What are the limitations of dimensional analysis?

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Q. What are the limitations of dimensional analysis?

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Concept understanding — Dimensional Analysis

The dimensions of a physical quantity record how it is built from the base quantities — how many powers of mass, length, time (and, where needed, electric current, temperature, amount, luminous intensity) it contains. Written in square brackets, force is [M L T⁻²] and energy is [M L² T⁻²]. This single idea — that an equation must be dimensionally consistent — is one of the most powerful cross-checks in physics, and JEE Main mines it heavily.

1 — Dimensional formulae. Every derived quantity has a dimensional formula obtained from its defining relation: velocity [LT⁻¹], acceleration [LT⁻²], force [MLT⁻²], work/energy/torque [ML²T⁻²], power [ML²T⁻³], pressure/stress [ML⁻¹T⁻²], momentum/impulse [MLT⁻¹]. For constants, isolate the constant in its equation and read off its dimensions — from F = Gm₁m₂/r², [G] = [M⁻¹L³T⁻²]; from E = hν, [h] = [ML²T⁻¹]; from PV = nRT, [R] = [ML²T⁻²K⁻¹mol⁻¹]. Electrical quantities carry the base dimension of current [A]: charge [AT], potential [ML²T⁻³A⁻¹], resistance [ML²T⁻³A⁻²].

2 — The principle of homogeneity. In any valid equation, every additive term has the same dimensions. This lets you (a) test whether a given equation can be correct, (b) find a missing exponent by matching the powers of M, L and T on both sides, and (c) reject a proposed formula that is dimensionally inconsistent. A crucial corollary: the argument of any sin, cos, log or exponential — and any exponent — must be dimensionless. So in y = A sin(ωt), ωt is dimensionless, forcing [ω] = [T⁻¹].

3 — Deriving a relation. When a quantity depends on a few others, assume a power-law y = k·x₁ᵃ x₂ᵇ x₃ᶜ, write the dimensions of both sides, and equate the exponents of M, L, T to solve for a, b, c. This recovers the form of many results — the pendulum's T ∝ √(L/g), the speed of a wave on a string v ∝ √(T/μ), Stokes' drag F ∝ ηrv. The dimensionless constant k (like the 2π in the pendulum) is what dimensional analysis cannot supply.

4 — Converting between systems of units. Because a physical quantity is unit-independent, n₁u₁ = n₂u₂. Using the dimensional formula, n₂ = n₁ [M₁/M₂]ᵃ [L₁/L₂]ᵇ [T₁/T₂]ᶜ. This is how 1 N = 10⁵ dyne, 1 J = 10⁷ erg, and how the numerical value of a constant like G changes from SI to CGS. The same machinery lets you express a quantity when a new set of quantities (say force, velocity, time) is chosen as fundamental. …

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Step 1. Dimensional analysis gives no information about purely numerical (dimensionless) constants in a formula, such as $1,2,\pi,e$ -- these must come from elsewhere (a full derivation or experiment).

Step 2. It cannot decide whether a given quantity in an equation is a vector or a scalar -- both have the same dimensional formula if built the same way (e.g. torque and energy).

Step 3. It is not suitable for deriving relations that involve trigonometric, exponential, or logarithmic functions, since the argument of such a function must itself be dimensionless, and the method offers no way to determine the function's internal form.

Step 4. It cannot be applied if a relationship connects MORE THAN THREE physical quantities, because matching powers of $M,L,T$ only ever supplies three independent equations -- not enough to solve for four or more unknown exponents. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 50 on this concept.

CBSE 2026Set 55/2/11 markMCQ
$\phi$
›Reveal solutionSolution
The key idea is dimensional consistency: each term in $\phi = 5At^2 + Bt - 2C$ must have the same unit as $\phi$ (Wb). This forces $A$ to have units $\text{Wb s}^{-2}$ and $B$ to have units $\text{Wb s}^{-1}$ , so the correct option is (C).

The problem gives you a relation between magnetic flux $\phi$ (in webers) and time $t$ (in seconds):
$\phi = 5At^2 + Bt - 2C.$
You’re asked for the SI units of $A$ and $B$ . The constants $5$ and $2$ are pure numbers — they have no units. So the only way this equation makes physical sense is if every term on the right-hand side has the same unit as $\phi$ , which is the weber (Wb). This is the principle of dimensional homogeneity, and it’s the entire foundation of the solution.

Let’s apply it term by term.

First term: $5At^2$ Since $5$ is dimensionless, the unit of $5At^2$ is the unit of $A$ multiplied by the unit of $t^2$ . Time $t$ is in seconds, so $t^2$ has unit $\text{s}^2$ . For this term to equal a flux in Wb, we need:

$\text{unit of }A \times \text{s}^2 = \text{Wb}.$
Therefore:
$\text{unit of }A = \frac{\text{Wb}}{\text{s}^2} = \text{Wb s}^{-2}.$

Second term: $Bt$ Here $B$ is multiplied by $t$ (unit s). So:

$\text{unit of }B \times \text{s} = \text{Wb}.$
Hence:
$\text{unit of }B = \frac{\text{Wb}}{\text{s}} = \text{Wb s}^{-1}.$

Third term: $-2C$ …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following physical quantity has no dimensions?$
(1) Angular Velocity
(2) Angular Acceleration
(3) Angular Displacement
(4) Stress
›Reveal solutionSolution
Angular displacement is dimensionless because it is defined as the ratio of two lengths (arc / radius).

Check each option's dimensional formula:

Angular velocity ω = dθ/dt → dimension [T^-1] (angle is dimensionless, so only 1/time survives).

Angular acceleration α = dω/dt → dimension [T^-2].

