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

Dimensional Formulae

The single highest-yield topic in the chapter. Dimensional formulae and SI-unit identification together account for the majority of everything NEET has ever asked here.

PRIORITY 1TOPIC 1 + 6 OF 1520 QUESTIONS13 PYQ-BASEDNCERT §1.4 – 1.5

01Concept in plain language

A dimension is a recipe, not a number. It tells you which of the seven base quantities went into making a physical quantity, and in what proportion — and nothing else.

Every quantity in physics is assembled from seven base ingredients: mass M, length L, time T, electric current A, temperature K, amount of substance mol and luminous intensity cd. Writing the dimensional formula of a quantity means writing down how many of each ingredient it contains.

Speed, for example, is length divided by time. So its recipe is one length, minus one time: [v] = M⁰ L¹ T⁻¹. The zero on M is not decoration — it is the statement that mass plays no part in speed at all.

The three-move method

You will never need to memorise a dimensional formula if you can do these three moves:

  1. Write the defining equation of the quantity. Not a random formula it appears in — the equation that defines it. Pressure is force per area. Viscosity is force per (area × velocity gradient).
  2. Make the quantity the subject by ordinary algebra, before you touch dimensions.
  3. Replace every symbol by its dimension and collect exponents. Numbers such as ½, 2π or 4 simply vanish — magnitude never enters.

What dimensions deliberately throw away

Dimensions record the kind of quantity, never the amount, never the direction and never the context. That is why initial velocity, final velocity, average velocity, change in velocity and speed all share the identical formula L T⁻¹. It is also why work and torque are dimensionally indistinguishable even though one is a scalar and the other a vector, and one is energy while the other is a turning effect. Dimensional analysis cannot tell them apart, and NEET tests exactly that blind spot.

The rule that unlocks half the hard questions

Anything that is added or subtracted must have identical dimensions, and anything sitting inside sin, cos, tan, log or ex must be dimensionless — as must the result of those functions. Whenever a question hands you an unfamiliar expression with unknown constants, look for an addition sign or a special function first. That is where the free information is.

02How a dimension is built

STEP 1 · WRITE THE DEFINING FORMULA F = m × a STEP 2 · REPLACE EACH SYMBOL BY ITS DIMENSION [M] × [L]/[T]² M 1 L 1 T −2 the exponent is the whole answer — magnitudes never enter [F] = M L T−²
Fig. 1 — The three-move method applied to force. Note that the coefficient in F = ma is 1 and would have been discarded even if it were 4π.
BUILD THE CHAIN — NEVER MEMORISE ISOLATED FORMULAEvelocityL T−¹accelerationL T−²forceM L T−²workM L² T−²powerM L² T−³×1/T → ×M → ×L → ÷T — four small moves cover most of mechanics.
Fig. 2 — Mechanics is one chain, not twenty separate facts. Multiply by 1/T to go from velocity to acceleration, by M to reach force, by L to reach work, and divide by T to reach power.
PRINCIPLE OF HOMOGENEITY · EVERY TERM MUST CARRY THE SAME DIMENSION ½ m v² M L² T−² m g h M L² T−² balanced → dimensionally correct. But balance alone does not prove the equation is right. a tipped beam, however, proves it wrong — every time.
Fig. 3 — The principle of homogeneity. A balanced equation may still be wrong; an unbalanced one is certainly wrong. NCERT states this asymmetry explicitly and NEET has tested it as a statement question.

03Master formula catalogue

Every entry below was generated from its defining relation by a dimension-algebra check, not typed from memory. Learn the Built from column and the third column comes free.

MECHANICS

QuantityBuilt fromDimensional formulaSI unit
Areal × b[L2]m²
Volumel × b × h[L3]m³
Densitym / V[M L-3]kg m⁻³
Velocity / speeds / t[L T-1]m s⁻¹
Accelerationv / t[L T-2]m s⁻²
Forcem × a[M L T-2]N
Momentumm × v[M L T-1]kg m s⁻¹
ImpulseF × t[M L T-1]N s
Work / energy / torque / heatF × s[M L2 T-2]J
PowerW / t[M L2 T-3]W
Pressure / stress / all moduliF / A[M L-1 T-2]Pa
Energy densityE / V[M L-1 T-2]J m⁻³
StrainΔl / l[M0 L0 T0]—
Surface tension / surface energyF / l[M T-2]N m⁻¹
Spring (force) constantF / x[M T-2]N m⁻¹
Coefficient of viscosityF / [A(dv/dx)][M L-1 T-1]Pa s
Frequency / angular velocity / decay constant1 / t[T-1]s⁻¹
Angular accelerationω / t[T-2]rad s⁻²
Moment of inertiam r²[M L2]kg m²
Angular momentumm v r[M L2 T-1]kg m² s⁻¹
Gravitational constant GF r² / (m₁m₂)[M-1 L3 T-2]N m² kg⁻²
Gravitational potentialW / m[L2 T-2]J kg⁻¹
Gravitational field intensityF / m[L T-2]N kg⁻¹
Planck's constant hE / ν[M L2 T-1]J s

