Animated Solution for Physics - Physics and Measurement: Match List-I with List-II. $\begin{array}{ll}\text{List-I}& \text{List-II}\\ \text{A. }{R}_H\text{ (Rydberg constant)}& {\text{1. kg m}}^{-1}{\text{s}}^{-1}\\ \text{B. }h\text{ (Planck’s constant)}& {\text{2. kg m}}^2{\text{s}}^{-1}\\ \text{C. }{\mu }_B\text{ (Magnetic field energy density)}& {\text{3. m}}^{-1}\\ \text{D. }\eta \text{ (Coefficient of viscosity)}& {\text{4. kg m}}^{-1}{\text{s}}^{-2}\end{array}$ Choose the most appropriate answer from the options given below.

Dimensional analysis and unit matching questions are some of the most scoring problems in JEE. They don't require complex calculations; instead, they test your fundamental grasp of physics formulas. Let's break down this matrix match question step-by-step by deriving the SI units for each physical quantity from scratch.

The Rydberg constant is a key player in atomic physics, specifically when calculating the wavelengths of spectral lines. The Rydberg formula is given by:

Here, $n_1$ and $n_2$ are principal quantum numbers, which are purely dimensionless integers. Therefore, the unit of the Rydberg constant $R_H$ must be identical to the unit of $\frac{1}{\lambda }$. Since wavelength $\lambda$ is a length measured in meters ($\text{m}$), the unit of $R_H$ is simply ${\text{m}}^{-1}$.

This gives us our first match: A $\to$ 3.

Planck's constant bridges the gap between the energy of a photon and its frequency. The famous relation is:

Rearranging this to solve for $h$, we get $h = \frac{E}{ u}$.

We know that energy ($E$) is measured in Joules ($\text{J}$) and frequency ($ u$) is measured in Hertz or inverse seconds (${\text{s}}^{-1}$). Thus, the unit of $h$ is $\text{J}\cdot \text{s}$.

To express this in fundamental SI units, we recall that Work (Energy) = Force $\times$ Displacement. Force is mass $\times$ acceleration ($\text{kg}\cdot \text{m}\cdot {\text{s}}^{-2}$), so Energy is $\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-2}$.

Substituting this back:

This gives us our second match: B $\to$ 2.

The term "energy density" is a dead giveaway. Whether it is electric field energy density or magnetic field energy density, it always represents energy per unit volume.

We just derived the fundamental units of energy as $\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-2}$. The unit of volume is simply ${\text{m}}^3$. Dividing the two:

This gives us our third match: C $\to$ 4.

Finally, we look at the coefficient of viscosity. The most straightforward way to find its units is through Newton's law of viscous friction, which states that the viscous force $F$ between fluid layers is proportional to the area $A$ and the velocity gradient $\frac{dv}{dx}$:

Rearranging for $\eta$:

Let's plug in the fundamental units for each term: - Force ($F$): $\text{kg}\cdot \text{m}\cdot {\text{s}}^{-2}$ - Area ($A$): ${\text{m}}^2$ - Velocity gradient ($\frac{dv}{dx}$): $\frac{\text{m}\cdot {\text{s}}^{-1}}{\text{m}}={\text{s}}^{-1}$

Substituting these into our rearranged equation:

This gives us our final match: D $\to$ 1.

Putting all our derived matches together: - A $\to$ 3 - B $\to$ 2 - C $\to$ 4 - D $\to$ 1

Looking at the given options, this corresponds perfectly to Option (b). By relying on fundamental formulas rather than rote memorization, you can confidently and accurately solve any dimensional analysis problem!