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

Determine the dimensions of each quantity in LIST-I.

A. Boltzmann constant (k): Derived from the ideal gas law, \( PV = NkT \). Given P is pressure (\( \text{ML}^{-1}\text{T}^{-2} \)), V is volume (\( \text{L}^3 \)), N is the number of particles (dimensionless), and T is temperature (K). The dimension of k is calculated as \( k = \frac{PV}{NT} = \frac{(\text{ML}^{-1}\text{T}^{-2})(\text{L}^3)}{(1)(\text{K})} = \text{ML}^2\text{T}^{-2}\text{K}^{-1} \). 
Therefore, A corresponds to III. 

B. Coefficient of viscosity (\( \eta \)): Derived from the viscous force formula \( F = 6\pi \eta r v \). Given F is force (\( \text{MLT}^{-2} \)), r is radius (L), and v is velocity (\( \text{LT}^{-1} \)). The dimension of \( \eta \) is calculated as \( \eta = \frac{F}{6\pi r v} = \frac{\text{MLT}^{-2}}{(1)(\text{L})(\text{LT}^{-1})} = \frac{\text{MLT}^{-2}}{\text{L}^2\text{T}^{-1}} = \text{ML}^{-1}\text{T}^{-1} \). 
Therefore, B corresponds to IV. 

C. Planck's constant (h): Derived from the energy of a photon equation \( E = hf \). Given E is energy (\( \text{ML}^2\text{T}^{-2} \)) and f is frequency (\( \text{T}^{-1} \)). The dimension of h is calculated as \( h = \frac{E}{f} = \frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-1}} = \text{ML}^2\text{T}^{-1} \). 
Therefore, C corresponds to I. 

D. Thermal conductivity (K): Derived from the rate of heat flow equation \( \frac{dQ}{dt} = -KA \frac{dT}{dx} \). Given \( \frac{dQ}{dt} \) is power (\( \text{ML}^2\text{T}^{-3} \)), A is area (\( \text{L}^2 \)), and \( \frac{dT}{dx} \) is temperature gradient (\( \text{KL}^{-1} \)). The dimension of K is calculated as \( K = \frac{(dQ/dt) dx}{A dT} = \frac{(\text{ML}^2\text{T}^{-3})(\text{L})}{(\text{L}^2)(\text{K})} = \frac{\text{ML}^3\text{T}^{-3}}{\text{L}^2\text{K}} = \text{MLT}^{-3}\text{K}^{-1} \). 
Therefore, D corresponds to II. 

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