HEAT & THERMODYNAMICS

QuantityBuilt fromDimensional formulaSI unit
Temperaturebase quantity[K]K
Specific heat capacityQ / (m ΔT)[L2 T-2 K-1]J kg⁻¹ K⁻¹
Molar specific heatQ / (n ΔT)[M L2 T-2 K-1 mol-1]J mol⁻¹ K⁻¹
Latent heatQ / m[L2 T-2]J kg⁻¹
Thermal conductivityQ L / (A t ΔT)[M L T-3 K-1]W m⁻¹ K⁻¹
Boltzmann constant kE / T[M L2 T-2 K-1]J K⁻¹
Universal gas constant RPV / (nT)[M L2 T-2 K-1 mol-1]J mol⁻¹ K⁻¹
EntropyQ / T[M L2 T-2 K-1]J K⁻¹
Stefan's constant σ(P/A) / T⁴[M T-3 K-4]W m⁻² K⁻⁴
Wien's constant bλm T[L K]m K
Coefficient of linear expansion αΔl / (l ΔT)[K-1]K⁻¹

ELECTRICITY & MAGNETISM

QuantityBuilt fromDimensional formulaSI unit
Electric chargeI × t[T A]C
Current densityI / A[L-2 A]A m⁻²
Electric potential / emfW / q[M L2 T-3 A-1]V
Electric fieldF / q[M L T-3 A-1]N C⁻¹
ResistanceV / I[M L2 T-3 A-2]Ω
ResistivityR A / l[M L3 T-3 A-2]Ω m
Conductivity1 / ρ[M-1 L-3 T3 A2]S m⁻¹
Capacitanceq / V[M-1 L-2 T4 A2]F
InductanceV t / I[M L2 T-2 A-2]H
Permittivity ε₀q² / (F r²)[M-1 L-3 T4 A2]C² N⁻¹ m⁻²
Permeability μ₀2πd(F/l) / (I₁I₂)[M L T-2 A-2]T m A⁻¹
Magnetic field BF / (q v)[M T-2 A-1]T
Magnetic flux φB × A[M L2 T-2 A-1]Wb
Electric dipole momentq × 2a[L T A]C m
Magnetic dipole momentI × A[L2 A]A m²

OPTICS, WAVES & MODERN

QuantityBuilt fromDimensional formulaSI unit
Wave number / Rydberg constant / power of a lens1 / λ[L-1]m⁻¹
Intensity / solar constantP / A[M T-3]W m⁻²
Refractive indexc / v[M0 L0 T0]—
Work functionhν₀[M L2 T-2]J
Avogadro number NAN / n[mol-1]mol⁻¹
Activity (radioactivity)−dN/dt[T-1]Bq
Hubble constant H₀v / r[T-1]s⁻¹

04Same-dimension pairs

NEET's favourite question shape in this chapter is "which pair has the same dimensions?" The table below is the complete working list. Read it once a week until the groupings are automatic.

quantities NEET loves to pair up
Shared dimensionEverything that shares it
[M L2 T-2]Work · energy · heat · torque · moment of force · work function
[M L-1 T-2]Pressure · stress · Young's / bulk / shear modulus · energy density · ½ε₀E² · B²/2μ₀
[M L2 T-1]Planck's constant · angular momentum · action
[M L T-1]Impulse · linear momentum
[M T-2]Surface tension · surface energy per unit area · spring constant · energy per unit area
[T-1]Frequency · angular velocity · velocity gradient · decay constant · activity · Hubble constant
[L2 T-2]Latent heat · gravitational potential · (velocity)² · specific heat × temperature
[L-1]Wave number · Rydberg constant · power of a lens · propagation constant k
[L T-1]Velocity · 1/√(μ₀ε₀) · E/B · ω/k · √(γP/ρ)
[T]RC · L/R · √(LC) · L₁√(μ₀ε₀)

Same dimensions does not mean same quantity

Work and torque share M L² T⁻², but joule and newton-metre are kept typographically distinct precisely because the physics differs: work is a scalar transfer of energy, torque is a vector turning effect. If a question asks 'are they the same quantity?', the answer is no; if it asks 'do they have the same dimensions?', the answer is yes. Read the verb.

05Exceptions and traps

Dimensionless is not the same as unitless

Plane angle is dimensionless (M⁰L⁰T⁰) yet carries the unit radian. Solid angle is dimensionless and carries the steradian. The two ideas are independent.

Fractional exponents are legal

Nothing requires an exponent to be a whole number. In the NEET 2024 re-examination item, [B] came out as L1/2. If your working produces a half-power, that is a signal you are probably on the right track, not a mistake.

The same symbol means different things

T is both time and temperature. L is both length and inductance. R is both radius and resistance. In thermal questions write Θ for temperature; in circuit questions read L as inductance. A large share of lost marks in this chapter are symbol collisions, not physics.

Quantities that look derived but are base

Temperature, amount of substance and luminous intensity feel derived but are base quantities with their own dimensions K, mol and cd. Molar quantities keep mol⁻¹ in the formula; per-mass quantities do not keep M.

The four limitations of the method

  1. It cannot find dimensionless constants — the 2π in T = 2π√(l/g) must come from elsewhere.
  2. It cannot distinguish quantities with identical dimensions (work vs torque).
  3. It fails for equations containing sums of several terms, and for trigonometric, logarithmic or exponential relations.
  4. It works only when the quantity depends on at most three others, because there are only three equations (M, L, T) to solve in mechanics.

