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	$\left[ MT ^{-3} K ^{-4}\right]$
B	Stefan's constant	2	$\left[ M ^{-1} L ^{-2} T ^{3} K \right]$
C	Wien's constant	3	$\left[ ML ^{2} T ^{-3}\right]$
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 $=\frac{L}{K A}$
Dimensions $=\frac{[ L ]}{\left[\frac{ ML ^{2} T ^{-2}}{ T } \times \frac{ L }{ L ^{2} K }\right] L ^{2}}$
$=\frac{[ L ]}{\left[ MLT ^{-3} K ^{-1}\right] L ^{2}}$
$=\left[ M ^{-1} L ^{-2} T ^{3} K \right]$
B. Stefan-Boltzmann law,
Heat radiated per unit time by body
$=\frac{\Delta Q}{\Delta t}=e \sigma A T^{4}$
$\Rightarrow \sigma=\frac{\Delta Q / \Delta t}{e A T^{4}}$
$e=$ emissivity (dimensionless constant)
Dimensions of $\sigma$ (Stefan-Boltzmann constant)
$=\frac{\left[ ML ^{2} T ^{-2}\right]}{\left[ TL ^{2} K ^{4}\right]}=\left[ MT ^{-3} K ^{-4}\right]$
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, $H=\frac{\Delta Q}{\Delta t}=\frac{K A\left(T_{h}-T_{l}\right)}{L}$
Dimensions of $H$ is $\left[ ML ^{2} T ^{-3}\right]$.
Hence, $A \rightarrow 2, B \rightarrow 1, C \rightarrow 4$ and $D \rightarrow 3$.