note that (ρ − σ) is a legal subtraction because both are densities, and the result still has the dimensions of a density.
Formula
v = 2r²(ρ − σ)g / 9η
Baby steps
  1. Numerator: L² × (M L⁻³) × (L T⁻²) = M L⁰ T⁻².
  2. Denominator: [η] = M L⁻¹ T⁻¹.
  3. Divide: (M L⁰ T⁻²) / (M L⁻¹ T⁻¹).
  4. M cancels; L: 0 − (−1) = 1; T: −2 − (−1) = −1.
  5. Result = L T⁻¹, a velocity. Consistent.
Answer
(c) dimensionally correct
Why not others
(a)It reduces exactly to L T⁻¹.
(b)σ is a density like ρ, so the subtraction is legal for any value of σ.
(c)Correct.
(d)Subtracting two densities is perfectly legal — they have identical dimensions.
Shortcut
A subtraction inside a formula is a free consistency check: if the two quantities being subtracted did not match, the formula would already be wrong.
Where it goes wrong
Discarding the 2/9 or worrying about it. Pure numbers never enter a dimensional check.
Q08Checking

The escape velocity from a planet of mass M and radius R is given as v = √(2GM/R). Checking dimensionally, this expression:

  1. is incorrect; it gives L² T⁻²
  2. is correct; it gives L T⁻¹
  3. is incorrect; it gives M L T⁻¹
  4. is correct only if G is dimensionless
Show full solution
Given
[G] = M⁻¹ L³ T⁻².
Asked
Whether √(2GM/R) has the dimensions of a velocity.
Concept
G is the only awkward term; substitute its dimensional formula and let the M cancel against the planet's mass.
Formula
v = √(2GM/R)
Baby steps
  1. GM = (M⁻¹ L³ T⁻²)(M) = L³ T⁻².
  2. GM/R = L³ T⁻² / L = L² T⁻².
  3. Take the square root: L T⁻¹.
  4. Consistent with a velocity. The 2 contributes nothing.
Answer
(b) is correct; it gives L T⁻¹
Why not others
(a)That is GM/R before the square root.
(b)Correct.
(c)The two masses cancel exactly.
(d)G has definite dimensions, M⁻¹ L³ T⁻², and the expression works precisely because of them.
Shortcut
Any gravitational expression containing GM will have that mass cancelled by the M⁻¹ inside G. Expect GM to reduce to L³ T⁻² every time.
Where it goes wrong
Treating G as a pure number because it is called a constant. Dimensional constants such as G, h and k carry dimensions; only pure numbers such as 2π do not.
Q09PYQ · NCERT Ex. 1.13Placing a constant

A boy recalls the relativistic mass relation almost correctly but forgets where the constant c belongs, writing m = m₀ / (1 − v²)1/2. The correct placement is:

  1. m = m₀ / (1 − v²/c)1/2
  2. m = m₀ / (1 − v/c²)1/2
  3. m = m₀c / (1 − v²)1/2
  4. m = m₀ / (1 − v²/c²)1/2
Show full solution
Given
m = m₀ / (1 − v²)1/2, with c missing.
Asked
Where the constant c belongs.
Concept
The 1 in the bracket is a pure number, so whatever is subtracted from it must also be dimensionless. That single requirement fixes the placement uniquely.
Formula
1 − X must be dimensionless → [X] = M⁰ L⁰ T⁰
Baby steps
  1. v² has dimensions L² T⁻² — not dimensionless.
  2. To cancel them, divide by something with the same dimensions.
  3. c² = (L T⁻¹)² = L² T⁻². So v²/c² is dimensionless. ✓
  4. Hence m = m₀/(1 − v²/c²)1/2.
Answer
(d) m = m₀ / (1 − v²/c²)1/2
Why not others
(a)v²/c gives L T⁻¹, still not dimensionless.
(b)v/c² gives L⁻¹ T, not dimensionless.
(c)This makes m have the dimensions of momentum per unit... in any case m and m₀ must have identical dimensions, so no stray c may multiply the numerator.
(d)Correct — the standard Lorentz factor.
Shortcut
Wherever a pure number appears in a sum, everything added to or subtracted from it must be dimensionless. This is the fastest route through every 'where does the constant go' question.
Where it goes wrong
Also check the numerator. Since m and m₀ are both masses, the prefactor must be exactly 1 — which eliminates option (c) without any work on the bracket.
Q10Limitations

