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 :

JEE Main 2026 — Physics Mechanics

2026mcqmedium

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 :

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

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

Did you get this right?

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

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List-I

List-II

A. Meter (L)

\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}}

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

JEE Main 2026 — Physics Mechanics

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JEE Main 2026 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics