Q51 / 75 · JEE Main 2026 (05 April Shift 2) · Physics

Match List-I with List-II.
List-IList-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 :

  • A
    A-II, B-IV, C-I, D-III
  • Correct answer
    B
    A-IV, B-II, C-I, D-III
  • C
    A-IV, B-I, C-II, D-III
  • D
    A-III, B-I, C-II, D-IV

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: $\sqrt{\dfrac{hc}{G}} = \sqrt{\dfrac{(M L^2 T^{-1})(L T^{-1})}{M^{-1} L^3 T^{-2}}} = \sqrt{M^2} = M$ (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

Hence, the correct answer is option: 'B'.

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