The method of dimensions cannot be used to determine:

  1. the dimensional formula of a derived quantity
  2. whether a given equation is dimensionally consistent
  3. the conversion factor between two systems of units
  4. the value of a dimensionless constant in a formula
Show full solution
Given
Four possible uses of dimensional analysis.
Asked
The one it cannot do.
Concept
Dimensions record only the exponents of the base quantities. A pure number has no exponents at all, so it is invisible to the method.
Formula
T = k√(l/g) — the method gives the form, never k
Baby steps
  1. Finding a dimensional formula: yes, from the defining equation.
  2. Checking consistency: yes, that is the principle of homogeneity.
  3. Converting between systems: yes, via n₁u₁ = n₂u₂.
  4. Finding a dimensionless constant: no — 2π and 6π and 4/3 leave no trace in a dimensional equation.
Answer
(d) the value of a dimensionless constant in a formula
Why not others
(a)This is a standard use of the method.
(b)This is the principle of homogeneity.
(c)This is the conversion application.
(d)Correct.
Shortcut
The four limitations, worth learning as a list: no dimensionless constants; cannot distinguish quantities of the same dimension; fails for equations that are sums of several terms or contain trigonometric, logarithmic or exponential functions; and works only when the quantity depends on at most three others.
Where it goes wrong
Note the third limitation carefully: the method can check s = ut + ½at², but it can never derive it, because a product-type assumption cannot produce a sum of terms.
Q11Limitations

Which of the following statements is correct?

  1. A dimensionally correct equation is always physically correct
  2. A dimensionally incorrect equation may still be physically correct
  3. A dimensionally correct equation need not be physically correct, but a dimensionally incorrect one must be wrong
  4. Dimensional analysis can distinguish between work and torque
Show full solution
Given
Four statements about the power of the dimensional test.
Asked
The correct one.
Concept
The test is one-directional. Passing it is necessary but not sufficient; failing it is conclusive.
Formula
consistent ⇒ possibly right ; inconsistent ⇒ certainly wrong
Baby steps
  1. T = 2π√(l/g) and T = 5√(l/g) are both dimensionally consistent, yet only one is correct. So consistency does not guarantee correctness.
  2. If any term disagrees, no choice of numerical factors can repair it.
  3. So inconsistency is conclusive proof of error.
  4. NCERT states this asymmetry in exactly these words: if an equation fails this consistency test it is proved wrong, but if it passes, it is not proved right.
Answer
(c) A dimensionally correct equation need not be physically correct, but a dimensionally incorrect one must be wrong
Why not others
(a)Numerical factors are invisible to the test, so correctness cannot be guaranteed.
(b)An incorrect equation cannot be rescued; dimensional failure is decisive.
(c)Correct.
(d)Work and torque share M L² T⁻²; the method cannot separate them.
Shortcut
One sentence covers this whole section: dimensional analysis can prove an equation wrong but never prove it right.
Where it goes wrong
This asymmetry appears constantly as an assertion–reason item. Both parts are usually true; the question is whether the reason explains the assertion.
Q12Deriving

The time period of oscillation of a liquid drop depends on its density ρ, radius r and surface tension S. Dimensional analysis gives T proportional to:

