Match List-(I) with List-(II).
List-(I)                                                                List-(II)
(a) R$$_H$$ (Rydberg constant)                          (i) kg m$$^{-1}$$ s$$^{-1}$$
(b) $$h$$ (Planck's constant)                            (ii) kg m$$^2$$ s$$^{-1}$$
(c) $$\mu_B$$ (Magnetic field energy density)     (iii) m$$^{-1}$$
(d) $$\eta$$ (coefficient of viscosity)                     (iv) kg m$$^{-1}$$ s$$^{-2}$$
Choose the most appropriate answer from the options given below:

We start with the Rydberg constant $$R_H$$. In the Balmer‐type formula for spectral lines we write

$$\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right).$$

Here the right-hand factor in brackets is dimensionless, so the left-hand side and $$R_H$$ must have the same dimension. Because $$\lambda$$ is a length, $$1/\lambda$$ has the dimension $$L^{-1}$$. Therefore, in SI units,

$$[R_H]=\text{metre}^{-1}=m^{-1},$$

which is entry (iii) in List-(II).

Now we look at Planck’s constant $$h$$. The fundamental relation is

$$E=h\nu,$$

where the energy $$E$$ has dimension $$ML^{2}T^{-2}$$ (joule) and frequency $$\nu$$ has dimension $$T^{-1}$$. Solving for $$h$$ we obtain

$$[h]=\frac{[E]}{[\nu]}=\frac{ML^{2}T^{-2}}{T^{-1}}=ML^{2}T^{-1},$$

that is $$\text{kilogram metre}^{2}\ \text{second}^{-1}=kg\,m^{2}\,s^{-1}$$, which corresponds to entry (ii).

Next we examine the magnetic-field energy density, denoted here by $$\mu_B$$. For a magnetic field of magnitude $$B$$ we use the formula

$$u_B=\frac{B^{2}}{2\mu_0}.$$

Energy density is energy per unit volume, so its dimension is

$$\frac{\text{joule}}{\text{metre}^{3}}=\frac{ML^{2}T^{-2}}{L^{3}}=ML^{-1}T^{-2},$$

which in SI becomes $$kg\,m^{-1}\,s^{-2}$$. This matches entry (iv).

Finally, we consider the coefficient of viscosity $$\eta$$. In fluid mechanics the viscous force is given by

$$F=\eta\,A\,\frac{dv}{dy},$$

where $$F$$ is force, $$A$$ area, and $$dv/dy$$ the velocity gradient. Rearranging,

$$\eta=\frac{F}{A}\left(\frac{dv}{dy}\right)^{-1}.$$

Force has dimension $$MLT^{-2}$$, area has $$L^{2}$$, and the velocity gradient has $$T^{-1}$$. Substituting these we get

$$[\eta]=\frac{MLT^{-2}}{L^{2}}\times T=ML^{-1}T^{-1},$$

namely $$kg\,m^{-1}\,s^{-1}$$, which corresponds to entry (i).

Collecting our findings:

$$(a)\;R_H\ \longrightarrow\ (iii),\qquad (b)\;h\ \longrightarrow\ (ii),\qquad (c)\;\mu_B\ \longrightarrow\ (iv),\qquad (d)\;\eta\ \longrightarrow\ (i).$$

This exact pairing appears in Option C.

Hence, the correct answer is Option C.