Question

Match the LIST-I with LIST-II

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

Choose the correct answer from the options given below :

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

Hint:

To find the dimensions of a physical quantity, use the fundamental formulas relating it to other quantities whose dimensions are known. Remember the fundamental dimensions of mass (M), length (L), time (T), and temperature (K). Derive the dimensions step by step using the definitions of the quantities involved.

Answer 1

The Correct Option is A

Let's find the dimensions of each quantity in LIST-I. 

A. Boltzmann constant (k): From the ideal gas law, \( PV = NkT \), where P is pressure 
(\( \text{ML}^{-1}\text{T}^{-2} \)), V is volume (\( \text{L}^3 \)), N is the number of particles (dimensionless), k is the Boltzmann constant, and T is temperature (K). 
So, \( k = \frac{PV}{NT} = \frac{(\text{ML}^{-1}\text{T}^{-2})(\text{L}^3)}{(1)(\text{K})} = \text{ML}^2\text{T}^{-2}\text{K}^{-1} \) 
Thus, A matches with III. 

B. Coefficient of viscosity (\( \eta \)): From viscous force \( F = 6\pi \eta r v \), where F is force (\( \text{MLT}^{-2} \)), r is radius (L), and v is velocity (\( \text{LT}^{-1} \)). 
So, \( \eta = \frac{F}{6\pi r v} = \frac{\text{MLT}^{-2}}{(1)(\text{L})(\text{LT}^{-1})} = \frac{\text{MLT}^{-2}}{\text{L}^2\text{T}^{-1}} = \text{ML}^{-1}\text{T}^{-1} \) 
Thus, B matches with IV. 

C. Planck's constant (h): From the energy of a photon \( E = hf \), where E is energy (\( \text{ML}^2\text{T}^{-2} \)) and f is frequency (\( \text{T}^{-1} \)). So, \( h = \frac{E}{f} = \frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-1}} = \text{ML}^2\text{T}^{-1} \) 
Thus, C matches with I. 

D. Thermal conductivity (K): From the rate of heat flow \( \frac{dQ}{dt} = -KA \frac{dT}{dx} \), where \( \frac{dQ}{dt} \) is power (\( \text{ML}^2\text{T}^{-3} \)), A is area (\( \text{L}^2 \)), and \( \frac{dT}{dx} \) is temperature gradient (\( \text{KL}^{-1} \)). 
So, \( K = \frac{(dQ/dt) dx}{A dT} = \frac{(\text{ML}^2\text{T}^{-3})(\text{L})}{(\text{L}^2)(\text{K})} = \frac{\text{ML}^3\text{T}^{-3}}{\text{L}^2\text{K}} = \text{MLT}^{-3}\text{K}^{-1} \) 
Thus, D matches with II. 

The correct matching is A-III, B-IV, C-I, D-II, which corresponds to option (A).

Answer 2

To solve this problem, we need to match the physical quantities in LIST-I with their correct dimensional formulas in LIST-II. Let's analyze each option step-by-step:

1. Boltzmann Constant: The dimensional formula for the Boltzmann constant (\( k \)) is given by \([k] = \text{ML}^2\text{T}^{-2}\text{K}^{-1}\). This dimensional formula corresponds to the option III in LIST-II. Therefore, Boltzmann constant is matched with III.
2. Coefficient of Viscosity: The coefficient of viscosity (\( \eta \)) has the dimensional formula \([\eta] = \text{ML}^{-1}\text{T}^{-1}\). This matches with option IV in LIST-II. Therefore, the coefficient of viscosity is matched with IV.
3. Planck's Constant: Planck's constant (\( h \)) has the dimensional formula \([h] = \text{ML}^2\text{T}^{-1}\). This corresponds to option I in LIST-II. Therefore, Planck's constant is matched with I.
4. Thermal Conductivity: The dimensional formula for thermal conductivity (\( \kappa \)) is \([\kappa] = \text{MLT}^{-3}\text{K}^{-1}\). This matches with option II in LIST-II. Therefore, thermal conductivity is matched with II.

Based on the above analysis, the correct matching is:

• A - III
• B - IV
• C - I
• D - II

This corresponds to the correct answer: A-III, B-IV, C-I, D-II. Therefore, the solution is validated by matching the physical quantities with their correct dimensional formulas.