JEE PYQ: Units & Measurements - Question ID d5526a41bbf6 (JEE Main 2025)

ID: d5526a41bbf6JEE Main 2025Single Correct MCQ

 Match the LIST-I with LIST-II \text { Match the LIST-I with LIST-II }

LIST-I

LIST-II
A. <br> Boltzmann constant <br><br>\text { Boltzmann constant }<br>
I <br>ML2 T1<br><br>\mathrm{ML}^2 \mathrm{~T}^{-1}<br>
B <br> Coefficient of viscosity <br><br>\text { Coefficient of viscosity }<br>
II <br>MLT3 K1<br><br>\mathrm{MLT}^{-3} \mathrm{~K}^{-1}<br>
C <br> Planck’s constant <br><br>\text { Planck's constant }<br>
III <br>ML2 T2 K1<br><br>\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}<br>
D <br> Thermal conductivity <br><br>\text { Thermal conductivity }<br>
IV <br>ML1 T1<br><br>\mathrm{ML}^{-1} \mathrm{~T}^{-1}<br>
Choose the correct answer from the options given below:

Select Option

Step-by-step Explanation

Core Formula & Concept:

In dimensional analysis, every physical quantity can be expressed in terms of the fundamental dimensions: mass (MM), length (LL), time (TT), and temperature (KK or Θ\Theta). The key idea is to write the defining formula for each quantity in LIST-I and then express its dimensions in the form MaLbTcKdM^a L^b T^c K^d.

Step-by-Step Derivation:

1. Boltzmann constant (kBk_B)
Definition: kB=ETk_B = \dfrac{E}{T}, where EE is energy and TT is absolute temperature.
Energy has dimensions ML2T2ML^2T^{-2}; temperature has dimension KK.
Therefore, [kB]=ML2T2K=ML2T2K1.[k_B] = \frac{ML^2T^{-2}}{K} = ML^2T^{-2}K^{-1}. This matches LIST-II entry III.

2. Coefficient of viscosity (η\eta)
Definition from Newton’s law of viscosity: τ=ηdvdx\tau = \eta\,\dfrac{dv}{dx}.
Shear stress τ\tau has dimensions ML1T2ML^{-1}T^{-2}; velocity gradient dvdx\dfrac{dv}{dx} has dimensions T1T^{-1}.
Therefore, [η]=ML1T2T1=ML1T1.[\eta] = \frac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}. This matches LIST-II entry IV.

3. Planck’s constant (hh)
Definition: E=hνE = h\nu, where EE is energy and ν\nu is frequency.
Energy has dimensions ML2T2ML^2T^{-2}; frequency has dimension T1T^{-1}.
Therefore, [h]=ML2T2T1=ML2T1.[h] = \frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}. This matches LIST-II entry I.

4. Thermal conductivity (kk)
Definition from Fourier’s law: dQdt=kAdTdx\dfrac{dQ}{dt} = kA\,\dfrac{dT}{dx}.
Heat flow rate dQdt\dfrac{dQ}{dt} has dimensions ML2T3ML^2T^{-3}; area AA has dimension L2L^2; temperature gradient dTdx\dfrac{dT}{dx} has dimension KL1KL^{-1}.
Therefore, [k]=ML2T3L2KL1=MLT3K1.[k] = \frac{ML^2T^{-3}}{L^2 \cdot KL^{-1}} = MLT^{-3}K^{-1}. This matches LIST-II entry II.

Matching LIST-I to LIST-II gives: A → III, B → IV, C → I, D → II.

Common Traps & Exam Tip:

• Students often confuse the dimensions of Boltzmann’s constant with those of the gas constant RR, forgetting that R=NAkBR = N_A k_B and so [R]=ML2T2K1[R] = ML^2T^{-2}K^{-1} as well. • Planck’s constant is sometimes mistakenly assigned the dimensions of angular momentum (which is also ML2T1ML^2T^{-1}), but in this question it must be matched to the given options. • Thermal conductivity’s dimensions can be misremembered if one forgets that the temperature gradient introduces a K1K^{-1} factor.

The correct option is therefore A: A-III, B-IV, C-I, D-II.

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