JEE Main 2026 — Units and Dimensions Question with Solution

The dimensional formulas for the given fundamental constants are:
$[h]=M{L}^2{T}^{-1}$
$[c]=L{T}^{-1}$
$[G]=M^{-1}{L}^3{T}^{-2}$

Evaluating the dimensions of the expressions in List-II:

For I: $\sqrt{\frac{hc}{G}}=\sqrt{\frac{(M{L}^2{T}^{-1})(L{T}^{-1})}{M^{-1}{L}^3{T}^{-2}}}=\sqrt{M^2}=M$ (Kilogram).
Thus, C $\to$ I.

For II: $\sqrt{\frac{Gh}{c^5}}=\sqrt{\frac{(M^{-1}{L}^3{T}^{-2})(M{L}^2{T}^{-1})}{(L{T}^{-1}{)}^5}}=\sqrt{\frac{L^5{T}^{-3}}{L^5{T}^{-5}}}=\sqrt{T^2}=T$ (Second).
Thus, B $\to$ II.

For IV: $\sqrt{\frac{Gh}{c^3}}=\sqrt{\frac{(M^{-1}{L}^3{T}^{-2})(M{L}^2{T}^{-1})}{(L{T}^{-1}{)}^3}}=\sqrt{\frac{L^5{T}^{-3}}{L^3{T}^{-3}}}=\sqrt{L^2}=L$ (Meter).
Thus, A $\to$ IV.

For III: Substituting $L^2=\frac{Gh}{c^3}$ from IV into the expression gives $\sqrt{\frac{K^2{L}^2{c}^3}{Gh}}=\sqrt{K^2\left(\frac{Gh}{c^3}\right)\frac{c^3}{Gh}}=\sqrt{K^2}=K$ (Kelvin).
Thus, D $\to$ III.

The correct matching is A-IV, B-II, C-I, D-III.

Answer: A-IV, B-II, C-I, D-III