Dimensional Analysis JEE 2023: Young’s Modulus Solution

What is the correct answer to the dimensional analysis JEE 2023 Young’s modulus question?

Option A is correct for the dimensional analysis JEE 2023 question on Young’s modulus. The source is JEE Advanced 2023, Paper 2, Physics, Units and Measurements, not JEE Main.

The task is to determine the exponents when Young’s modulus is expressed using the speed of light, Planck’s constant and the gravitational constant: $Y = c^{α} h^{β} G^{γ}$

The options give the exponents in this order: $(α, β, γ)$

This is a single-correct question, rated medium on the question bank’s scale. Its bank-listed target time is 90 seconds, not an official time limit.

How do you write the dimensions of Young’s modulus, Planck’s constant and the gravitational constant?

Start with Young’s modulus as stress divided by strain. Strain is a ratio of lengths, so it is dimensionless; Young’s modulus therefore has the dimensions of stress.

Y = \frac{stress}{strain}, [strain] = \frac{L}{L} = 1

Since stress is force per unit area:

[Y] = \frac{[force]}{[area]} = \frac{M L T^{- 2}}{L^{2}} = M L^{- 1} T^{- 2}

The speed of light has the dimensions of any speed: $[c] = L T^{- 1}$

For Planck’s constant, use energy equals Planck’s constant times frequency. Divide the dimensions of energy by those of frequency: $E = h ν$

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

For the gravitational constant, rearrange Newton’s gravitational-force equation before taking dimensions:

F = \frac{G m_{1} m_{2}}{r^{2}}

[G] = \frac{[F] [r]^{2}}{[m_{1}] [m_{2}]} = \frac{(M L T^{- 2}) L^{2}}{M^{2}} = M^{- 1} L^{3} T^{- 2}

These are dimensions, not numerical SI values. No values of the constants are needed.

How do you match the mass, length and time powers separately?

Substitute the dimensions, collect powers of each base dimension, then equate matching exponents. Keep the time equation unchanged until all three equations have been written to avoid an unnecessary sign change during setup.

Taking dimensions on both sides gives:

M L^{- 1} T^{- 2} = (L T^{- 1})^{α} (M L^{2} T^{- 1})^{β} (M^{- 1} L^{3} T^{- 2})^{γ}

Collect the right-hand powers:

M L^{- 1} T^{- 2} = M^{β - γ} L^{α + 2 β + 3 γ} T^{- α - β - 2 γ}

Match mass powers: $β - γ = 1$

Match length powers: $α + 2 β + 3 γ = - 1$

Match time powers: $- α - β - 2 γ = - 2$

Now multiply the entire time equation by negative one. Every term changes sign, including the right-hand side: $α + β + 2 γ = 2$

How do you solve the three equations and verify option A?

Use the mass equation to eliminate the Planck’s-constant exponent, then subtract the two remaining equations. This follows the official solution’s sequence and gives option A without testing the options.

From the mass equation:

β - γ = 1 \Rightarrow β = 1 + γ

Substitute into the length equation: $α + 2 (1 + γ) + 3 γ = - 1$ $α + 2 + 5 γ = - 1$ $α + 5 γ = - 3$

Substitute into the rewritten time equation: $α + (1 + γ) + 2 γ = 2$ $α + 3 γ = 1$

Subtract the latter equation from the former. Write both right-hand sides explicitly to protect the signs:

(α + 5 γ) - (α + 3 γ) = - 3 - 1

2 γ = - 4 \Rightarrow γ = - 2

Back-substitute into the mass relation: $β = 1 - 2 = - 1$

Then use the reduced time equation: $α + 3 (- 2) = 1$

α - 6 = 1 \Rightarrow α = 7

The answer is option A:

(α, β, γ) = (7, - 1, - 2)

For an independent audit, check the dimensions of the resulting combination:

\frac{c^{7}}{h G^{2}}

The mass, length and time powers are respectively:

M : & - 1 + 2 = 1 L : & 7 - 2 - 6 = - 1 T : & - 7 + 1 + 4 = - 2

Thus:

[\frac{c^{7}}{h G^{2}}] = M L^{- 1} T^{- 2}

All three powers agree with Young’s modulus. This check returns to the original dimensions rather than reusing the elimination steps.

