Dimensions of Physical Quantities & Dimensional Formulae of Derived Quantities | JEE Physics

Concept Card: Dimensions of Physical Quantities & Dimensional Formulae

1. Definition of Dimensions:
The dimensions of a physical quantity are the powers (or exponents) to which the fundamental base quantities must be raised to represent that quantity.

The standard dimensional expression is represented in terms of the seven base dimensions:
$[Q] = [M^a L^b T^c A^d K^e \text{mol}^f \text{cd}^g]$
where $a, b, c, d, e, f, g$ are the dimensions of mass, length, time, electric current, temperature, amount of substance, and luminous intensity respectively.

2. Classification of Physical Quantities:

• Dimensional Variables: Quantities possessing physical dimensions that assume variable values.
  Examples: Velocity $[L T^{-1}]$, Acceleration $[L T^{-2}]$, Force $[M L T^{-2}]$, Energy $[M L^2 T^{-2}]$.
• Dimensionless Variables: Quantities possessing no physical dimensions ($[M^0 L^0 T^0]$) whose numerical values vary.
  Examples: Strain ($\Delta L / L$), Specific Gravity, Refractive Index, Relative Permittivity, Dielectric Constant.
• Dimensional Constants: Quantities possessing constant numerical values while having definite physical dimensions.
  Examples: Universal Gravitational Constant $G$, Planck’s Constant $h$, Speed of Light $c$, Boltzmann Constant $k_B$, Permittivity of Free Space $\varepsilon_0$, Permeability of Free Space $\mu_0$.
• Dimensionless Constants: Pure numerical values or mathematical constants lacking both units and dimensions.
  Examples: $\pi$, Euler’s number $e$, pure geometric ratios, integer factors.

3. Comprehensive Reference Table of Derived Quantities:

Physical Quantity	Governing Physical Relation	Dimensional Formula
Linear Momentum ($p$)	$p = m v$	$[M L T^{-1}]$
Force ($F$)	$F = m a$	$[M L T^{-2}]$
Work / Energy ($W, E$)	$W = \vec{F}\cdot\vec{s}$	$[M L^2 T^{-2}]$
Torque ($\tau$)	$\vec{\tau} = \vec{r}\times\vec{F}$	$[M L^2 T^{-2}]$
Power ($P$)	$P = W / t$	$[M L^2 T^{-3}]$
Pressure / Stress	$P = F / A$	$[M L^{-1} T^{-2}]$
Modulus of Elasticity ($Y, B, \eta$)	$\text{Stress} / \text{Strain}$	$[M L^{-1} T^{-2}]$
Surface Tension ($T$)	$T = F / L$	$[M T^{-2}]$
Coefficient of Viscosity ($\eta$)	$F = 6\pi \eta r v$	$[M L^{-1} T^{-1}]$
Universal Gravitational Constant ($G$)	$F = G m_1 m_2 / r^2$	$[M^{-1} L^3 T^{-2}]$
Planck’s Constant ($h$)	$E = h \nu$	$[M L^2 T^{-1}]$
Boltzmann Constant ($k_B$)	$E = \frac{3}{2} k_B T$	$[M L^2 T^{-2} K^{-1}]$
Thermal Conductivity ($K$)	$\frac{dQ}{dt} = K A \frac{\Delta T}{L}$	$[M L T^{-3} K^{-1}]$
Stefan’s Constant ($\sigma$)	$E = \sigma T^4$	$[M T^{-3} K^{-4}]$
Electric Potential / EMF ($V$)	$V = W / q$	$[M L^2 T^{-3} A^{-1}]$
Capacitance ($C$)	$C = q / V$	$[M^{-1} L^{-2} T^4 A^2]$
Resistance ($R$)	$R = V / I$	$[M L^2 T^{-3} A^{-2}]$
Inductance ($L$)	$\mathcal{E} = -L \frac{dI}{dt}$	$[M L^2 T^{-2} A^{-2}]$
Magnetic Field Induction ($B$)	$F = q v B$	$[M T^{-2} A^{-1}]$
Magnetic Flux ($\Phi_B$)	$\Phi_B = B A$	$[M L^2 T^{-2} A^{-1}]$
Permittivity ($\varepsilon_0$)	$F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2}$	$[M^{-1} L^{-3} T^4 A^2]$
Permeability ($\mu_0$)	$F = \frac{\mu_0 I_1 I_2 L}{2\pi r}$	$[M L T^{-2} A^{-2}]$

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Solved Examples

Example 1 (Derivation of Coefficient of Viscosity & Planck’s Constant):
Derive the dimensional formulae for:
(a) Coefficient of viscosity $\eta$ using Stokes’ Law $F = 6\pi \eta r v$, where $F$ is viscous drag, $r$ is sphere radius, and $v$ is terminal speed.
(b) Planck’s constant $h$ using Einstein’s relation $E = h \nu$, where $E$ is photon energy and $\nu$ is radiation frequency.

