Question

Match the Column I (quantity) with Column II (dimension) and select the correct answer from the codes given below.

Column I		Column II
A	Thermal resistance	1	$\left[M{T}^{-3}{K}^{-4}\right]$
B	Stefan's constant	2	$\left[M^{-1}{L}^{-2}{T}^{3}K\right]$
C	Wien's constant	3	$\left[M{L}^2{T}^{-3}\right]$
D	Heat current	4	$[LK]$

• A. A - (1) , B - (2) , C - (3) , D - (4)
• B. A - (1) , B - (4) , C - (3) , D - (2)
• C. A - (2) , B - (1) , C - (4) , D - (3)
• D. A - (2) , B - (1) , C - (3) , D - (4)

Answer 1

Answer: (C)

A. Thermal resistance $=\frac{L}{KA}$
Dimensions $=\frac{[L]}{\left[\frac{M{L}^2{T}^{-2}}{T}\times \frac{L}{L^{2}K}\right]L^2}$
$=\frac{[L]}{\left[ML{T}^{-3}{K}^{-1}\right]L^2}$
$=\left[M^{-1}{L}^{-2}{T}^{3}K\right]$
B. Stefan-Boltzmann law,
Heat radiated per unit time by body
$=\frac{\Delta Q}{\Delta t}=e\sigma A{T}^4$
$\Rightarrow \sigma =\frac{\Delta Q/\Delta t}{eA{T}^4}$
$e=$ emissivity (dimensionless constant)
Dimensions of $\sigma$ (Stefan-Boltzmann constant)
$=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[T{L}^2{K}^4\right]}=\left[M{T}^{-3}{K}^{-4}\right]$
C. From Wien's displacement law, ${\lambda }_{m}T=$ constant
Here, 'constant' is called Wien's constant, its value is $2.9\times 10^{-3}m-K$ and dimensions [LK].
D. Heat current, $H=\frac{\Delta Q}{\Delta t}=\frac{KA\left(T_h-T_l\right)}{L}$
Dimensions of $H$ is $\left[M{L}^2{T}^{-3}\right]$.
Hence, $A\to 2,B\to 1,C\to 4$ and $D\to 3$.