Dimensional Analysis JEE 2024: Advanced Paper 1 Solution

What is the dimensional analysis question from JEE Advanced 2024?

Option A is correct for this JEE Advanced 2024 dimensional analysis question. The source is JEE Advanced 2024, Paper 1, Physics, Units and Measurements, not JEE Main.

This is a single-correct question tagged medium by the question bank. Its suggested solving time is 120 seconds, not an official time limit.

Choose the exponent tuple that makes this product dimensionless: $e^{α} ε_{0}^{β} h^{γ} c^{δ}$

The constants are electronic charge, vacuum permittivity, Planck’s constant and the speed of light, respectively. The tuple order and the question’s non-zero integer restriction are:

(α, β, γ, δ), n \in Z, n \neq 0

The supplied choices are:

How do you derive the dimensions of all four constants?

Use charge equals current multiplied by time, Coulomb’s law, the energy-frequency relation and speed equals distance divided by time. These give every dimensional input without a memorised expression for vacuum permittivity. Pay particular attention to permittivity: dividing by force reverses its negative time power.

Use these fundamental dimension symbols:

M : mass, L : length, T : time, A : electric current

The dimension symbol for electric current is distinct from answer option A. Since charge equals current multiplied by time: $[e] = [A T]$

Start from Coulomb’s law:

F = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}

The numerical factor is dimensionless: $[4 π] = 1$

Rearranging dimensionally gives:

[ε_{0}] = \frac{[q]^{2}}{[F] [r]^{2}} = \frac{A^{2} T^{2}}{(M L T^{- 2}) L^{2}} = M^{- 1} L^{- 3} T^{4} A^{2}

The fourth power of time is positive: the numerator contributes two powers, and dividing by force contributes two more. Do not carry the force’s negative time power unchanged into the answer.

Energy equals Planck’s constant multiplied by frequency. Dividing energy dimensions by frequency dimensions gives:

E = h ν, [h] = \frac{M L^{2} T^{- 2}}{T^{- 1}} = M L^{2} T^{- 1}

Speed is distance divided by time: $[c] = L T^{- 1}$

How do you set up all four dimensional equations?

Substitute the dimensions, collect like powers and set each net exponent to zero. A dimensionless product must cancel every fundamental dimension, not just electric current. Follow the official solution’s order: current first, then mass, length and time.

The full substitution is:

[e^{α} ε_{0}^{β} h^{γ} c^{δ}] = (A T)^{α} (M^{- 1} L^{- 3} T^{4} A^{2})^{β} (M L^{2} T^{- 1})^{γ} (L T^{- 1})^{δ}

Collecting powers gives:

M^{- β + γ} L^{- 3 β + 2 γ + δ} T^{α + 4 β - γ - δ} A^{α + 2 β}

Dimensionless means: $M^{0} L^{0} T^{0} A^{0}$

Each net exponent must therefore vanish:

Electric current: $α + 2 β = 0$
Mass:

- β + γ = 0 \Rightarrow γ = β

Length, after substituting the mass relation: $- 3 β + 2 γ + δ = 0$

- 3 β + 2 β + δ = 0 \Rightarrow δ = β

Time: $α + 4 β - γ - δ = 0$

Do not assume these are four independent constraints merely because there are four equations. Check whether substitution makes one equation repeat another.

Why is one exponent free, and how does that give option A?

The time equation repeats the electric-current equation, leaving three independent constraints for four exponents. One exponent therefore remains free. The repeated equation is a consistency check, not a fourth independent constraint.

Substitute the mass and length relations into the time equation:

α + 4 β - β - β = α + 2 β = 0

The complete set of relations is:

α = - 2 β, γ = β, δ = β

Use the question’s parametrisation: $β = - n$

Then:

α = 2 n, γ = - n, δ = - n

The answer is option A:

(α, β, γ, δ) = (2 n, - n, - n, - n)

The product becomes:

e^{2 n} ε_{0}^{- n} h^{- n} c^{- n} = {(\frac{e^{2}}{ε_{0} h c})}^{n}

Verify numerator and denominator directly:

[ε_{0} h c] = (M^{- 1} L^{- 3} T^{4} A^{2}) (M L^{2} T^{- 1}) (L T^{- 1}) = A^{2} T^{2} = [e^{2}]

For a numerical exponent check, take:

n = 1, (α, β, γ, δ) = (2, - 1, - 1, - 1)

All four net powers vanish:

M : & 1 - 1 = 0 L : & 3 - 2 - 1 = 0 T : & 2 - 4 + 1 + 1 = 0 A : & 2 - 2 = 0

A zero parameter gives the trivial product:

n = 0 \Rightarrow e^{0} ε_{0}^{0} h^{0} c^{0} = 1

The question excludes that case. Dimensional analysis establishes exponent ratios; it does not itself require integer exponents.

