JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Units & Measurement

With $[h]=M{L}^2{T}^{-1}$, $[c]=L{T}^{-1}$ and $[G]=M^{-1}{L}^3{T}^{-2}$, each combination is reduced to the base dimensions. For $\sqrt{\frac{Gh}{c^3}}$, $\frac{Gh}{c^3}=\frac{(M^{-1}{L}^3{T}^{-2})(M{L}^2{T}^{-1})}{L^3{T}^{-3}}=L^2$, giving the dimension of length, the metre. For $\sqrt{\frac{Gh}{c^5}}$, $\frac{Gh}{c^5}=\frac{L^5{T}^{-3}}{L^5{T}^{-5}}=T^2$, giving time, the second. For $\sqrt{\frac{hc}{G}}$, $\frac{hc}{G}=\frac{(M{L}^2{T}^{-1})(L{T}^{-1})}{M^{-1}{L}^3{T}^{-2}}=M^2$, giving mass, the kilogram. The remaining combination $\sqrt{\frac{K^2{L}^2{c}^3}{Gh}}$ carries the temperature unit and gives the kelvin. Hence the metre corresponds to $\sqrt{\frac{Gh}{c^3}}$, the second to $\sqrt{\frac{Gh}{c^5}}$, the kilogram to $\sqrt{\frac{hc}{G}}$ and the kelvin to $\sqrt{\frac{K^2{L}^2{c}^3}{Gh}}$.