06Numbers to remember

learn the value and the dimension together — NEET asks both
ConstantSymbolValueDimensional formula
Speed of light in vacuumc3.00 × 10⁸ m s⁻¹[L T-1]
Planck's constanth6.63 × 10⁻³⁴ J s[M L2 T-1]
Gravitational constantG6.67 × 10⁻¹¹ N m² kg⁻²[M-1 L3 T-2]
Boltzmann constantk1.38 × 10⁻²³ J K⁻¹[M L2 T-2 K-1]
Universal gas constantR8.314 J mol⁻¹ K⁻¹[M L2 T-2 K-1 mol-1]
Avogadro numberNA6.022 × 10²³ mol⁻¹[mol-1]
Stefan's constantσ5.67 × 10⁻⁸ W m⁻² K⁻⁴[M T-3 K-4]
Wien's constantb2.9 × 10⁻³ m K[L K]
Permittivity of free spaceε₀8.85 × 10⁻¹² C² N⁻¹ m⁻²[M-1 L-3 T4 A2]
Permeability of free spaceμ₀4π × 10⁻⁷ T m A⁻¹[M L T-2 A-2]
Elementary chargee1.6 × 10⁻¹⁹ C[T A]
Rydberg constantRH1.097 × 10⁷ m⁻¹[L-1]

07Scientists to remember

high-frequency names in statement and matching questions
NameWhat to attach to the name
Joseph FourierIntroduced the idea of dimensional homogeneity in his 1822 work on heat. The founding figure of dimensional analysis.
Lord Rayleigh (J. W. Strutt)Developed the method of dimensions into a working tool for deducing relations — the technique used in NCERT Example 1.5.
Edgar BuckinghamThe π-theorem, which formalises how many dimensionless groups a problem has. Named in passing, not examined numerically.
Isaac NewtonLaw of gravitation, which is where G comes from; the unit of force.
Henry CavendishFirst laboratory measurement of G, using a torsion balance.
Max Planckh, the quantum of action — dimensionally identical to angular momentum.
Ludwig Boltzmannk, energy per kelvin per molecule; with Josef Stefan, the T⁴ radiation law.
Josef StefanEstablished the fourth-power radiation law experimentally in 1879.
Wilhelm WienDisplacement law λmT = b, so Wien's constant has dimensions L K.
Amedeo AvogadroNA, the only common constant whose dimension is mol⁻¹ alone.
Johannes RydbergRH, dimensionally a reciprocal length.
James Clerk Maxwellc = 1/√(μ₀ε₀), the relation behind Q11 below.

08Twenty worked questions

About the PYQ labels

Items tagged PYQ follow the wording and structure of questions that have appeared in NEET/AIPMT papers. Year labels are indicative and worth cross-checking against the official NTA paper before quoting them in a test report. The physics and the answer key have been verified independently.

Q01PYQ · NEET (repeat item)Direct lookup

The dimensional formula of Planck's constant h is the same as that of:

  1. angular momentum
  2. linear momentum
  3. torque
  4. moment of inertia
Show full solution
Given
Planck's constant h, defined by E = hν.
Asked
Which listed quantity shares its dimensional formula.
Concept
Get [h] from its defining equation, then compare with each option. Energy divided by frequency.
Formula
[h] = [E] / [ν] = (M L² T⁻²) / (T⁻¹)
Baby steps
  1. Energy: [E] = M L² T⁻².
  2. Frequency: [ν] = T⁻¹.
  3. Divide: [h] = M L² T⁻² ÷ T⁻¹ = M L² T⁻¹.
  4. Angular momentum L = mvr → M × L T⁻¹ × L = M L² T⁻¹. Identical.
Answer
(a) angular momentum
Why not others
(a)Angular momentum mvr gives M L² T⁻¹ — the exact match.
(b)Linear momentum mv is M L T⁻¹: one power of L short.
(c)Torque is M L² T⁻² — that is energy, not action.
(d)Moment of inertia mr² is M L² with no time at all.
Shortcut
h and angular momentum are both called action. Remember one word, get both formulae. Bohr's condition mvr = nh/2π only makes sense if the two sides match.
Where it goes wrong
Students see 'J s' and think energy. J s is energy × time, so the T exponent is −2 + 1 = −1, not −2.
Q02PYQ · NEET 2022Direct lookup

The dimensional formula of the universal gravitational constant G is:

  1. [M⁻¹ L³ T⁻¹]
  2. [M L³ T⁻²]
  3. [M⁻¹ L² T⁻²]
  4. [M⁻¹ L³ T⁻²]
Show full solution
Given
Newton's law of gravitation, F = G m₁m₂ / r².
Asked
Dimensional formula of G.
Concept
Make G the subject first, then substitute dimensions term by term.
Formula
G = F r² / (m₁ m₂)
Baby steps
  1. Rearrange: G = F r² / (m₁m₂).
  2. [F] = M L T⁻², [r²] = L², [m₁m₂] = M².
  3. G = (M L T⁻² × L²) / M² = M¹⁻² L³ T⁻².
  4. So [G] = M⁻¹ L³ T⁻².
Answer
(d) [M⁻¹ L³ T⁻²]
Why not others
(a)T exponent wrong — force already carries T⁻² and nothing cancels it.
(b)M exponent has the wrong sign; dividing by M² must make it negative.
(c)L exponent is 2, but r² adds to the single L inside force, giving 3.
(d)Correct: M⁻¹ L³ T⁻², matching the SI unit N m² kg⁻².
Shortcut
Read the SI unit straight off: N m² kg⁻² = (M L T⁻²)(L²)(M⁻²) = M⁻¹ L³ T⁻². The unit is the answer in disguise.
Where it goes wrong
Writing G = F r² m₁m₂ instead of dividing. Always rearrange the defining equation on paper before substituting.
Q03PYQ · NEET 2021Direct lookup

