Match List-I with List-II.

List-I	List-II
A. Meter (L)	I. \(\sqrt{\dfrac{hc}{G}}\)
B. Second (S)	II. \(\sqrt{\dfrac{Gh}{c^5}}\)
C. Kilogram (M)	III. \(\sqrt{\dfrac{K^2L^2c^3}{Gh}}\)
D. Kelvin (K)	IV. \(\sqrt{\dfrac{Gh}{c^3}}\)

where \(h\) (Planck's constant), \(G\) (gravitational constant) and \(c\) (speed of light in vacuum) as fundamental units.
Choose the correct answer from the options given below :

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

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

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

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

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

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