Match List-I with List-II, where h , c , k_B , σ_s , b , and T...

Let us find the dimensional formula for each expression in List-I.

A. $\dfrac{h c}{k_B T}$ :
$[h c] = [\text{M}\text{L}^2\text{T}^{-1}] \times [\text{L}\text{T}^{-1}] = [\text{M}\text{L}^3\text{T}^{-2}]$
$[k_B T] = \text{Energy} = [\text{M}\text{L}^2\text{T}^{-2}]$
\left[\dfrac{h c}{k_B T}\right] = \dfrac{[\text{M}\text{L}^3\text{T}^{-2}]}{[\text{M}\text{L}^2\text{T}^{-2}]} = [\text{L}]$(Matches IV)<br><br>B.$\dfrac{\sigma_s T^4}{c}$:<br>$[\sigma_s T^4] = \text{Power per unit area} = \dfrac{[\text{M}\text{L}^2\text{T}^{-3}]}{[\text{L}^2]} = [\text{M}\text{T}^{-3}]$<br>$[c] = [\text{L}\text{T}^{-1}]$<br>\left[\dfrac{\sigma_s T^4}{c}\right] = \dfrac{[\text{M}\text{T}^{-3}]}{[\text{L}\text{T}^{-1}]} = [\text{M}\text{L}^{-1}\text{T}^{-2}]$ (Matches I)

C. $\dfrac{b k_B}{h}$ :
$[b] = [\text{L}\text{K}]$
$[k_B] = [\text{M}\text{L}^2\text{T}^{-2}\text{K}^{-1}]$
$[b k_B] = [\text{M}\text{L}^3\text{T}^{-2}]$
$[h] = [\text{M}\text{L}^2\text{T}^{-1}]$
\left[\dfrac{b k_B}{h}\right] = \dfrac{[\text{M}\text{L}^3\text{T}^{-2}]}{[\text{M}\text{L}^2\text{T}^{-1}]} = [\text{L}\text{T}^{-1}]$(Matches II)<br><br>D.$\dfrac{\sigma_s b^4}{h c^2}$:<br>$[\sigma_s] = [\text{M}\text{T}^{-3}\text{K}^{-4}]$<br>$[b^4] = [\text{L}^4\text{K}^4]$<br>$[\sigma_s b^4] = [\text{M}\text{L}^4\text{T}^{-3}]$<br>$[h c^2] = [\text{M}\text{L}^2\text{T}^{-1}] \times [\text{L}^2\text{T}^{-2}] = [\text{M}\text{L}^4\text{T}^{-3}]$<br>\left[\dfrac{\sigma_s b^4}{h c^2}\right] = \dfrac{[\text{M}\text{L}^4\text{T}^{-3}]}{[\text{M}\text{L}^4\text{T}^{-3}]} = [\text{M}^0\text{L}^0\text{T}^0]$ (Matches III)

Therefore, the correct matching is A-IV, B-I, C-II, D-III.

Answer: A-IV, B-I, C-II, D-III