The dimensions of the coefficient of viscosity η are:

  1. [M L⁻¹ T⁻²]
  2. [M L T⁻¹]
  3. [M L⁻¹ T⁻¹]
  4. [M L⁻² T⁻¹]
Show full solution
Given
Newton's law of viscous flow, F = η A (dv/dx).
Asked
Dimensional formula of η.
Concept
Velocity gradient dv/dx is velocity per length, so it is simply T⁻¹.
Formula
η = F / [A × (dv/dx)]
Baby steps
  1. [dv/dx] = (L T⁻¹) / L = T⁻¹.
  2. [A] = L².
  3. η = (M L T⁻²) / (L² × T⁻¹).
  4. = M L¹⁻² T⁻²⁺¹ = M L⁻¹ T⁻¹.
Answer
(c) [M L⁻¹ T⁻¹]
Why not others
(a)That is pressure. Viscosity is pressure × time, so one more T is needed.
(b)This is momentum, unrelated.
(c)Correct — matches the SI unit Pa s (also called poiseuille; 1 poise = 0.1 Pa s).
(d)L exponent wrong; only one power of L is lost.
Shortcut
η = pressure × time. Since [P] = M L⁻¹ T⁻², just add one T: M L⁻¹ T⁻¹.
Where it goes wrong
Treating dv/dx as having dimensions of velocity. It does not — the length in the denominator cancels the L in velocity.
Q04PYQ · NEET 2024Direct lookup

The dimensional formula of surface tension is:

  1. [M L T⁻²]
  2. [M L⁰ T⁻²]
  3. [M L⁻¹ T⁻²]
  4. [M L T⁻¹]
Show full solution
Given
Surface tension S = force per unit length of the free surface.
Asked
Dimensional formula of S.
Concept
Force divided by length — one power of L cancels completely, leaving zero L.
Formula
S = F / l (equivalently, surface energy per unit area, W/A)
Baby steps
  1. [F] = M L T⁻², [l] = L.
  2. S = M L T⁻² / L = M L⁰ T⁻².
  3. Cross-check with energy per area: (M L² T⁻²)/L² = M L⁰ T⁻². Same.
  4. So [S] = M T⁻², written M L⁰ T⁻².
Answer
(b) [M L⁰ T⁻²]
Why not others
(a)That is force itself; you forgot to divide by length.
(b)Correct — SI unit N m⁻¹ = J m⁻².
(c)That is pressure; you divided by area instead of by length.
(d)T exponent wrong.
Shortcut
Surface tension shares its dimensions with the spring constant k (also F/x). If a question offers 'force constant' as an option, it is a match.
Where it goes wrong
Confusing N m⁻¹ with N m⁻². Surface tension is per length, pressure is per area.
Q05PYQ · NEET 2018Direct lookup

Young's modulus of a wire has the same dimensions as:

  1. pressure
  2. force
  3. torque
  4. surface tension
Show full solution
Given
Y = longitudinal stress / longitudinal strain.
Asked
Which quantity shares its dimensional formula.
Concept
Strain is a ratio of two lengths and is therefore dimensionless, so any modulus reduces to stress, which is a pressure.
Formula
Y = (F/A) / (Δl/l)
Baby steps
  1. [strain] = L/L = M⁰ L⁰ T⁰ (dimensionless).
  2. So [Y] = [stress] = [F]/[A] = (M L T⁻²)/L².
  3. = M L⁻¹ T⁻².
  4. That is exactly the dimensional formula of pressure.
Answer
(a) pressure
Why not others
(a)Correct. All three elastic moduli, stress, pressure and energy density share M L⁻¹ T⁻².
(b)Force is M L T⁻² — two powers of L too many.
(c)Torque is M L² T⁻², the same as energy.
(d)Surface tension is M T⁻².
Shortcut
Any 'modulus' (Young, bulk, shear) = pressure dimensions. Compressibility, being 1/K, is the reciprocal: M⁻¹ L T².
Where it goes wrong
Forgetting that strain is dimensionless and trying to give it dimensions of length.
Q06PYQ · NEET 2020Thermal

The dimensional formula of Boltzmann's constant k is:

  1. [M L² T⁻² K]
  2. [M L² T⁻¹ K⁻¹]
  3. [M L T⁻² K⁻¹]
  4. [M L² T⁻² K⁻¹]
Show full solution
Given
Average kinetic energy of a gas molecule = (3/2) kT.
Asked
Dimensional formula of k.
Concept
k converts a temperature into an energy, so it must be energy per unit temperature.
Formula
k = E / T
Baby steps
  1. [E] = M L² T⁻².
  2. [temperature] = K.
  3. k = (M L² T⁻²) / K.
  4. = M L² T⁻² K⁻¹. SI unit J K⁻¹.
Answer
(d) [M L² T⁻² K⁻¹]
Why not others
(a)K exponent has the wrong sign — temperature is in the denominator.
(b)Time exponent wrong; energy carries T⁻².
(c)Only one power of L — that would be force per kelvin.
(d)Correct. Note this equals the dimensions of entropy and of heat capacity too.
Shortcut
k, entropy and heat capacity all read 'energy per kelvin'. Universal gas constant R is the same but with an extra mol⁻¹, since R = NAk.
Where it goes wrong
Mixing up k (per molecule, J K⁻¹) with R (per mole, J mol⁻¹ K⁻¹). The mol⁻¹ is the only difference.
Q07PYQ · NEET 2017Electromagnetism

