Dimensional Analysis — Quick Verification

Why Dimensional Analysis Matters

When we forget a formula in the middle of a JEE question, dimensional analysis is the parachute. It cannot give us the exact numerical constants, but it can tell us whether $T = 2\pi\sqrt{L/g}$ or $T = 2\pi\sqrt{g/L}$ — and that is usually the difference between right and wrong. Every physicist, from the JEE topper to Nobel laureates, uses dimensional analysis as a sanity check before submitting an answer.

The core idea: every physical quantity has a unique combination of base dimensions — mass [M], length [L], time [T], current [A], temperature [K], amount [mol], luminous intensity [cd]. Equations in physics must have the same dimensions on both sides. If they don’t, the equation is wrong, period.

For Class 11, JEE Main, and NEET, dimensional analysis appears as direct questions (find the dimensions of viscosity), as derivation questions (derive the time period of a pendulum), and as elimination tricks in MCQs.

Key Terms & Definitions

Base dimensions: [M], [L], [T], [A], [K], [mol], [cd] — the seven SI base quantities.

Derived dimension: combination of base dimensions, like force $[MLT^{-2}]$ or pressure $[ML^{-1}T^{-2}]$ .

Dimensional formula: the symbolic expression of a quantity’s dimensions, e.g. $[F] = [MLT^{-2}]$ .

Dimensionless quantity: ratio of two quantities with the same dimensions, like strain, refractive index, angles in radians.

Principle of homogeneity: every term in a physical equation must have the same dimensions.

Methods & Concepts

Use 1: Verifying equations

Given a candidate equation, check both sides have the same dimensions. If not, the equation is wrong.

Example: Check $v = u + at$ .

LHS: $[LT^{-1}]$ . RHS: $[LT^{-1}] + [LT^{-2}][T] = [LT^{-1}] + [LT^{-1}]$ . Both terms match LHS. Equation is dimensionally consistent.

Use 2: Deriving relations (with one unknown formula)

If a quantity $Q$ depends on three other quantities $a$ , $b$ , $c$ , assume $Q = k\,a^x b^y c^z$ where $k$ is dimensionless. Equate dimensions on both sides to get three linear equations in $x$ , $y$ , $z$ .

$Q \propto a^x b^y c^z \Rightarrow [Q] = [a]^x [b]^y [c]^z$

Example: Find how the time period $T$ of a simple pendulum depends on length $L$ , mass $m$ , and $g$ .

Assume $T = k m^a L^b g^c$ . Then $[T] = [M]^a [L]^b [LT^{-2}]^c = [M]^a [L]^{b+c} [T]^{-2c}$ .

Match: $a = 0$ , $b + c = 0$ , $-2c = 1$ → $c = -1/2$ , $b = 1/2$ . Hence $T \propto \sqrt{L/g}$ .

The dimensionless constant $k = 2\pi$ cannot be found this way — only the form of the dependence.

Use 3: Converting units

The numerical value times the unit is the same in any system. So if we want to convert $1$ N (SI) to dynes (CGS): $1\,\text{N} = 1\,kg\cdot m/s^2 = 10^3 g \cdot 10^2 cm / s^2 = 10^5\,g\cdot cm/s^2 = 10^5$ dynes.

Use 4: Eliminating wrong MCQ options

If a JEE option has the wrong dimensions, eliminate it without solving the problem. This trick has won countless ranks.

Solved Examples

Example 1 (CBSE — Easy)

Find the dimensions of universal gravitation constant $G$ from $F = Gm_1m_2/r^2$ .

$[G] = [F][L]^2/[M]^2 = [MLT^{-2}][L^2]/[M^2] = [M^{-1}L^3T^{-2}]$ .

Example 2 (JEE Main — Medium)

The viscous force on a sphere of radius $r$ moving at speed $v$ through a fluid of viscosity $\eta$ is $F = k\eta r v$ . Verify dimensionally and find $[\eta]$ .

$[\eta] = [F]/([r][v]) = [MLT^{-2}]/([L][LT^{-1}]) = [ML^{-1}T^{-1}]$ .

This is the SI unit pascal-second (Pa·s).

Example 3 (JEE Advanced — Hard)

The energy $E$ of a photon depends on its frequency $\nu$ and Planck’s constant $h$ . Find the form of $E(\nu, h)$ from dimensions, knowing $[h] = [ML^2T^{-1}]$ .

Assume $E = k h^a \nu^b$ . $[E] = [ML^2T^{-2}]$ .

$[ML^2T^{-2}] = [M^a L^{2a} T^{-a}][T^{-b}] = [M^a L^{2a} T^{-a-b}]$ .

