Match the LIST-I with LIST-II LIST-I A. Boltzmann constant B. Coefficient of viscosity C. Planck's constant D. Thermal conductivity LIST-II I. ML^2 T^-1 II. MLT^-3 K^-1 III. ML^2 T^-2 K^-1 IV. ML^-1 T^-1 Choose the correct answer from the options given below:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q26 – JEE-MAIN 2025

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Home Physics Units & MeasurementsQuestion #33

Match the LIST-I with LIST-II

\begin{array}{lcl} LIST-I & LIST-II \\ A. Boltzmann constant & I. M L^{2} T^{- 1} \\ B. Coefficient of viscosity & II. M L T^{- 3} K^{- 1} \\ C. Planck's constant & III. M L^{2} T^{- 2} K^{- 1} \\ D. Thermal conductivity & IV. M L^{- 1} T^{- 1} \end{array}

Choose the correct answer from the options given below:

Clue 1 of 3: Dimensional Analysis

Recall the fundamental definitions and associated formulas for each physical constant to derive their dimensions.

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Ninja Strategy•Systematic Elimination by Constant

Derive the dimensions for one constant at a time and eliminate options that do not match, progressively narrowing down to the correct answer.

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Step 1: Determine dimensions for Boltzmann constant and Planck's constant✦ Active

The Boltzmann constant $k_{B}$ relates energy $E$ to temperature $T$ via $E = k_{B} T$ . Planck's constant $h$ relates energy $E$ to frequency $ν$ via $E = h ν$ .

[k_{B}] = \frac{[E]}{[T]} = \frac{M L^{2} T^{- 2}}{K} = M L^{2} T^{- 2} K^{- 1} ⟹ A - III

[h] = \frac{[E]}{[ν]} = \frac{M L^{2} T^{- 2}}{T^{- 1}} = M L^{2} T^{- 1} ⟹ C - I

Step 2: Determine dimensions for coefficient of viscosity and thermal conductivity○ Expand

The coefficient of viscosity $η$ is derived from Stokes' law $F = 6 π η r v$ . Thermal conductivity $K$ is derived from Fourier's law of heat conduction $\frac{d Q}{d t} = - K A \frac{d T}{d x}$ , where $\frac{d Q}{d t}$ is power.

[η] = \frac{[F]}{[r] [v]} = \frac{M L T^{- 2}}{L \cdot L T^{- 1}} = M L^{- 1} T^{- 1} ⟹ B - IV

[K] = \frac{[P o w e r]}{[A] [\frac{d T}{d x}]} = \frac{M L^{2} T^{- 3}}{L^{2} \cdot \frac{K}{L}} = M L T^{- 3} K^{- 1} ⟹ D - II

Step 3: Match the constants to their dimensions○ Expand

Based on the derived dimensions, the correct matches are: A - III, B - IV, C - I, D - II. Comparing this with the given options, option 1 is the correct answer.

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Physics Question 26 – JEE-MAIN 2025

Clue 1 of 3: Dimensional Analysis

🧠 Full Solution Path