Match the LIST-I with LIST-II.

	LIST-I		LIST-II
A.	Boltzmann constant	I.	$$ML^2T^{-1}$$
B.	Coefficient of viscosity	II.	$$MLT^{-3}K^{-1}$$
C.	Planck's constant	III.	$$ML^2T^{-2}K^{-1}$$
D.	Thermal conductivity	IV.	$$ML^{-1}T^{-1}$$

Choose the correct answer:

The dimensional analysis will be carried out for each physical quantity in LIST-I and then compared with the four alternatives given in LIST-II.

The average translational energy of one molecule in an ideal gas is $$E = \tfrac{3}{2}\,k_B\,T$$.
Hence $$k_B = \dfrac{E}{T}$$.
Energy has dimension $$ML^{2}T^{-2}$$ and temperature contributes the factor $$K$$ in the denominator.
Therefore $$[k_B] = \dfrac{ML^{2}T^{-2}}{K} = ML^{2}T^{-2}K^{-1}$$, which corresponds to LIST-II entry III.

Newton’s law of viscous flow states $$\tau = \eta \,\dfrac{dv}{dy}$$, where $$\tau$$ is shear stress.
Shear stress $$\tau = \dfrac{\text{Force}}{\text{Area}}$$ has dimension $$\dfrac{MLT^{-2}}{L^{2}} = ML^{-1}T^{-2}$$.
Velocity gradient $$\dfrac{dv}{dy}$$ has dimension $$T^{-1}$$.
Thus $$[\eta] = \dfrac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}$$, matching LIST-II entry IV.

The photon energy is $$E = h\nu$$, and frequency $$\nu$$ has dimension $$T^{-1}$$.
Hence $$h = \dfrac{E}{\nu}$$.
Taking $$[E] = ML^{2}T^{-2}$$ and dividing by $$T^{-1}$$ gives $$[h] = ML^{2}T^{-1}$$, i.e. LIST-II entry I.

Fourier’s law of heat conduction is $$\dfrac{Q}{t} = -K\,A\,\dfrac{\Delta T}{L}$$.
Heat conducted per unit time $$\dfrac{Q}{t}$$ is power with dimension $$ML^{2}T^{-3}$$.
Area $$A$$ contributes $$L^{2}$$ and temperature gradient $$\dfrac{\Delta T}{L}$$ contributes $$K\,L^{-1}$$ in the denominator.
Therefore

$$[K] = \dfrac{ML^{2}T^{-3}}{L^{2}\,K\,L^{-1}} = \dfrac{ML^{2}T^{-3}}{L^{1}K} = MLT^{-3}K^{-1}$$, corresponding to LIST-II entry II.

Collecting the results:
A → III, B → IV, C → I, D → II

This is exactly the pairing given in Option A.

Hence, the correct answer is Option A.