The dimensional formula of the permittivity of free space ε₀ is:

  1. [M⁻¹ L⁻³ T² A²]
  2. [M⁻¹ L⁻² T⁴ A²]
  3. [M⁻¹ L⁻³ T⁴ A²]
  4. [M L³ T⁻⁴ A⁻²]
Show full solution
Given
Coulomb's law, F = q₁q₂ / (4πε₀ r²).
Asked
Dimensional formula of ε₀.
Concept
Make ε₀ the subject. Remember charge is current × time, so [q] = A T.
Formula
ε₀ = q₁q₂ / (4π F r²)
Baby steps
  1. [q²] = (A T)² = A² T².
  2. [F r²] = (M L T⁻²)(L²) = M L³ T⁻².
  3. ε₀ = A² T² / (M L³ T⁻²).
  4. = M⁻¹ L⁻³ T⁴ A² (T exponent: 2 − (−2) = 4).
Answer
(c) [M⁻¹ L⁻³ T⁴ A²]
Why not others
(a)T² instead of T⁴ — the subtraction of the −2 was missed.
(b)L⁻² instead of L⁻³; r² contributes two L on top of the one inside force.
(c)Correct. SI unit C² N⁻¹ m⁻² (= F m⁻¹).
(d)Every sign is inverted — this is 1/ε₀.
Shortcut
Easier route: ε₀ = C·(length)⁻¹, i.e. farad per metre. Take [C] = M⁻¹ L⁻² T⁴ A² and divide by L.
Where it goes wrong
Subtracting exponents in the wrong direction. T² ÷ T⁻² = T⁴, not T⁰.
Q08PYQ · NEET 2019Electromagnetism

The dimensions of magnetic field B are:

  1. [M T⁻¹ A⁻¹]
  2. [M T⁻² A⁻¹]
  3. [M L² T⁻² A⁻¹]
  4. [M L T⁻² A⁻¹]
Show full solution
Given
Lorentz force on a moving charge, F = q v B (with v ⊥ B).
Asked
Dimensional formula of B.
Concept
Rearrange for B and substitute [q] = A T.
Formula
B = F / (q v)
Baby steps
  1. [q v] = (A T)(L T⁻¹) = A L.
  2. [F] = M L T⁻².
  3. B = (M L T⁻²) / (A L) — the L cancels.
  4. = M T⁻² A⁻¹. SI unit tesla, T = N A⁻¹ m⁻¹.
Answer
(b) [M T⁻² A⁻¹]
Why not others
(a)T⁻¹ instead of T⁻².
(b)Correct. Also obtainable from F = BIl → B = F/(I l).
(c)That is magnetic flux φ = BA, two powers of L larger.
(d)L should cancel completely.
Shortcut
Flux φ = BA, so [φ] = [B] × L² = M L² T⁻² A⁻¹. Learn one, derive the other.
Where it goes wrong
Using F = BIl but forgetting the l, which leaves a stray L.
Q09PYQ · NEET 2024 (Re-examination)Exponent extraction

The potential energy of a particle moving along the x-direction varies as V = A x² / (√x + B). The dimensions of A²/B are:

  1. [M² L1/2 T⁻⁴]
  2. [M² L3/2 T⁻⁴]
  3. [M L1/2 T⁻²]
  4. [M² L5/2 T⁻⁴]
Show full solution
Given
V = A x² / (√x + B), where V is a potential energy and x a length.
Asked
Dimensions of the combination A²/B.
Concept
Principle of homogeneity: only like quantities can be added, so B must match √x. Then force the whole right-hand side to be an energy.
Formula
[B] = [√x] = L1/2 ; [A][x²]/[√x] = [V]
Baby steps
  1. Addition rule: B must have the dimensions of √x, so [B] = L1/2.
  2. [V] = M L² T⁻². So [A] × L² / L1/2 = M L² T⁻².
  3. [A] × L3/2 = M L² T⁻² → [A] = M L1/2 T⁻².
  4. [A²] = M² L T⁻⁴. Divide by [B] = L1/2:
  5. [A²/B] = M² L1/2 T⁻⁴.
Answer
(a) [M² L1/2 T⁻⁴]
Why not others
(a)Correct: L exponent 1 − ½ = ½.
(b)Obtained by forgetting to divide by B at the end.
(c)This is [A] itself, not A²/B.
(d)Comes from adding ½ instead of subtracting it.
Shortcut
Two-step reflex for every question of this shape: (1) anything added must match, (2) the whole expression must equal the stated quantity. Never try to guess A directly.
Where it goes wrong
Ignoring the √x + B addition and setting [B] = [x] = L. That single slip changes every subsequent exponent.
Q10PYQ · NEET 2021Electromagnetism

The dimensional formula of magnetic flux is:

  1. [M L T⁻² A⁻¹]
  2. [M L² T⁻¹ A⁻¹]
  3. [M L² T⁻² A⁻²]
  4. [M L² T⁻² A⁻¹]
Show full solution
Given
φ = B × A, with [B] = M T⁻² A⁻¹.
Asked
Dimensional formula of magnetic flux.
Concept
Flux is field × area. Alternatively use Faraday's law, e = −dφ/dt, so φ = e × t.
Formula
φ = B A = emf × time
Baby steps
  1. Route 1: [φ] = (M T⁻² A⁻¹)(L²) = M L² T⁻² A⁻¹.
  2. Route 2 (check): [emf] = M L² T⁻³ A⁻¹; multiply by T.
  3. = M L² T⁻³⁺¹ A⁻¹ = M L² T⁻² A⁻¹. Both agree.
  4. So [φ] = M L² T⁻² A⁻¹. SI unit weber = V s.
Answer
(d) [M L² T⁻² A⁻¹]
Why not others
(a)This is B itself multiplied by only one L.
(b)T exponent wrong.
(c)A⁻² belongs to inductance, not flux.
(d)Correct. Note inductance = φ/I adds one more A⁻¹.
Shortcut
Weber = volt × second. Reading the unit aloud gives the dimensional formula immediately.
Where it goes wrong
Mixing flux (Wb) with flux density B (T). B is flux per unit area.
Q11PYQ · AIPMT (repeat item)Combination

The dimensions of (μ₀ε₀)−1/2 are those of:

  1. length × time
  2. time / length
  3. speed
  4. acceleration
Show full solution
Given
μ₀ = M L T⁻² A⁻² and ε₀ = M⁻¹ L⁻³ T⁴ A².
Asked
What kind of quantity is (μ₀ε₀)−1/2.
Concept
Multiply the two dimensional formulae first; the electrical dimensions cancel completely, which is the whole point of the result.
Formula
c = 1 / √(μ₀ε₀)
Baby steps
  1. μ₀ε₀ = (M L T⁻² A⁻²)(M⁻¹ L⁻³ T⁴ A²).
  2. M: 1−1 = 0. A: −2+2 = 0. L: 1−3 = −2. T: −2+4 = 2.
  3. So μ₀ε₀ = L⁻² T², i.e. 1/(speed)².
  4. Raise to −½: (L⁻²T²)−1/2 = L T⁻¹ — a speed.
Answer
(c) speed
Why not others
(a)L T, which is neither.
(b)T L⁻¹ is the reciprocal of speed — the sign of the exponent was dropped.
(c)Correct: this combination is the speed of light in vacuum, 3.00 × 10⁸ m s⁻¹.
(d)Acceleration is L T⁻².
Shortcut
Maxwell's relation c = 1/√(μ₀ε₀) is on the formula sheet. If you recall the physics, you never need the algebra.
Where it goes wrong
Forgetting the negative half-power and answering with the dimensions of μ₀ε₀ itself.
Q12Same-dimension pair

Which of the following pairs has the same dimensional formula?

  1. work and power
  2. impulse and linear momentum
  3. force and surface tension
  4. torque and moment of inertia
Show full solution
Given
Four candidate pairs.
Asked
The pair that matches dimensionally.
Concept
Work out both members of each pair rather than trusting intuition. The impulse–momentum theorem guarantees one pair by construction.
Formula
J = FΔt = Δp
Baby steps
  1. Impulse = F × t = (M L T⁻²)(T) = M L T⁻¹.
  2. Momentum = m v = M × L T⁻¹ = M L T⁻¹. Match.
  3. Work M L² T⁻² vs power M L² T⁻³ — differ by one T.
  4. Force M L T⁻² vs surface tension M T⁻² — differ by one L.
Answer
(b) impulse and linear momentum
Why not others
(a)Power is work per unit time; they cannot match.
(b)Correct — and it must be so, because impulse equals change of momentum.
(c)Surface tension is force per unit length.
(d)Torque is M L² T⁻²; moment of inertia is M L² with no T.
Shortcut
Any physical law of the form 'X = Y' hands you a matched pair for free: impulse = Δp, work = ΔKE, torque = Iα.
Where it goes wrong
Assuming force and surface tension match because both are 'about force'. Always finish the division.
Q13Dimensionless quantities

Which set contains only dimensionless quantities?

  1. strain, refractive index, relative density
  2. strain, stress, relative density
  3. plane angle, torque, refractive index
  4. Poisson's ratio, surface tension, strain
Show full solution
Given
Four candidate sets.
Asked
The set in which every member is dimensionless.
Concept
A quantity is dimensionless when it is the ratio of two quantities of the same kind. Test each member for that pattern.
Formula
[dimensionless] = M⁰ L⁰ T⁰
Baby steps
  1. Strain = Δl/l — length over length. Dimensionless.
  2. Refractive index = c/v — speed over speed. Dimensionless.
  3. Relative density = ρsubstance/ρwater — density over density. Dimensionless.
  4. Stress, torque and surface tension all carry dimensions, so the other sets fail.
Answer
(a) strain, refractive index, relative density
Why not others
(a)Correct — each is a ratio of like quantities.
(b)Stress is M L⁻¹ T⁻².
(c)Torque is M L² T⁻².
(d)Surface tension is M T⁻².
Shortcut
Full NEET list of dimensionless quantities: strain, Poisson's ratio, refractive index, relative density, specific gravity, coefficient of friction, dielectric constant, angle (plane and solid), all trigonometric / logarithmic / exponential values, and pure numbers such as 2π.
Where it goes wrong
Calling plane angle dimensionless but then denying it a unit. It is dimensionless and it still has the unit radian. Dimensionless does not mean unitless.
Q14Constants inside an equation

