MHT CET · Physics · Units and Dimensions

The SI unit an dimension of Stefan’s constants $σ$ in case of Stefan’s law of radiation is

A $\frac{J}{m^{3} s^{4}}, [M^{1} L^{0} T^{- 3} K^{- 4}]$
B $\frac{J}{m^{2} s^{4} K}, [M^{1} L^{0} T^{- 3} K^{3}]$
C $\frac{J}{m^{3} s K^{4}}, [M^{1} L^{0} T^{- 3} K^{4}]$
D $\frac{J}{m^{2} s K^{4}}, [M^{1} L^{0} T^{- 3} K^{- 4}]$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{J}{m^{2} s K^{4}}, [M^{1} L^{0} T^{- 3} K^{- 4}]$

Step-by-step Solution

Detailed explanation

According to Stefan’s law, energy emitted by a body per unit area per second is proportional to fourth power of the absolute temperature.

E = σ T^{4}

where, E = energy emitted/areas/second,
T = absolute temperature in kelvin
And

σ

= Stefan’s constant

\Rightarrow σ = \frac{E}{T^{4}}

Unit of Stefan’s constant

\Rightarrow \frac{\frac{J}{m^{2} s}}{K^{4}} = \frac{J}{m^{2} s K^{4}}

Dimensions of Stefan’s constant,

σ \frac{[E n e r g y]}{[A r e a] \times [T i m e] \times {[T e m p e r a t u r e]}^{4}}

= \frac{[M L^{2} T^{- 2}]}{[L^{2} T \times K^{4}]} = [M^{1} L^{0} T^{- 3} K^{- 4}]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The SI unit an dimension of Stefan’s constants σ in case of Stefan’s law of radiation is

Step-by-step Solution

The SI unit an dimension of Stefan’s constants $σ$ in case of Stefan’s law of radiation is