JEE MAIN - Physics (2025 - 3rd April Evening Shift - No. 14)
$$ \text { Match the LIST-I with LIST-II } $$
LIST-I |
LIST-II |
||
---|---|---|---|
A. | $$ \text { Boltzmann constant } $$ |
I | $$ \mathrm{ML}^2 \mathrm{~T}^{-1} $$ |
B | $$ \text { Coefficient of viscosity } $$ |
II | $$ \mathrm{MLT}^{-3} \mathrm{~K}^{-1} $$ |
C | $$ \text { Planck's constant } $$ |
III | $$ \mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1} $$ |
D | $$ \text { Thermal conductivity } $$ |
IV | $$ \mathrm{ML}^{-1} \mathrm{~T}^{-1} $$ |
Explanation
Explanation of Dimensional Analysis
In order to match the quantities from LIST-I with their respective dimensions in LIST-II, we need to analyze their dimensional formulas:
(A) Boltzmann Constant [k]:
The equation relating pressure (P), volume (V), number of moles (N), and temperature (T) gives us:
$ [\mathrm{k}] = \frac{\mathrm{PV}}{\mathrm{NT}} = \frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~K}} = \mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1} $
This corresponds to option III.
(B) Coefficient of Viscosity [η]:
The coefficient of viscosity can be defined using the relation:
$ [\eta] = \frac{\mathrm{F}}{6 \pi \mathrm{rv}} = \frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \mathrm{~T}^{-1}} = \mathrm{ML}^{-1} \mathrm{~T}^{-1} $
This corresponds to option IV.
(C) Planck's Constant [h]:
The relationship between energy (E) and frequency (f) gives the dimensional formula:
$ [\mathrm{h}] = \frac{\mathrm{E}}{\mathrm{f}} = \frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}} = \mathrm{ML}^2 \mathrm{~T}^{-1} $
This corresponds to option I.
(D) Thermal Conductivity [k]:
From the heat conduction equation, we get:
$ \mathrm{k} = \frac{\left(\mathrm{ML}^2 \mathrm{~T}^{-3}\right) \mathrm{L}}{\mathrm{~L}^2 \cdot \mathrm{~K}} = \mathrm{MLT}^{-3} \mathrm{~K}^{-1} $
This corresponds to option II.
To conclude, the correct matches are:
A - III
B - IV
C - I
D - II
This analysis ensures each physical quantity is aligned with its correct dimensional representation.
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