  1. √(ρr/S)
  2. √(ρr³/S)
  3. √(S/ρr³)
  4. √(ρr²/S)
Show full solution
Given
T = k ρa rb Sc; [ρ] = M L⁻³, [S] = M T⁻².
Asked
The dependence of T on ρ, r and S.
Concept
Standard exponent matching. Since both ρ and S carry M, the M equation links a and c rather than fixing one outright.
Formula
[T] = (M L⁻³)a Lb (M T⁻²)c
Baby steps
  1. RHS = Ma+c L−3a+b T−2c.
  2. Match T: −2c = 1 → c = −½.
  3. Match M: a + c = 0 → a = +½.
  4. Match L: −3a + b = 0 → b = 3a = 3/2.
  5. So T ∝ ρ½ r3/2 S−½ = √(ρr³/S).
Answer
(b) √(ρr³/S)
Why not others
(a)b would have to be ½, which contradicts the L equation.
(b)Correct.
(c)This is the reciprocal — the signs of all three exponents are inverted.
(d)b = 1 does not satisfy −3a + b = 0 with a = ½.
Shortcut
Once the half-powers appear, read the result as a square root of a single fraction. ρ½ r3/2 S−½ is √(ρr³/S) — all three exponents are half of 1, 3 and −1.
Where it goes wrong
Option (c) is the same expression upside down. After solving, check one exponent against the physics: a bigger surface tension should pull the drop back faster, so a larger S must give a smaller T. S therefore belongs in the denominator.
Q13Checking

The frequency n of a stretched string depends on its length l, tension F and mass per unit length μ. The relation n = (1/2l)√(F/μ) is:

  1. dimensionally correct
  2. dimensionally incorrect; it gives T⁻²
  3. dimensionally incorrect; it gives L T⁻¹
  4. correct only in SI units
Show full solution
Given
[F] = M L T⁻²; [μ] = M L⁻¹.
Asked
Whether the expression has the dimensions of a frequency.
Concept
Reduce F/μ first, take the root, then divide by the length.
Formula
n = (1/2l)√(F/μ)
Baby steps
  1. F/μ = (M L T⁻²)/(M L⁻¹) = L² T⁻².
  2. √(F/μ) = L T⁻¹ — a speed, which is the wave speed on the string.
  3. Divide by l: (L T⁻¹)/L = T⁻¹.
  4. That is a frequency, so the relation is dimensionally correct.
Answer
(a) dimensionally correct
Why not others
(a)Correct.
(b)The division by l removes exactly one L, leaving T⁻¹, not T⁻².
(c)That is √(F/μ) before the division by l.
(d)Dimensional consistency is independent of the system of units — that is one of its advantages.
Shortcut
Read the structure: √(F/μ) is a speed, and speed divided by length is a frequency. Recognising the sub-expressions is faster than grinding out all the exponents.
Where it goes wrong
Dimensional consistency holds in every system of units. NCERT notes this as an advantage of dimensions over units — you need not commit to any particular choice.
Q14Checking

In the photoelectric relation E = hν, checking dimensions confirms that h has the dimensions of:

  1. energy
  2. power
  3. force × length
  4. angular momentum
Show full solution
Given
E = hν, with ν a frequency.
Asked
What h shares its dimensions with.
Concept
Rearranging gives h = E/ν, i.e. energy × time, which is the dimension of action.
Formula
h = E/ν = (M L² T⁻²)/(T⁻¹)
Baby steps
  1. [h] = M L² T⁻² × T = M L² T⁻¹.
  2. Angular momentum = mvr = M × L T⁻¹ × L = M L² T⁻¹. Match. ✓
  3. Energy is M L² T⁻² — one power of T away.
  4. So h has the dimensions of angular momentum, also called action.
Answer
(d) angular momentum
Why not others
(a)Energy is M L² T⁻²; h carries an extra time.
(b)Power is M L² T⁻³.
(c)Force × length is M L² T⁻², i.e. energy again.
(d)Correct — which is why Bohr's condition mvr = nh/2π is dimensionally sound.
Shortcut
Whenever a constant appears in a relation, make it the subject and read off its dimensions. Every dimensional constant in physics can be recovered this way.
Where it goes wrong
The unit J s makes h look like an energy. J s is energy × time, so the T exponent is −1, not −2.
Q15Limitations

Dimensional analysis cannot be used to derive which of the following relations?