Match: $a = 1$ , $2a = 2$ ✓, $-a-b = -2 \Rightarrow b = 1$ . So $E \propto h\nu$ . (Numerical $k = 1$ .)

Exam-Specific Tips

For JEE Main, dimensional formulas appear in 1-2 MCQs every year. Memorise the dimensions of: viscosity $[ML^{-1}T^{-1}]$ , surface tension $[MT^{-2}]$ , Planck’s constant $[ML^2T^{-1}]$ , gravitational constant $[M^{-1}L^3T^{-2}]$ .

For NEET, dimensional analysis is mostly used to eliminate options. Always check option dimensions before computing.

For CBSE 11, the derivation of pendulum period is a 3-mark question. Practise the method until the algebra is mechanical.

Common Mistakes to Avoid

1. Adding quantities of different dimensions. $v + at$ is fine ( $v$ and $at$ both $[LT^{-1}]$ ). $v + s$ is not. Watch for this in board exam derivations.

2. Forgetting that trigonometric, exponential, and logarithmic functions take dimensionless arguments. $\sin(kx)$ requires $kx$ to be dimensionless, so $[k] = [L^{-1}]$ .

3. Treating dimensional constants as dimensionless. Constants like $G$ , $h$ , $k_B$ have dimensions and cannot be ignored.

4. Assuming dimensional analysis can fix any equation. It cannot determine numerical factors like $2\pi$ , $1/2$ , or $4\pi$ . It also fails for equations involving more than three variables (more unknowns than equations).

5. Confusing units with dimensions. Joule and erg are units; energy has dimensions $[ML^2T^{-2}]$ in any unit system.

Practice Questions

Q1. Find $[k]$ in $F = -kx$ for a spring.

$[k] = [F]/[L] = [MLT^{-2}]/[L] = [MT^{-2}]$ . Same as surface tension dimensions, by coincidence.

Q2. Find $[B]$ in $F = qvB$ .

$[B] = [F]/([q][v]) = [MLT^{-2}]/([AT][LT^{-1}]) = [MA^{-1}T^{-2}]$ .

Q3. Verify $E = \tfrac{1}{2}mv^2$ dimensionally.

$[E] = [ML^2T^{-2}]$ . RHS: $[M][LT^{-1}]^2 = [ML^2T^{-2}]$ . Consistent.

Q4. What is the dimension of $RC$ (resistance times capacitance)?

$RC$ is the time constant of an RC circuit, so $[RC] = [T]$ .

Q5. Find dimensions of $a$ and $b$ in van der Waals equation $(P + a/V^2)(V - b) = nRT$ .

$b$ has dimensions of volume $[L^3]$ . $a/V^2$ has dimensions of pressure $[ML^{-1}T^{-2}]$ , so $[a] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$ .

Q6. The radius of orbit of an electron in a hydrogen atom depends on $\hbar$ , $m_e$ , $e$ , $\varepsilon_0$ . Use dimensional analysis to find the dependence.

This is the Bohr radius derivation. $r \propto 4\pi\varepsilon_0 \hbar^2 / (m_e e^2)$ . Dimensions check via $[\varepsilon_0] = [M^{-1}L^{-3}T^4 A^2]$ .

Q7. Why does $E = mc^2$ pass dimensional check?

$[mc^2] = [M][LT^{-1}]^2 = [ML^2T^{-2}] = [E]$ . ✓

Q8. Can $y = A\sin(kx)$ work if $[k] = [L]$ ?

No. $\sin$ requires dimensionless argument, so $[k] = [L^{-1}]$ .

FAQs

Q: Can dimensional analysis prove an equation is correct?

No. It can only show that an equation is consistent. A wrong equation (like $E = mc$ ) might still pass if you missed a factor.

Q: What about angles — are they dimensionless?

Yes, angles in radians are dimensionless (length divided by length). Same for steradians.

Q: How do I find dimensions of new quantities like impedance?

Use the defining equation. Impedance: $V = IZ$ , so $[Z] = [V]/[I]$ — same as resistance.

Q: Why do we need dimensions if we have units?

Dimensions are unit-system-independent. They tell you the physical nature of a quantity. Units convert between systems, but dimensions are fundamental.

Q: Can I use dimensional analysis to find the coefficient of friction?

No. Coefficient of friction is dimensionless, so dimensional analysis tells us nothing about its value.

Q: What are the limits of dimensional analysis?

It cannot find dimensionless constants, cannot handle equations with more than three variables (without additional information), and cannot derive purely additive constants.

Q: Is “dimensional formula” the same as “dimensional equation”?

Roughly yes. The “dimensional formula” is the symbolic expression $[ML^2T^{-2}]$ . The “dimensional equation” is the full statement $[E] = [ML^2T^{-2}]$ .