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).