Question

Match List - I with List - II. 

Choose the most appropriate answer from the options given below : 
 

• A. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
• B. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
• C. (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
• D. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)

Hint:

For dimensional analysis questions, it's crucial to remember the fundamental formulas connecting the physical quantities. Even if you forget a specific formula, try to recall any formula involving that quantity (e.g., for Planck's constant, E=hf or angular momentum L=nh/2π). For viscosity, the force equation is most direct.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question requires matching physical quantities from List-I with their corresponding SI units from List-II. This is a dimensional analysis problem.
Step 2: Detailed Explanation:
Let's find the units for each quantity in List-I.
(a) Rydberg constant (R\(_H\)):
The Rydberg formula is \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \).
Here, \( \lambda \) is wavelength (unit: m) and \( n_1, n_2 \) are dimensionless integers.
So, the unit of \( R_H \) is the same as the unit of \( \frac{1}{\lambda} \), which is \( \text{m}^{-1} \).
Therefore, (a) matches with (iii).
(b) Planck's constant (h):
The formula for the energy of a photon is \( E = hf \), where \( E \) is energy and \( f \) is frequency.
The unit of energy (E) is Joule (J), which is \( \text{kg m}^2\text{s}^{-2} \).
The unit of frequency (f) is Hertz (Hz), which is \( \text{s}^{-1} \).
So, the unit of \( h \) is \( \frac{\text{unit of E}}{\text{unit of f}} = \frac{\text{kg m}^2\text{s}^{-2}}{\text{s}^{-1}} = \text{kg m}^2\text{s}^{-1} \).
Therefore, (b) matches with (ii).
(c) Magnetic field energy density (\(u_B\)):
Energy density is defined as energy per unit volume.
Unit of Energy = Joule (J) = \( \text{kg m}^2\text{s}^{-2} \).
Unit of Volume = \( \text{m}^3 \).
Unit of energy density = \( \frac{\text{kg m}^2\text{s}^{-2}}{\text{m}^3} = \text{kg m}^{-1}\text{s}^{-2} \).
Therefore, (c) matches with (iv).
(d) Coefficient of viscosity (\(\eta\)):
From Stokes' law, the viscous force is given by \( F = 6\pi\eta rv \), or more generally from Newton's law of viscosity, \( F = \eta A \frac{dv}{dx} \).
Using \( F = \eta A \frac{dv}{dx} \), we can find the units of \( \eta \).
\( \eta = \frac{F \cdot dx}{A \cdot dv} \).
Unit of Force (F) = Newton (N) = \( \text{kg m s}^{-2} \).
Unit of Area (A) = \( \text{m}^2 \).
Unit of velocity (dv) = \( \text{m s}^{-1} \).
Unit of distance (dx) = m.
Unit of \( \eta \) = \( \frac{(\text{kg m s}^{-2}) \cdot (\text{m})}{(\text{m}^2) \cdot (\text{m s}^{-1})} = \frac{\text{kg m}^2\text{s}^{-2}}{\text{m}^3\text{s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1} \).
Therefore, (d) matches with (i).
Step 3: Final Answer:
The correct matching is:
(a) - (iii)
(b) - (ii)
(c) - (iv)
(d) - (i)
This corresponds to option (D).