Dimensional matching fixes these exponents, not a dimensionless prefactor. It does not establish this combination as the actual Young’s modulus of every material.

How can a sign error produce option B?

Start from the correctly derived relation and gravitational-constant exponent. Reversing the subtraction order while isolating the remaining exponent produces option B:

α + 3 γ = 1, γ = - 2

The incorrect rearrangement is:

α = 3 γ - 1 = 3 (- 2) - 1 = - 7

Keeping the other two exponents correct then produces option B. Isolating the remaining exponent requires subtracting the same term from both sides, not reversing the subtraction order: $α + 3 γ - 3 γ = 1 - 3 γ$

α = 1 - 3 γ = 1 - 3 (- 2) = 7

A dimensional audit rejects option B:

[c^{- 7} h^{- 1} G^{- 2}] = (L^{- 7} T^{7}) (M^{- 1} L^{- 2} T) (M^{2} L^{- 6} T^{4})

[c^{- 7} h^{- 1} G^{- 2}] = M L^{- 15} T^{12} \neq M L^{- 1} T^{- 2}

The mass power passes, but length and time fail. Write the subtraction explicitly and check all three base dimensions, not just mass.

How do you apply the method to two related Units and Measurements questions?

Energy density gives the same exponents as Young’s modulus, while constructing a length gives fractional exponents. Both questions below are original practice questions, not additional verified PYQs. Attempt each setup before reading its worked check.

Question 1: If energy per unit volume is written in the following form, determine the three exponents:

\frac{energy}{volume} = c^{p} h^{q} G^{r}

Its dimensions are:

[\frac{energy}{volume}] = \frac{M L^{2} T^{- 2}}{L^{3}} = M L^{- 1} T^{- 2}

These match Young’s modulus, so the same three exponent equations apply:

q - r = 1, p + 2 q + 3 r = - 1, - p - q - 2 r = - 2

Hence:

(p, q, r) = (7, - 1, - 2)

Equal dimensions do not make energy density and Young’s modulus the same physical quantity. One measures energy per volume; the other relates stress to strain.

Question 2: Construct a quantity with dimensions of length using: $c^{p} h^{q} G^{r}$

Matching against length gives:

q - r = 0, p + 2 q + 3 r = 1, - p - q - 2 r = 0

The mass and time equations give:

q = r, p = - 3 r

Substitute into the length equation: $- 3 r + 5 r = 1$

r = q = \frac{1}{2}, p = - \frac{3}{2}

Therefore:

c^{- 3 / 2} h^{1 / 2} G^{1 / 2} = \sqrt{\frac{h G}{c^{3}}}

Check it directly:

[\frac{h G}{c^{3}}] = \frac{(M L^{2} T^{- 1}) (M^{- 1} L^{3} T^{- 2})}{L^{3} T^{- 3}} = L^{2}

Taking the square root gives length. Now cover both checks and rebuild the exponent equations from the target dimensions alone.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Keep going with How to Study Differentiation JEE: Rules and Practice.

Frequently asked questions

What is the answer to the JEE 2023 dimensional analysis question on Young’s modulus?

Option A is correct: the exponents in Y = c^α h^β G^γ are (α, β, γ) = (7, -1, -2). The question appeared in JEE Advanced 2023, Paper 2, Physics, not JEE Main.

What are the dimensions of Young’s modulus, Planck’s constant and the gravitational constant?

Young’s modulus has dimensions ML^-1 T^-2 because it is stress divided by dimensionless strain. Planck’s constant has dimensions ML^2 T^-1, obtained from E = hν. The gravitational constant has dimensions M^-1 L^3 T^-2, obtained from Newton’s gravitational-force equation.

Why is option B wrong in the JEE 2023 Young’s modulus question?

Option B gives (-7, -1, -2), which produces dimensions ML^-15 T^12 rather than ML^-1 T^-2. It can result from incorrectly rearranging α + 3γ = 1 as α = 3γ - 1. The correct rearrangement is α = 1 - 3γ, giving α = 7 when γ = -2.

Does dimensional analysis prove that Young’s modulus equals c^7/(hG^2)?

No: dimensional matching fixes the exponents but cannot determine a dimensionless prefactor. The combination c^7/(hG^2) has the dimensions of Young’s modulus, but this does not establish it as the actual Young’s modulus of every material.