  1. T = 2π√(l/g)
  2. s = ut + ½at²
  3. F = 6πηrv
  4. v = √(E/ρ)
Show full solution
Given
Four physical relations.
Asked
Which one the method cannot produce.
Concept
The method assumes a single product of powers. It can never generate a relation that is a sum of two or more terms.
Formula
method assumes X = k Aẅ Bʸ CẔ — a product, never a sum
Baby steps
  1. T = 2π√(l/g) is a single product of powers — derivable in form.
  2. F = 6πηrv is a single product — derivable in form.
  3. v = √(E/ρ) is a single product — derivable in form.
  4. s = ut + ½at² is a sum of two terms. No product assumption can produce it, so the method can only check it, never derive it.
Answer
(b) s = ut + ½at²
Why not others
(a)Derivable in form; only the 2π must come from elsewhere.
(b)Correct.
(c)Derivable in form; only the 6π must come from elsewhere.
(d)Derivable in form.
Shortcut
Ask one question: is the target a single product of powers? If yes, the method may derive its form. If it is a sum, the method can only check it.
Where it goes wrong
In all three derivable cases the method still fails to supply the numerical constant. 'Derivable' here means the form, not the complete formula.
Q16Deriving

A quantity is formed as √(Gh/c³), where G is the gravitational constant, h Planck's constant and c the speed of light. Its dimensions are those of:

  1. time
  2. length
  3. mass
  4. energy
Show full solution
Given
[G] = M⁻¹ L³ T⁻², [h] = M L² T⁻¹, [c] = L T⁻¹.
Asked
The dimensions of √(Gh/c³).
Concept
Multiply and divide the dimensional formulae, then halve every exponent for the square root. Expect M to cancel between G and h.
Formula
√(Gh/c³)
Baby steps
  1. Gh = (M⁻¹ L³ T⁻²)(M L² T⁻¹) = L⁵ T⁻³ — M cancels.
  2. c³ = L³ T⁻³.
  3. Gh/c³ = L⁵⁻³ T⁻³⁺³ = L².
  4. Square root: L — a length. This is the Planck length, about 1.6 × 10⁻³⁵ m.
Answer
(b) length
Why not others
(a)The Planck time is √(Gh/c⁵), with c to the fifth power.
(b)Correct.
(c)The Planck mass is √(hc/G).
(d)The Planck energy is the Planck mass times c².
Shortcut
Set the exponents out in a column — M, L and T on separate lines — and add them. Combining three dimensional formulae in one line is where errors appear.
Where it goes wrong
The three Planck quantities differ only in the power of c: length uses c³, time uses c⁵, and mass inverts G. Options are usually built from that same family.
Q17Checking

In the equation y = A sin(ωt − kx + φ), which of the following must be dimensionless?

  1. ωt, kx and φ
  2. A and y
  3. ω and k
  4. A, ω and k
Show full solution
Given
y = A sin(ωt − kx + φ).
Asked
Which quantities must be dimensionless.
Concept
Two rules apply at once: the argument of a sine must be dimensionless, and every term added inside that argument must match the others — that is, all must be dimensionless.
Formula
argument of sin, cos, tan, log, ex → always dimensionless
Baby steps
  1. The whole bracket is the argument of a sine, so it must be dimensionless.
  2. Since the three terms are added, each must separately be dimensionless.
  3. So ωt, kx and φ are all dimensionless — giving [ω] = T⁻¹ and [k] = L⁻¹.
  4. A and y are not dimensionless; they share the dimensions of the displacement, usually L.
Answer
(a) ωt, kx and φ
Why not others
(a)Correct.
(b)A and y have the dimensions of the displacement being described.
(c)ω is T⁻¹ and k is L⁻¹ — both carry dimensions.
(d)A carries the dimensions of y.
Shortcut
One rule solves an entire question family: whatever sits inside sin, cos, tan, log or an exponential is dimensionless, and so is the value that comes out.
Where it goes wrong
The output of the sine is dimensionless too, which is why A must carry all of y's dimensions. If y is a length, A is a length.
Q18Checking

In Bernoulli's equation P + ½ρv² + ρgh = constant, the terms ½ρv² and ρgh have the dimensions of:

  1. energy
  2. force
  3. power
  4. pressure
Show full solution
Given
Bernoulli's equation.
Asked
The dimensions of the second and third terms.
Concept
By homogeneity, every term must match the first, which is a pressure. Verify at least one by direct substitution.
Formula
all terms in a sum must share dimensions
Baby steps
  1. [P] = M L⁻¹ T⁻².
  2. ½ρv² = (M L⁻³)(L T⁻¹)² = M L⁻¹ T⁻². ✓
  3. ρgh = (M L⁻³)(L T⁻²)(L) = M L⁻¹ T⁻². ✓
  4. All three terms are pressures — and equally, energy densities, since J m⁻³ = Pa.
Answer
(d) pressure
Why not others
(a)Energy is M L² T⁻² — these terms are energy per unit volume.
(b)Force is M L T⁻².
(c)Power is M L² T⁻³.
(d)Correct.
Shortcut
When one term of a sum is recognisable, the rest are settled by homogeneity. You need check only one to be confident about all.
Where it goes wrong
Calling these terms 'energy' is half right and fully wrong for the exam: they are energy per unit volume, which is dimensionally a pressure.
Q19Deriving

The energy E of a photon is assumed to depend on Planck's constant h and the frequency ν as E = k haνb. Dimensional analysis gives:

  1. a = 2, b = 1
  2. a = 1, b = 2
  3. a = 1, b = 1
  4. a = ½, b = ½
Show full solution
Given
[E] = M L² T⁻², [h] = M L² T⁻¹, [ν] = T⁻¹.
Asked
The exponents a and b.
Concept
Only h carries M and L, so those two equations fix a immediately; the T equation then gives b.
Formula
M L² T⁻² = (M L² T⁻¹)a (T⁻¹)b
Baby steps
  1. Match M: a = 1.
  2. Match L: 2a = 2 → a = 1, consistent. ✓
  3. Match T: −a − b = −2 → −1 − b = −2 → b = 1.
  4. So E = k hν, and the physics gives k = 1: E = hν.
Answer
(c) a = 1, b = 1
Why not others
(a)a = 2 contradicts the M equation.
(b)b = 2 contradicts the T equation.
(c)Correct.
(d)Half-powers contradict both the M and L equations.
Shortcut
Two independent equations giving the same value for a is a built-in check. If M and L disagree about a, you have substituted something wrong.
Where it goes wrong
Here k happens to equal 1, so dimensional analysis produces the complete formula. That is a coincidence, not a rule — for the pendulum k is 2π and for Stokes' law it is 6π.
Q20Limitations

Two physical quantities have the same dimensional formula M L² T⁻². Dimensional analysis can tell you:

  1. nothing that distinguishes them
  2. that one is a scalar and the other a vector
  3. that they are the same physical quantity
  4. which of the two is measured in joules
Show full solution
Given
Two quantities sharing M L² T⁻².
Asked
What the method can say about the difference between them.
Concept
Dimensions record only the exponents of the base quantities. Vector or scalar nature, physical meaning and unit names lie entirely outside what a dimensional formula encodes.
Formula
[work] = [torque] = [heat] = M L² T⁻²
Baby steps
  1. Work, torque, heat, energy and the work function all share M L² T⁻².
  2. Work is a scalar; torque is a vector. Dimensions cannot tell them apart.
  3. Work is measured in joules, torque in newton metres, though the two are dimensionally identical.
  4. So the method can say nothing that distinguishes them — this is one of its four standard limitations.
Answer
(a) nothing that distinguishes them
Why not others
(a)Correct.
(b)Vector nature is not encoded in a dimensional formula at all.
(c)Sharing a dimensional formula does not make two quantities the same.
(d)The unit name is a convention chosen to keep the two apart; dimensions cannot supply it.
Shortcut
Whenever a question asks what dimensional analysis cannot do, work and torque is the standard illustration. Keep that example ready.
Where it goes wrong
The distinction between 'same dimensions' and 'same quantity' is tested constantly. Read the verb in the question: 'have the same dimensions' is true, 'are the same quantity' is false.