In van der Waals' equation (P + a/V²)(V − b) = RT, the dimensions of a/b are:

  1. [M L⁵ T⁻²]
  2. [L³]
  3. [M L⁻¹ T⁻²]
  4. [M L² T⁻²]
Show full solution
Given
(P + a/V²)(V − b) = RT, with P a pressure and V a volume.
Asked
Dimensions of the ratio a/b.
Concept
Homogeneity applied twice: a/V² must match P, and b must match V.
Formula
[a] = [P][V²] ; [b] = [V]
Baby steps
  1. From V − b: [b] = [V] = L³.
  2. From P + a/V²: [a] = [P][V²] = (M L⁻¹ T⁻²)(L⁶) = M L⁵ T⁻².
  3. Divide: [a/b] = M L⁵ T⁻² / L³.
  4. = M L² T⁻² — the dimensions of energy.
Answer
(d) [M L² T⁻²]
Why not others
(a)That is [a] alone.
(b)That is [b] alone.
(c)That is pressure, i.e. [a/V²].
(d)Correct. a/b carries energy dimensions, which is why a/b relates to the depth of the molecular attraction.
Shortcut
Any bracket of the form (X + Y) hands you [X] = [Y] instantly. Attack additions before multiplications.
Where it goes wrong
If the equation is written per mole, a picks up mol⁻² and b picks up mol⁻¹, so a/b gains mol⁻¹. Read whether the paper writes RT or nRT before answering.
Q15Exponent extraction

The force acting on a particle is given by F = a√x + b t², where x is distance and t is time. The dimensions of a/b are:

  1. [L1/2 T⁻²]
  2. [L−1/2 T⁻²]
  3. [L−1/2 T²]
  4. [L T²]
Show full solution
Given
F = a√x + b t², F a force, x a length, t a time.
Asked
Dimensions of a/b.
Concept
Both terms on the right must separately equal a force. Solve for a and b independently, then divide.
Formula
[a] = [F]/[√x] ; [b] = [F]/[t²]
Baby steps
  1. [a] = (M L T⁻²)/L1/2 = M L1/2 T⁻².
  2. [b] = (M L T⁻²)/T² = M L T⁻⁴.
  3. a/b = (M L1/2 T⁻²)/(M L T⁻⁴).
  4. M cancels; L: ½ − 1 = −½; T: −2 − (−4) = +2.
  5. So [a/b] = L−1/2 T².
Answer
(c) [L−1/2 T²]
Why not others
(a)Sign of the L exponent flipped.
(b)Sign of the T exponent flipped — the double negative was mishandled.
(c)Correct. M cancels entirely, which is a useful sanity check.
(d)L exponent should be a half, not one.
Shortcut
When a ratio of two constants is asked, M almost always cancels — if it survives in your answer, re-check.
Where it goes wrong
Subtracting −4 as if it were +4. Write the exponents in a column and subtract carefully; this is a mechanical error, not a conceptual one.
Q16Waves

A progressive wave is described by y = A sin(ωt − kx). The dimensions of ω/k are:

  1. [L⁻¹ T]
  2. [L T⁻¹]
  3. [L T]
  4. dimensionless
Show full solution
Given
y = A sin(ωt − kx).
Asked
Dimensions of ω/k.
Concept
The argument of a sine must be dimensionless, so ωt and kx are each pure numbers.
Formula
[ωt] = M⁰L⁰T⁰ → [ω] = T⁻¹ ; [kx] = M⁰L⁰T⁰ → [k] = L⁻¹
Baby steps
  1. ωt dimensionless → [ω] = T⁻¹.
  2. kx dimensionless → [k] = L⁻¹.
  3. ω/k = T⁻¹ / L⁻¹ = L T⁻¹.
  4. Physically this is the wave speed v = ω/k = νλ. The dimensions confirm it.
Answer
(b) [L T⁻¹]
Why not others
(a)The reciprocal — k/ω, not ω/k.
(b)Correct: ω/k is the wave speed.
(c)Neither exponent is negative here.
(d)ω/k has dimensions; only the sine's argument is dimensionless.
Shortcut
Rule for every special function: whatever sits inside sin, cos, tan, log, or ex is dimensionless, and so is the output. This single rule solves most 'find the dimension of the constant' questions.
Where it goes wrong
Assuming A, the amplitude, is dimensionless. A has the same dimensions as y — usually length.
Q17Thermal

Stefan's constant σ, appearing in E = σAT⁴t, has dimensions:

  1. [M T⁻³ K⁻⁴]
  2. [M L² T⁻³ K⁻⁴]
  3. [M T⁻³ K⁻¹]
  4. [M L T⁻³ K⁻⁴]
Show full solution
Given
Stefan–Boltzmann law: energy radiated per unit area per unit time = σT⁴.
Asked
Dimensional formula of σ.
Concept
Write the law as power per unit area, then divide by T⁴ where T here is temperature (dimension K), not time.
Formula
σ = (P/A) / (temperature)⁴
Baby steps
  1. [P] = M L² T⁻³, [A] = L².
  2. P/A = M T⁻³ — both L cancel.
  3. Divide by K⁴: σ = M T⁻³ K⁻⁴.
  4. SI unit W m⁻² K⁻⁴, value 5.67 × 10⁻⁸.
Answer
(a) [M T⁻³ K⁻⁴]
Why not others
(a)Correct — note that intensity P/A on its own is M T⁻³.
(b)The L² from area was not cancelled.
(c)The fourth power of temperature was ignored.
(d)One stray L remains.
Shortcut
σ = intensity ÷ K⁴, and intensity (also the solar constant) is M T⁻³. Learn intensity once and this follows.
Where it goes wrong
Using the symbol T for both time and temperature in the same line. Write Θ or θ for temperature while solving.
Q18Thermal

The dimensional formula of the coefficient of thermal conductivity is:

  1. [M L² T⁻³ K⁻¹]
  2. [M L T⁻² K⁻¹]
  3. [M L⁻¹ T⁻³ K⁻¹]
  4. [M L T⁻³ K⁻¹]
Show full solution
Given
Q = K A (ΔΘ/L) t, the steady-state conduction equation.
Asked
Dimensional formula of K.
Concept
Rearrange for K, then substitute. The temperature gradient contributes K L⁻¹.
Formula
K = Q L / (A ΔΘ t)
Baby steps
  1. [Q] = M L² T⁻². Multiply by L: M L³ T⁻².
  2. Divide by [A] = L²: M L T⁻².
  3. Divide by [ΔΘ] = K and by [t] = T.
  4. K = M L T⁻³ K⁻¹. SI unit W m⁻¹ K⁻¹.
Answer
(d) [M L T⁻³ K⁻¹]
Why not others
(a)That is power per kelvin, missing the division by length.
(b)T exponent wrong — the division by t was forgotten.
(c)L exponent wrong in sign.
(d)Correct — read straight off the unit W m⁻¹ K⁻¹.
Shortcut
Write the SI unit first (W m⁻¹ K⁻¹), then translate: W = M L² T⁻³, divide by L, divide by K.
Where it goes wrong
Dropping the thickness L from the numerator. Conductivity is defined per unit thickness, unlike thermal conductance.
Q19Circuits

Which of the following combinations has the dimensions of time?

  1. RC only
  2. L/R only
  3. RC, L/R and √(LC), all three
  4. √(LC) only
Show full solution
Given
R resistance, C capacitance, L inductance.
Asked
Which combinations carry the dimension of time.
Concept
Each is the time constant of a standard circuit, so all three must reduce to T. Verify at least one by algebra.
Formula
τRC = RC ; τLR = L/R ; TLC = 2π√(LC)
Baby steps
  1. [R] = M L² T⁻³ A⁻², [C] = M⁻¹ L⁻² T⁴ A².
  2. RC: M and A cancel; L: 2−2 = 0; T: −3+4 = 1 → T. ✓
  3. [L] = M L² T⁻² A⁻², so L/R gives T⁻²⁺³ = T. ✓
  4. LC = (M L² T⁻² A⁻²)(M⁻¹ L⁻² T⁴ A²) = T², so √(LC) = T. ✓
Answer
(c) RC, L/R and √(LC), all three
Why not others
(a)True but incomplete.
(b)True but incomplete.
(c)Correct — all three are time constants, which is exactly why they appear in the decay and oscillation formulae.
(d)True but incomplete.
Shortcut
Any quantity that appears in an exponent e−t/τ must itself be a time. Spot τ in the physics and you have the dimensions free.
Where it goes wrong
Reading L as length instead of inductance. In circuit questions, L is inductance — check the symbol list before substituting.
Q20PYQ · NEET 2024Trap item

The dimensional formula of specific heat capacity is:

  1. [M L² T⁻² K⁻¹]
  2. [M⁰ L² T⁻² K⁻¹]
  3. [M L² T⁻² K⁻¹ mol⁻¹]
  4. [M⁰ L² T⁻¹ K⁻¹]
Show full solution
Given
Q = m s ΔΘ.
Asked
Dimensional formula of specific heat capacity s.
Concept
'Specific' means per unit mass. That single word removes the M from the formula.
Formula
s = Q / (m ΔΘ)
Baby steps
  1. [Q] = M L² T⁻².
  2. Divide by [m] = M: L² T⁻².
  3. Divide by [ΔΘ] = K.
  4. s = M⁰ L² T⁻² K⁻¹. SI unit J kg⁻¹ K⁻¹.
Answer
(b) [M⁰ L² T⁻² K⁻¹]
Why not others
(a)That is heat capacity (for the whole body) or Boltzmann's constant — not the specific value.
(b)Correct: mass cancels, so the M exponent is zero.
(c)That is molar specific heat, which carries mol⁻¹.
(d)T exponent wrong.
Shortcut
Decode the prefix: 'specific' = per kg, 'molar' = per mole, plain = for the whole body. Three closely related quantities, three different dimensional formulae.
Where it goes wrong
This is the single most-missed item in the chapter. Latent heat is L² T⁻² (per kg, no K); specific heat is L² T⁻² K⁻¹; molar specific heat and R add mol⁻¹; Boltzmann's k keeps M and drops mol.