Q138 / 180 · NEET 2026 (Re-NEET) · Physics

Consider that $\sigma_s$, $k_B$, b represent Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of $\sigma_s k_B^{-1}b$ is :

  • A
    $[\text{L}^{-1}\text{T}^{-1}\text{K}^{-4}]$
  • Correct answer
    B
    $[\text{L}^{-1}\text{T}^{-1}\text{K}^{-2}]$
  • C
    $[\text{L}^{-1}\text{K}^{-2}]$
  • D
    $[\text{L}^{-1}\text{T}^{-1}\text{K}^{-3}]$

Solution

The dimensional formula for the Stefan-Boltzmann constant $\sigma_s$ is obtained from the relation $E = \sigma_s A t T^4$, where $E$ is energy, $A$ is area, $t$ is time, and $T$ is temperature:

$[\sigma_s] = \dfrac{[\text{M}\text{L}^2\text{T}^{-2}]}{[\text{L}^2][\text{T}][\text{K}^4]} = [\text{M}\text{T}^{-3}\text{K}^{-4}]$

The dimensional formula for the Boltzmann constant $k_B$ is obtained from $E = k_B T$:

$[k_B] = \dfrac{[\text{M}\text{L}^2\text{T}^{-2}]}{[\text{K}]} = [\text{M}\text{L}^2\text{T}^{-2}\text{K}^{-1}]$

The dimensional formula for Wien's displacement constant $b$ is obtained from $\lambda_{\text{max}} T = b$:

$[b] = [\text{L}][\text{K}] = [\text{L}\text{K}]$

Now, substituting these dimensions into the given expression $\sigma_s k_B^{-1} b$:

$[\sigma_s k_B^{-1} b] = \dfrac{[\sigma_s][b]}{[k_B]}$

$[\sigma_s k_B^{-1} b] = \dfrac{[\text{M}\text{T}^{-3}\text{K}^{-4}][\text{L}\text{K}]}{[\text{M}\text{L}^2\text{T}^{-2}\text{K}^{-1}]}$

$[\sigma_s k_B^{-1} b] = \dfrac{[\text{M}\text{L}\text{T}^{-3}\text{K}^{-3}]}{[\text{M}\text{L}^2\text{T}^{-2}\text{K}^{-1}]}$

$[\sigma_s k_B^{-1} b] = [\text{L}^{-1}\text{T}^{-1}\text{K}^{-2}]$

Answer: $[\text{L}^{-1}\text{T}^{-1}\text{K}^{-2}]$

Hence, the correct answer is option: 'B'.

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