Explanation

This question is based on units and dimensions, which is a very important topic for JEE Main, JEE Advanced and IIT JEE.

Spring constant (k):

From Hooke’s law,

$$ F = kx $$

$$ k = \frac{F}{x} $$

Dimensions of force:

$$ [F] = MLT^{-2} $$

Therefore,

$$ [k] = \frac{MLT^{-2}}{L} = ML^0T^{-2} $$

So, A → II.

Thermal conductivity (K):

Heat conduction equation:

$$ \frac{Q}{t} = K A \frac{dT}{dx} $$

Rewriting,

$$ K = \frac{Q}{t} \cdot \frac{1}{A} \cdot \frac{dx}{dT} $$

Dimensions of heat:

$$ [Q] = ML^2T^{-2} $$

Substituting,

$$ [K] = \frac{ML^2T^{-2}}{T} \cdot \frac{1}{L^2} \cdot \frac{L}{K} = MLT^{-3}K^{-1} $$

So, B → IV.

Boltzmann constant ($k_B$):

From relation,

$$ E = k_B T $$

$$ k_B = \frac{E}{T} $$

Dimensions of energy:

$$ [E] = ML^2T^{-2} $$

Therefore,

$$ [k_B] = ML^2T^{-2}K^{-1} $$

So, C → I.

Inductive reactance ($X_L$):

Inductive reactance is given by:

$$ X_L = \omega L $$

Dimensions:

$$ [\omega] = T^{-1} $$

$$ [L] = \frac{V}{A \cdot \omega} $$

Using electrical dimensions,

$$ [X_L] = ML^2T^{-3}A^{-2} $$

So, D → III.

Hence, the correct matching is:

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

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