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PhyMedMain 2026 • 5th Apr - Shift 2

Match Physical Units with Fundamental Constants Expressions

Match List - I with List - II.

List - IList - IIA. Meter (L)I. hcGB. Second (S)II. Ghc5C. Kilogram (M)III. K2L2c3GhD. Kelvin (K)IV. Ghc3\begin{array}{ll} \mathbf{\text{List - I}} & \mathbf{\text{List - II}} \\ A. \text{ Meter (L)} & \text{I. } \sqrt{\frac{hc}{G}} \\ B. \text{ Second (S)} & \text{II. } \sqrt{\frac{Gh}{c^5}} \\ C. \text{ Kilogram (M)} & \text{III. } \sqrt{\frac{K^2 L^2 c^3}{Gh}} \\ D. \text{ Kelvin (K)} & \text{IV. } \sqrt{\frac{Gh}{c^3}} \end{array}

where hh (Planck's constant), GG (gravitational constant) and cc (speed of light in vacuum) as fundamental units.

Choose the correct answer from the options given below :

Options

A

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

B

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

Correct
C

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

D

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

Topics & Concepts

Units and MeasurementsUnits and Measurements

Step-by-Step Solution

To determine the correct matching between List - I (Physical Units) and List - II (Expressions in terms of fundamental constants), we analyze the dimensional formulas of the fundamental constants involved:

  1. Speed of light (cc): [c]=[LT−1][c] = [\text{L} \text{T}^{-1}]

  2. Planck's constant (hh): [h]=[ML2T−1][h] = [\text{M} \text{L}^2 \text{T}^{-1}]

  3. Gravitational constant (GG): [G]=[M−1L3T−2][G] = [\text{M}^{-1} \text{L}^3 \text{T}^{-2}]

Step 1: Matching Meter (Length, [L][\text{L}])

Consider expression IV: Ghc3\sqrt{\frac{Gh}{c^3}}

Substituting the dimensions: [Ghc3]=(M−1L3T−2)(ML2T−1)(LT−1)3=L5T−3L3T−3=L2=[L]\left[ \sqrt{\frac{Gh}{c^3}} \right] = \sqrt{\frac{(\text{M}^{-1} \text{L}^3 \text{T}^{-2}) (\text{M} \text{L}^2 \text{T}^{-1})}{(\text{L} \text{T}^{-1})^3}} = \sqrt{\frac{\text{L}^5 \text{T}^{-3}}{\text{L}^3 \text{T}^{-3}}} = \sqrt{\text{L}^2} = [\text{L}]

Thus, Meter (A) matches with IV.

Step 2: Matching Second (Time, [T][\text{T}])

Consider expression II: Ghc5\sqrt{\frac{Gh}{c^5}}

Substituting the dimensions: [Ghc5]=(M−1L3T−2)(ML2T−1)(LT−1)5=L5T−3L5T−5=T2=[T]\left[ \sqrt{\frac{Gh}{c^5}} \right] = \sqrt{\frac{(\text{M}^{-1} \text{L}^3 \text{T}^{-2}) (\text{M} \text{L}^2 \text{T}^{-1})}{(\text{L} \text{T}^{-1})^5}} = \sqrt{\frac{\text{L}^5 \text{T}^{-3}}{\text{L}^5 \text{T}^{-5}}} = \sqrt{\text{T}^2} = [\text{T}]

Thus, Second (B) matches with II.

Step 3: Matching Kilogram (Mass, [M][\text{M}])

Consider expression I: hcG\sqrt{\frac{hc}{G}}

Substituting the dimensions: [hcG]=(ML2T−1)(LT−1)M−1L3T−2=ML3T−2M−1L3T−2=M2=[M]\left[ \sqrt{\frac{hc}{G}} \right] = \sqrt{\frac{(\text{M} \text{L}^2 \text{T}^{-1}) (\text{L} \text{T}^{-1})}{\text{M}^{-1} \text{L}^3 \text{T}^{-2}}} = \sqrt{\frac{\text{M} \text{L}^3 \text{T}^{-2}}{\text{M}^{-1} \text{L}^3 \text{T}^{-2}}} = \sqrt{\text{M}^2} = [\text{M}]

Thus, Kilogram (C) matches with I.

Step 4: Matching Kelvin (Temperature)

By process of elimination, Kelvin (D) matches with III.

Conclusion

The correct matching sequence is: A−IV,B−II,C−I,D−III\mathbf{A - IV, \quad B - II, \quad C - I, \quad D - III}

This corresponds to Option B.

Match Physical Units with Fundamental Constants Expressions | Physics PYQ Solution - JEE Challenger