Solution:
(a) Rearranging Stokes’ Law:
$\eta = \frac{F}{6\pi r v}$
Since $6\pi$ is a dimensionless number ($[6\pi] = [1]$):
$[\eta] = \frac{[F]}{[r][v]} = \frac{[M L T^{-2}]}{[L][L T^{-1}]} = \frac{[M L T^{-2}]}{[L^2 T^{-1}]} = [M L^{-1} T^{-1}]$.
(b) From Einstein’s equation:
$h = \frac{E}{\nu}$
Dimensions of Energy $[E] = [M L^2 T^{-2}]$ and frequency $[\nu] = [T^{-1}]$:
$[h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$.
Notice that Planck’s constant possesses the same dimensions as angular momentum ($L = m v r \implies [M L^2 T^{-1}]$).

Example 2 (Determining Dimensions of Constants via Homogeneity Principle):
In the relation $P = \frac{a – t^2}{b x}$, where $P$ is pressure, $x$ is displacement, and $t$ is time, find the dimensional formula of the ratio $\frac{a}{b}$.

Solution:
1. By the principle of homogeneity, quantities added or subtracted must possess identical dimensions:
$[a] = [t^2] = [T^2]$.
2. Analyzing the entire equation:
$[P] = \frac{[a – t^2]}{[b][x]} = \frac{[T^2]}{[b][L]}$
Since Pressure has dimensions $[P] = [M L^{-1} T^{-2}]$:
$[M L^{-1} T^{-2}] = \frac{[T^2]}{[b][L]} \implies [b] = \frac{[T^2]}{[M L^{-1} T^{-2}][L]} = \frac{[T^2]}{[M T^{-2}]} = [M^{-1} T^4]$.
3. Calculating the dimensional formula of $\frac{a}{b}$:
$\left[\frac{a}{b}\right] = \frac{[a]}{[b]} = \frac{[T^2]}{[M^{-1} T^4]} = [M T^{-2}] = [M^1 L^0 T^{-2}]$.

Example 3 (Dimensional Analysis of Electromagnetic Combinations):
Find the dimensional formula of the following fundamental electromagnetic combinations:
(a) $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
(b) The characteristic wave impedance of free space $Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}$
(c) The ratio of electric field to magnetic field $\frac{E}{B}$.

Solution:
(a) From Maxwell’s electromagnetic theory, $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ represents the speed of light in vacuum ($c$):
$\left[\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right] = [c] = [L T^{-1}]$.
(b) For impedance of free space $Z_0$:
$[\mu_0] = [M L T^{-2} A^{-2}]$ and $[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$
$\left[\frac{\mu_0}{\varepsilon_0}\right] = \frac{[M L T^{-2} A^{-2}]}{[M^{-1} L^{-3} T^4 A^2]} = [M^2 L^4 T^{-6} A^{-4}]$
$\left[Z_0\right] = \sqrt{[M^2 L^4 T^{-6} A^{-4}]} = [M L^2 T^{-3} A^{-2}]$.
Notice that $[M L^2 T^{-3} A^{-2}]$ is identical to the dimensions of electrical resistance ($[R]$). In vacuum, $Z_0 \approx 377\,\Omega$.
(c) From the Lorentz force relation $F_E = q E$ and $F_B = q v B$, when electric and magnetic forces balance:
$q E = q v B \implies \frac{E}{B} = v$
$\left[\frac{E}{B}\right] = [v] = [L T^{-1}]$.

Example 4 (Expressing Fundamental Quantities in Terms of Universal Constants):
If the speed of light $c$, Planck’s constant $h$, and the universal gravitational constant $G$ are taken as fundamental base quantities, express mass ($M$) in terms of $c, h, G$.

Solution:
Let $[M] = [c]^x [h]^y [G]^z$.
Substitute the respective dimensional formulas:
$[M^1 L^0 T^0] = [L T^{-1}]^x [M L^2 T^{-1}]^y [M^{-1} L^3 T^{-2}]^z$
$[M^1 L^0 T^0] = [M^{y – z} L^{x + 2y + 3z} T^{-x – y – 2z}]$
Equating exponents:
1. For $M$: $y – z = 1 \implies y = z + 1$
2. For $L$: $x + 2y + 3z = 0$
3. For $T$: $-x – y – 2z = 0 \implies x = -y – 2z$
Substitute (3) into (2):
$(-y – 2z) + 2y + 3z = 0 \implies y + z = 0 \implies y = -z$
Equating with (1):
$-z = z + 1 \implies 2z = -1 \implies z = -\frac{1}{2}$
$y = -z = \frac{1}{2}$
$x = -y – 2z = -\frac{1}{2} – 2\left(-\frac{1}{2}\right) = -\frac{1}{2} + 1 = \frac{1}{2}$
Hence, $[M] = [c^{1/2} h^{1/2} G^{-1/2}] = \sqrt{\frac{c h}{G}}$ (known as Planck Mass).

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Worksheet: 10 Practice Problems

Problem 1 (JEE Main – Single Correct):
Which of the following pairs of physical quantities have the exact same dimensions?
(A) Torque and Angular momentum
(B) Torque and Work
(C) Linear momentum and Work
(D) Force and Torque

Problem 2 (JEE Main – Single Correct):
The dimensional formula for the ratio of Planck’s constant $h$ to elementary charge $e$ (i.e., $[h/e]$) is:
(A) $[M L^2 T^{-1} A^{-1}]$
(B) $[M L^2 T^{-2} A^{-1}]$
(C) $[M L T^{-1} A^{-2}]$
(D) $[M L^2 T^{-3} A^{-1}]$

Problem 3 (JEE Main – Single Correct):
The velocity $v$ of a particle at time $t$ is given by the relation $v = a t + \frac{b}{t + c}$. The dimensional formulae of $a, b$, and $c$ are respectively:
(A) $[L T^{-2}], [L], [T]$
(B) $[L^2], [T], [L T^{-2}]$
(C) $[L T^2], [L T], [T]$
(D) $[L], [L T], [T^2]$

Problem 4 (JEE Main – Single Correct):
Stefan’s law states that radiant energy emitted per unit area per unit time by a blackbody is $E = \sigma T^4$. The dimensional formula of Stefan’s constant $\sigma$ is:
(A) $[M L^0 T^{-3} K^{-4}]$
(B) $[M L^2 T^{-3} K^{-4}]$
(C) $[M L^{-1} T^{-2} K^{-4}]$
(D) $[M L T^{-3} K^{-2}]$

Problem 5 (JEE Main – Single Correct):
Which of the following physical quantities has the dimensional formula $[M L^{-1} T^{-1}]$?
(A) Surface Tension
(B) Modulus of Rigidity
(C) Coefficient of Viscosity
(D) Power

Problem 6 (JEE Advanced – One or More Correct):
Which of the following groups of physical quantities share identical dimensions?
(A) Pressure, Stress, Young’s Modulus
(B) Work, Heat, Energy, Torque
(C) Electric potential, Electromotive force (EMF), Potential difference
(D) Surface tension, Spring constant, Force per unit length

Problem 7 (JEE Advanced – One or More Correct):
In electrical circuit theory, if $L$ represents inductance, $C$ represents capacitance, and $R$ represents electrical resistance, which of the following combinations have the dimension of time $[T]$?
(A) $\frac{L}{R}$
(B) $R C$
(C) $\sqrt{L C}$
(D) $\frac{1}{R C}$

Problem 8 (JEE Advanced – One or More Correct):
Let $\varepsilon_0$ denote the permittivity of vacuum and $\mu_0$ denote the permeability of vacuum. Which of the following statements is/are correct?
(A) $[\mu_0 \varepsilon_0] = [L^{-2} T^2]$
(B) $\left[\sqrt{\frac{\mu_0}{\varepsilon_0}}\right] = [M L^2 T^{-3} A^{-2}]$
(C) $\left[\frac{1}{2} \varepsilon_0 E^2\right] = [M L^{-1} T^{-2}]$ (energy density)
(D) $\left[\frac{B^2}{2\mu_0}\right] = [M L^{-1} T^{-2}]$ (magnetic energy density)

Problem 9 (JEE Main / Advanced – Numerical Value Type):
The dimensional formula of the magnetic permeability of free space $\mu_0$ is expressed as $[M^a L^b T^c A^d]$. Determine the numerical value of $|a| + |b| + |c| + |d|$.

Problem 10 (JEE Main / Advanced – Numerical Value Type):
The capacitance $C$ of a parallel plate capacitor has the dimensional formula $[M^{-1} L^{-2} T^x A^y]$. Determine the integer value of $x + y$.

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Solutions & Explanations

Answer Key Summary:
1. (B) | 2. (B) | 3. (A) | 4. (A) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C) | 8. (A, B, C, D) | 9. 6 | 10. 6

Solution 1:
– Torque $\vec{\tau} = \vec{r} \times \vec{F} \implies [\tau] = [L][M L T^{-2}] = [M L^2 T^{-2}]$.
– Work $W = \vec{F} \cdot \vec{d} \implies [W] = [M L T^{-2}][L] = [M L^2 T^{-2}]$.
Both torque and work have identical dimensions $[M L^2 T^{-2}]$.
Correct Answer: (B)

