Explain the principle of homogeneity of dimensions. What are its uses? Give example.

III. Long Answer Questions · Q5

Q. Explain the principle of homogeneity of dimensions. What are its uses? Give example.

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Concept understanding — Dimensional Formula, Dimensional Constants and Homogeneity Principle

Every derived quantity reduces to some combination of powers of the seven fundamental (base) quantities -- length $[L]$ , mass $[M]$ , time $[T]$ , electric current $[A]$ , temperature $[K]$ , amount of substance $[\text{mol}]$ , luminous intensity $[\text{cd}]$ -- and this combination, written in square brackets, is called the quantity's dimensional formula. Writing that formula as an equation (e.g. $[v]=[M^0LT^{-1}]$ ) is the dimensional equation.

Worked derivation (velocity): $\text{velocity}=\dfrac{\text{displacement}}{\text{time}}=\dfrac{[L]}{[T]}=[M^0LT^{-1}]$ -- dimension $0$ in mass, $1$ in length, $-1$ in time.

Dimensional formulas of common quantities follow directly from their defining relation: force $=$ mass $\times$ acceleration $=[MLT^{-2}]$ ; work $=$ force $\times$ distance $=[ML^2T^{-2}]$ ; Planck's constant $h=$ energy/frequency $=[ML^2T^{-2}]/[T^{-1}]=[ML^2T^{-1}]$ ; the gravitational constant $G$ (from Newton's law $F=Gm_1m_2/r^2$ ) $=[\text{force}\times\text{distance}^2]/\text{mass}^2=[M^{-1}L^3T^{-2}]$ ; torque, being force $\times$ distance just like work, shares work's dimensional formula $[ML^2T^{-2}]$ -- so torque and energy, despite measuring physically different things, have the same dimensions.

Classifying by dimension. Every physical quantity falls into exactly one of four boxes:

Dimensional variables -- have dimensions, and take variable values (length, velocity, acceleration).
Dimensionless variables -- no dimensions, but variable values (specific gravity, strain, refractive index). …

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Step 1 (Statement). The principle of homogeneity of dimensions states that every term that is added, subtracted, or set equal in a physically valid equation must carry the SAME dimensions -- a length can never be added to a time, for instance.

Step 2 (Worked example). In the kinematic equation $v^2=u^2+2as$ : $[v^2]=[LT^{-1}]^2=[L^2T^{-2}]$ ; $[u^2]=[L^2T^{-2}]$ (same form); $[2as]=[LT^{-2}][L]=[L^2T^{-2}]$ (the exact constant 2 contributes no dimension). All three terms reduce to the identical $[L^2T^{-2}]$ , confirming the equation is homogeneous.

Step 3 (Use 1 -- checking correctness). Because every valid equation MUST be homogeneous, testing whether all terms share the same dimension is a fast way to catch an equation that cannot possibly be right. Example: $v=u+at$ : $[LT^{-1}]=[LT^{-1}]+[LT^{-2}][T]=[LT^{-1}]+[LT^{-1}]$ -- passes.

Step 4 (Use 2 -- unit conversion). Since $n_1[u_1]=n_2[u_2]$ for the same physical quantity in two unit systems, and each side must be dimensionally homogeneous with the other, a quantity's dimensional formula (exponents $a,b,c$ in mass, length, time) directly gives the conversion factor $n_2=n_1(M_1/M_2)^a(L_1/L_2)^b(T_1/T_2)^c$ between any two systems. …

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

CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following has the dimension of (mu0 epsilon0)^(-1/2)?$
(a) Velocity
(b) Length
(c) Force
(d) Time
›Reveal solutionSolution

(mu0 epsilon0)^(-1/2) is exactly the speed of light c, so its dimension is that of velocity, [M^0 L^1 T^-1].

Maxwell's electromagnetic theory gives the speed of an electromagnetic wave in vacuum as:

c = 1 / sqrt(mu0 epsilon0)

where mu0 is the permeability of free space and epsilon0 is the permittivity of free space.

Rearranging, (mu0 epsilon0)^(-1/2) = 1/sqrt(mu0 epsilon0) = c.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following pairs of physical quantities have same dimensions?$
(a) torque and power
(b) force and power
(c) force and torque
(d) torque and energy
›Reveal solutionSolution
Torque and energy share the same dimensional formula [ML^2T^-2], because both are computed as (force) x (distance), even though torque is a vector (cross product) and energy/work is a scalar (dot product).

Work out the dimensional formula of each quantity:

Force F = mass x acceleration -> [M][LT^-2] = [MLT^-2]

Torque tau = force x perpendicular distance -> [MLT^-2][L] = [ML^2T^-2]

Energy/Work W = force x displacement (along force) -> [MLT^-2][L] = [ML^2T^-2]

Power P = work/time -> [ML^2T^-2]/[T] = [ML^2T^-3]

Now compare the four given pairs:

torque [ML^2T^-2] vs power [ML^2T^-3] -> different (T exponent differs)

force [MLT^-2] vs power [ML^2T^-3] -> different …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. The Dimensional formula for strain :$
(a) ML^-2 T^-1
(b) M^0 L^0 T^0
(c) ML^-1 T^-2
(d) M^0 L T^0
›Reveal solutionSolution

Strain is always a ratio of two quantities of the same kind (for example, change in length to original length, or change in volume to original volume), so all units cancel out and the dimensional formula is M^0 L^0 T^0.

Definition of (longitudinal) strain:

Strain = (change in length, ΔL) / (original length, L)

Dimensionally:

[Strain] = [L] / [L] = [L^0] = dimensionless

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Which of the following pairs of physical quantities have the same dimensions ?$
(a) Torque and Power
(b) Force and Torque
(c) Force and Power
(d) Torque and Energy
›Reveal solutionSolution
Torque and Energy have identical dimensions [ML^2T^-2], even though they represent physically different concepts.

Dimensional formulas:

Torque, tau = F x r (perpendicular distance) -> [MLT^-2][L] = [ML^2T^-2]

Energy (work), W = F x d -> [MLT^-2][L] = [ML^2T^-2]

Power, P = W/t -> [ML^2T^-2]/[T] = [ML^2T^-3]

Force, F -> [MLT^-2]

Checking each option:

Torque [ML^2T^-2] and Power [ML^2T^-3] -- different.

Force [MLT^-2] and Torque [ML^2T^-2] -- different.

Force [MLT^-2] and Power [ML^2T^-3] -- different.

Torque [ML^2T^-2] and Energy [ML^2T^-2] -- SAME. …
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Q.Explain the principle of homogeneity of dimensions. What are its uses? Give example.

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