Dimensionless does not mean numerically equal to one. Dimensional analysis also cannot determine an additional dimensionless numerical multiplier.

Why can checking only electric current wrongly validate option D?

Option D passes the electric-current check but fails the mass check. Accepting an option after one cancellation is therefore an incomplete method. Passing any one dimensional check is necessary but not sufficient, and it does not uniquely select the answer.

For option D, the current equation gives:

α = 2 n, β = - n

$α + 2 β = 2 n - 2 n = 0$ Now test mass:

β = - n, γ = - 2 n

$- β + γ = n - 2 n = - n \neq 0$ The remaining mass power is non-zero because the parameter is non-zero. Hence D is not dimensionless, despite cancelling electric current.

Derive and check the complete constraint set. Use the repeated time equation as confirmation rather than stopping after the first successful cancellation.

How can you practise the same method on two related questions?

Solve the two original related practice questions below by balancing all four dimensions. These are not additional verified JEE PYQs. Attempt each before reading its worked answer, and keep numerical equality separate from dimensional agreement.

How do you give a product the dimensions of electric charge?

Match current first, then mass and length; check time last. Question 1: Determine the three exponents if this product has the dimensions of electric charge:

X = ε_{0}^{p} h^{q} c^{r}, [X] = A T

The balances give:

2 p = 1 \Rightarrow p = \frac{1}{2}

- p + q = 0 \Rightarrow q = \frac{1}{2}

- 3 p + 2 q + r = 0 \Rightarrow r = \frac{1}{2}

Check time:

4 p - q - r = 2 - \frac{1}{2} - \frac{1}{2} = 1

Therefore: $[\sqrt{ε_{0} h c}] = [e]$

The square root has charge dimensions. Dimensional agreement alone does not prove numerical equality with electronic charge.

Can the product be dimensionless without Planck’s constant?

No nontrivial solution exists: mass, current and length balances force every exponent to zero. Question 2: Can this product be dimensionless with at least one non-zero exponent? $e^{a} ε_{0}^{b} c^{d}$

The balances give:

- b = 0 \Rightarrow b = 0

a + 2 b = 0 \Rightarrow a = 0

- 3 b + d = 0 \Rightarrow d = 0

Time then also cancels: $a + 4 b - d = 0$

Removing Planck’s constant removes the required mass-dimension cancellation. Check mass first here, then carry the forced zero through the remaining equations.

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Frequently asked questions

What is the answer to the JEE Advanced 2024 dimensional analysis question?

Option A is correct: (α, β, γ, δ) = (2n, -n, -n, -n), where n is a non-zero integer. This makes e^α ε₀^β h^γ c^δ dimensionless because ε₀hc has the same dimensions as e². The question is from JEE Advanced 2024 Paper 1 Physics, not JEE Main.

How do I derive the dimensions of vacuum permittivity?

Rearrange Coulomb’s law to write [ε₀] = [q]²/([F][r]²). Substituting [q] = AT, [F] = MLT^-2 and [r] = L gives [ε₀] = M^-1 L^-3 T^4 A², where A denotes electric current. The time exponent is positive four because dividing by force contributes two additional powers of time.

Why is one exponent free in the JEE Advanced 2024 question?

There are only three independent dimensional constraints for four exponents. Mass and length give γ = β and δ = β, after which the time equation reduces to α + 2β = 0, identical to the current equation. Choosing β = -n therefore gives the complete tuple (2n, -n, -n, -n).

Why is option D wrong if electric current cancels?

For option D, (α, β, γ, δ) = (2n, -n, -2n, -2n), so the current exponent α + 2β is zero. However, the mass exponent -β + γ equals -n, which is non-zero under the question’s restriction. A dimensionless product must cancel every fundamental dimension, not just current.