Angular displacement θ = arc length / radius = a ratio of two lengths → the length dimensions cancel, leaving a pure number (dimensionless), measured in radians. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The dimensional formula of work done is the same as the dimensional formula of$
(a) Momentum
(b) Power
(c) Energy
(d) Torque
›Reveal solutionSolution
Work done and energy share the exact same dimensional formula [ML^2T^-2], because work is defined as a mode of energy transfer.

Work done W = Force x displacement = [MLT^-2] x [L] = [ML^2T^-2].

Check each option:

Momentum p = mv = [M][LT^-1] = [MLT^-1] -- different.

Power P = Work/time = [ML^2T^-2]/[T] = [ML^2T^-3] -- different.

Energy (kinetic or potential) = (1/2)mv^2 or mgh = [ML^2T^-2] -- matches exactly. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the dimensional formula of angular momentum?$
(a) [ML^2T^-1]
(b) [MLT^-2]
(c) [MLT^-1]
(d) [M^-1L^3T^2]
›Reveal solutionSolution
Angular momentum L = mvr, so its dimensions are mass x velocity x radius = [ML^2T^-1].

Angular momentum is defined as L = r x p = mvr (magnitude, for a point mass moving with linear velocity v at perpendicular distance r from the axis).

Dimensions:

Mass, m = [M] …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following physical quantities has the same dimensions as impulse?$
(a) Force
(b) Momentum
(c) Work
(d) Power
›Reveal solutionSolution
Impulse J = F x delta t has dimensions [MLT^-1], identical to momentum p = mv.

Impulse J = Force x time = [MLT^-2] x [T] = [MLT^-1].

Compare:

Force = [MLT^-2] -- different.

Momentum = mv = [M][LT^-1] = [MLT^-1] -- matches exactly.

Work = [ML^2T^-2] -- different.

Power = [ML^2T^-3] -- different. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The pair of physical quantities not having same dimension is:$
(a) Torque and Energy
(b) Surface Tension and Impulse
(c) Angular momentum and Planck's constant
(d) None of the above
›Reveal solutionSolution
Surface tension ( $MT^{-2}$ ) and impulse ( $MLT^{-1}$ ) have different dimensions; the other two pairs match.

Work out each pair's dimensional formula:

Torque $= r \times F$ , dimension $ML^2T^{-2}$ . Energy also has dimension $ML^2T^{-2}$ . Same.

Surface tension $=$ force/length, dimension $MLT^{-2}/L = MT^{-2}$ . Impulse $=$ force $\times$ time, dimension $MLT^{-2}\cdot T = MLT^{-1}$ . Different — an $L$ appears in impulse but not in surface tension. …
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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Dimensional formula of ____ is same as the dimensional formula of angular momentum.$
›Reveal solutionSolution

Planck's constant has the same dimensional formula, [M L² T⁻¹], as angular momentum.

Angular momentum L = mvr has dimensions [M][LT⁻¹][L] = [M L² T⁻¹].

Planck's constant is defined through E = hν, so h = E/ν. Energy E has dimensions [M L² T⁻²] and frequency ν has dimensions [T⁻¹], so:

h = [M L² T⁻²] / [T⁻¹] = [M L² T⁻¹]

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following has dimensional formula [ML^2T^-2] ?$
(a) Acceleration
(b) Force
(c) Work
(d) Linear momentum
›Reveal solutionSolution
[ML^2T^-2] is the dimension of energy/work, so the answer is (C) Work.

We find the dimensions of each option:

Acceleration = velocity/time = [LT^-2]

Force = mass x acceleration = [MLT^-2]

Work = force x displacement = [MLT^-2] x [L] = [ML^2T^-2]

Linear momentum = mass x velocity = [MLT^-1] …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following couples of quantities have same dimension? (A) Work and energy (B) Force and momentum (C) Force and power (D) Momentum and energy$
›Reveal solutionSolution
Work and energy share the same dimension $[ML^2T^{-2}]$ , both measured in joules.

Check each pair:

Work $W = Fd$ , dimension $[MLT^{-2}][L] = [ML^2T^{-2}]$ . Kinetic energy $= \tfrac{1}{2}mv^2$ , dimension $[M][LT^{-1}]^2 = [ML^2T^{-2}]$ . Same dimension.

Force $[MLT^{-2}]$ vs momentum $[MLT^{-1}]$ — different. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The velocity of a particle$
(v) at an instant t is given by v = at + bt^2. The dimension of b is
(a) [L]
(b) [LT^-1]
(c) [LT^-2]
(d) [LT^-3]
›Reveal solutionSolution

By the principle of dimensional homogeneity, every term added to v must have the dimension of velocity; this forces [b] = [LT^-3].

Given v = at + bt^2, where v is velocity, [v] = [LT^-1].

…

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CBSE 2025Set ANN1 markMCQ
$Q. Find out the fundamental quantity from among the physical quantities given below:$
(a) velocity
(b) temperature
(c) force
(d) density
›Reveal solutionSolution
Temperature is the fundamental quantity here; velocity, force and density are all derived from base quantities.

The seven SI base (fundamental) quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity. A derived quantity is built from these using multiplication or division.

Checking each option:

velocity = displacement / time = length / time -> derived. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The physical quantity having the dimensions [M¹T⁻²] is$
(a) (A) Density
(b) (B) Tension
(c) (C) Surface tension
(d) (D) Viscosity
›Reveal solutionSolution
[!TLDR]

(C) Surface tension

Why

Surface tension = force/length = (MLT⁻²)/L = MT⁻², exactly matching the g …
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Q.What are the limitations of dimensional analysis?

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