Solution 2:
Planck’s constant: $[h] = [M L^2 T^{-1}]$.
Elementary charge: $[e] = [q] = [A T]$.
Ratio: $\left[\frac{h}{e}\right] = \frac{[M L^2 T^{-1}]}{[A T]} = [M L^2 T^{-2} A^{-1}]$.
Notice this corresponds to magnetic flux ($\Phi_B = L I \implies [M L^2 T^{-2} A^{-1}]$).
Correct Answer: (B)

Solution 3:
From the principle of homogeneity:
1. In the denominator $(t + c)$, quantities added must have identical dimensions $\implies [c] = [t] = [T]$.
2. In $v = a t$, $[a t] = [v] \implies [a][T] = [L T^{-1}] \implies [a] = [L T^{-2}]$.
3. In $\frac{b}{t + c} = v$, $\frac{[b]}{[T]} = [L T^{-1}] \implies [b] = [L]$.
Thus, $[a] = [L T^{-2}]$, $[b] = [L]$, $[c] = [T]$.
Correct Answer: (A)

Solution 4:
Emissive power $E = \frac{\text{Energy}}{\text{Area} \times \text{Time}} \implies [E] = \frac{[M L^2 T^{-2}]}{[L^2][T]} = [M T^{-3}]$.
Stefan’s Law: $E = \sigma T^4 \implies \sigma = \frac{E}{T^4}$.
$[\sigma] = \frac{[M T^{-3}]}{[K^4]} = [M L^0 T^{-3} K^{-4}]$.
Correct Answer: (A)

Solution 5:
From Stokes’ Law, $F = 6\pi \eta r v \implies \eta = \frac{F}{6\pi r v}$.
$[\eta] = \frac{[M L T^{-2}]}{[L][L T^{-1}]} = [M L^{-1} T^{-1}]$.
– Surface tension: $[M T^{-2}]$
– Modulus of rigidity: $[M L^{-1} T^{-2}]$
– Power: $[M L^2 T^{-3}]$.
Correct Answer: (C)

Solution 6:
– (A) All represent force per unit area: $[M L^{-1} T^{-2}]$.
– (B) Work, Heat, Energy, Torque all share $[M L^2 T^{-2}]$.
– (C) Electric potential, EMF, and potential difference all represent work per unit charge: $[M L^2 T^{-3} A^{-1}]$.
– (D) Surface tension ($F/L$) and Spring constant ($F/x$) both have $[M T^{-2}]$.
All sets are correct.
Correct Answer: (A, B, C, D)

Solution 7:
– Inductive time constant: $\tau_L = \frac{L}{R} \implies [T]$.
– Capacitive time constant: $\tau_C = R C \implies [T]$.
– Resonant oscillation frequency $\omega = \frac{1}{\sqrt{L C}} \implies [\sqrt{L C}] = \left[\frac{1}{\omega}\right] = [T]$.
– $\frac{1}{R C}$ has dimensions of frequency $[T^{-1}]$.
Hence, (A), (B), and (C) have dimensions of time.
Correct Answer: (A, B, C)

Solution 8:
– (A) $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies \mu_0 \varepsilon_0 = \frac{1}{c^2} = [L^{-2} T^2]$. True.
– (B) $\sqrt{\frac{\mu_0}{\varepsilon_0}}$ is wave impedance of vacuum ($377\,\Omega$), which has dimensions of resistance: $[M L^2 T^{-3} A^{-2}]$. True.
– (C) $\frac{1}{2}\varepsilon_0 E^2$ is electrostatic energy density ($\text{Energy}/\text{Volume} = [M L^2 T^{-2}] / [L^3] = [M L^{-1} T^{-2}]$). True.
– (D) $\frac{B^2}{2\mu_0}$ is magnetic energy density ($\text{Energy}/\text{Volume} = [M L^{-1} T^{-2}]$). True.
All statements are correct.
Correct Answer: (A, B, C, D)

Solution 9:
From Biot-Savart Law / Ampere’s Force Law: $F = \frac{\mu_0 I_1 I_2 L}{2\pi r} \implies \mu_0 = \frac{2\pi r F}{I_1 I_2 L}$.
$[\mu_0] = \frac{[L][M L T^{-2}]}{[A^2][L]} = [M^1 L^1 T^{-2} A^{-2}]$.
Here $a = 1, b = 1, c = -2, d = -2$.
Sum of absolute values: $|a| + |b| + |c| + |d| = 1 + 1 + 2 + 2 = 6$.
Correct Answer: 6

Solution 10:
Capacitance $C = \frac{q}{V} = \frac{q^2}{W}$.
$[q] = [A T] \implies [q^2] = [A^2 T^2]$.
$[W] = [M L^2 T^{-2}]$.
$[C] = \frac{[A^2 T^2]}{[M L^2 T^{-2}]} = [M^{-1} L^{-2} T^4 A^2]$.
Comparing with $[M^{-1} L^{-2} T^x A^y]$:
$x = 4$ and $y = 2$.
$x + y = 4 + 2 = 6$.
Correct Answer: 6