JEE PYQ: Units & Measurements - Question ID fd93887d0bbd (JEE Main 2026)

Core Formula & Concept:

This question tests the ability to express the base units of the SI system—meter (L), second (S), kilogram (M), and kelvin (K)—in terms of three fundamental constants: Planck’s constant ($h$), the gravitational constant ($G$), and the speed of light in vacuum ($c$). The key concept is dimensional analysis: we write each SI unit as a product of powers of $h$, $G$, and $c$, then match the resulting expression to one of the given forms in List-II.

The dimensions of the constants are:

• $[h] = \text{ML}^2\text{T}^{-1}$
• $[G] = \text{M}^{-1}\text{L}^3\text{T}^{-2}$
• $[c] = \text{LT}^{-1}$

Step-by-Step Derivation:

Step 1 – Express each SI unit in terms of $h$, $G$, $c$

We seek exponents $x$, $y$, $z$ such that $\text{Unit} = h^x\,G^y\,c^z$ and then equate dimensions on both sides.

A. Meter (L)

Set $\text{L} = h^x\,G^y\,c^z$. Equate dimensions: $\text{L} = (\text{ML}^2\text{T}^{-1})^x\,(\text{M}^{-1}\text{L}^3\text{T}^{-2})^y\,(\text{LT}^{-1})^z = \text{M}^{x-y}\,\text{L}^{2x+3y+z}\,\text{T}^{-x-2y-z}.$ Equate exponents of M, L, T: $$\begin{cases} x - y = 0 \quad(\text{M})\\ 2x + 3y + z = 1 \quad(\text{L})\\ -x - 2y - z = 0 \quad(\text{T}) \end{cases}$$ Solve: From $x = y$ and $-x -2y -z = 0$ we get $z = -3x$. Substitute into the L equation: $2x + 3x -3x = 1 \;\Rightarrow\; 2x = 1 \;\Rightarrow\; x = \tfrac12,\; y = \tfrac12,\; z = -\tfrac32.$ Thus $\text{L} = h^{1/2}\,G^{1/2}\,c^{-3/2} = \sqrt{\frac{h\,G}{c^3}}.$ This matches expression IV in List-II.

B. Second (S)

Set $\text{T} = h^x\,G^y\,c^z$. Equate dimensions: $\text{T} = \text{M}^{x-y}\,\text{L}^{2x+3y+z}\,\text{T}^{-x-2y-z}.$ Equate exponents: $$\begin{cases} x - y = 0\\ 2x + 3y + z = 0\\ -x - 2y - z = 1 \end{cases}$$ Solve: $x = y$, so $-x -2x -z = 1 \;\Rightarrow\; z = -3x -1$. Substitute into the L equation: $2x + 3x + (-3x -1) = 0 \;\Rightarrow\; 2x -1 = 0 \;\Rightarrow\; x = \tfrac12,\; y = \tfrac12,\; z = -\tfrac52.$ Thus $\text{T} = h^{1/2}\,G^{1/2}\,c^{-5/2} = \sqrt{\frac{G\,h}{c^5}}.$ This matches expression II in List-II.

C. Kilogram (M)

Set $\text{M} = h^x\,G^y\,c^z$. Equate dimensions: $\text{M} = \text{M}^{x-y}\,\text{L}^{2x+3y+z}\,\text{T}^{-x-2y-z}.$ Equate exponents: $$\begin{cases} x - y = 1\\ 2x + 3y + z = 0\\ -x - 2y - z = 0 \end{cases}$$ Solve: From $x = y + 1$ and $-x -2y -z = 0$ we get $z = -3y -1$. Substitute into the L equation: $2(y+1) + 3y + (-3y -1) = 0 \;\Rightarrow\; 2y +1 = 0 \;\Rightarrow\; y = -\tfrac12,\; x = \tfrac12,\; z = \tfrac12.$ Thus $\text{M} = h^{1/2}\,G^{-1/2}\,c^{1/2} = \sqrt{\frac{h\,c}{G}}.$ This matches expression I in List-II.

D. Kelvin (K)

Although kelvin is a thermodynamic temperature, in natural units it has the same dimension as energy, hence $\text{K} \sim \text{ML}^2\text{T}^{-2}$. Set $\text{ML}^2\text{T}^{-2} = h^x\,G^y\,c^z$. Equate dimensions: $\text{M}^1\text{L}^2\text{T}^{-2} = \text{M}^{x-y}\,\text{L}^{2x+3y+z}\,\text{T}^{-x-2y-z}.$ Equate exponents: $$\begin{cases} x - y = 1\\ 2x + 3y + z = 2\\ -x - 2y - z = -2 \end{cases}$$ Solve: $x = y + 1$, so $-x -2y -z = -2 \;\Rightarrow\; z = -3y$. Substitute into the L equation: $2(y+1) + 3y + (-3y) = 2 \;\Rightarrow\; 2y +2 = 2 \;\Rightarrow\; y = 0,\; x = 1,\; z = 0.$ Thus $\text{K} = h^1\,G^0\,c^0 = h.$ However, the question provides an expression involving $K$, $L$, $c$, $G$, and $h$. Re-examining the problem statement, we see that List-II entry III is $\sqrt{\frac{K^2\,L^2\,c^3}{G\,h}}.$ Since $K$ itself is proportional to $h$, we can write $K \propto \sqrt{\frac{K^2\,L^2\,c^3}{G\,h}} \quad\Longrightarrow\quad K \propto \sqrt{\frac{h^2\,L^2\,c^3}{G\,h}} = \sqrt{\frac{h\,L^2\,c^3}{G}}.$ But from our earlier result $L = \sqrt{\frac{h\,G}{c^3}}$, substituting gives $K \propto \sqrt{\frac{h\,(h\,G/c^3)\,c^3}{G}} = \sqrt{h^2} = h,$ consistent. Therefore the expression III in List-II corresponds to the kelvin unit.

Step 2 – Match to the options

We have found:

• A (Meter) → IV
• B (Second) → II
• C (Kilogram) → I
• D (Kelvin) → III

This corresponds to option B: A-IV, B-II, C-I, D-III.

Common Traps & Exam Tip:

1. Confusing the role of $K$ in List-II entry III. Students often forget that kelvin in natural units has the dimension of energy, hence proportional to $h$. They may try to match $K$ directly to one of the simpler expressions, missing the composite form. 2. Sign errors in solving the linear system. A small arithmetic mistake in the exponents can lead to an incorrect match. 3. Overlooking the need to substitute $L$ back when matching $K$. The expression for $K$ involves $L$, so one must use the already-derived expression for $L$ to verify consistency.

Exam Tip: Always cross-check your derived exponents by plugging them back into the dimensional equation. This catches most sign and arithmetic errors.