Question:
The dimensional formula of emissivity of a body is:
The dimensional formula of emissivity of a body is:
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Any physical quantity defined as a ratio of two similar physical quantities (like refractive index, strain, emissivity) is always dimensionless.
Updated On: Jul 18, 2026
- \([M^1 L^0 T^{-3}]\)
- \([M^1 L^2 T^{-3}]\)
- \([M^0 L^0 T^0]\)
- \([M^1 L^2 T^{-2}]\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the concept of emissivity.
Emissivity is defined as the ratio of the emissive power of a given surface to the emissive power of a perfectly black body at the same temperature. Mathematically, it is written as: \[ e = \frac{E}{E_b} \] where \(E\) is emissive power of the body and \(E_b\) is emissive power of a black body. Since it is a ratio of two similar physical quantities, its dimensional nature depends on whether both quantities cancel each other or not.
Step 2: Dimensional formula of emissive power.
Emissive power is defined as energy emitted per unit area per unit time. Energy has dimension \([M L^2 T^{-2}]\), area is \([L^2]\), and time is \([T]\). Hence: \[ [E] = \frac{[M L^2 T^{-2}]}{[L^2][T]} = [M T^{-3}] \] So emissive power has dimensional formula \([M T^{-3}]\).
Step 3: Dimensional formula of black body emissive power.
A black body is an ideal emitter, but its emissive power still represents the same physical quantity as energy emitted per unit area per unit time. Therefore, its dimensional formula is also: \[ [E_b] = [M T^{-3}] \] This shows both numerator and denominator have identical dimensions.
Step 4: Finding the dimensional formula of emissivity.
Now substituting the dimensions into the formula: \[ e = \frac{[M T^{-3}]}{[M T^{-3}]} \] On dividing, all fundamental dimensions cancel completely: \[ e = [M^0 L^0 T^0] \] Thus emissivity is a dimensionless quantity.
Step 5: Analysis of options.
- Option (1) \([M^1 L^0 T^{-3}]\): This corresponds to emissive power, not emissivity.
- Option (2) \([M^1 L^2 T^{-3}]\): Dimensionally incorrect combination.
- Option (3) \([M^0 L^0 T^0]\): Correct, as emissivity is dimensionless.
- Option (4) \([M^1 L^2 T^{-2}]\): Represents energy, not emissivity.
Step 6: Final conclusion.
Since emissivity is a ratio of two identical physical quantities (emissive powers), all dimensions cancel out completely, making it a pure number. Therefore, the correct answer is: \[ \boxed{[M^0 L^0 T^0]} \]
Emissivity is defined as the ratio of the emissive power of a given surface to the emissive power of a perfectly black body at the same temperature. Mathematically, it is written as: \[ e = \frac{E}{E_b} \] where \(E\) is emissive power of the body and \(E_b\) is emissive power of a black body. Since it is a ratio of two similar physical quantities, its dimensional nature depends on whether both quantities cancel each other or not.
Step 2: Dimensional formula of emissive power.
Emissive power is defined as energy emitted per unit area per unit time. Energy has dimension \([M L^2 T^{-2}]\), area is \([L^2]\), and time is \([T]\). Hence: \[ [E] = \frac{[M L^2 T^{-2}]}{[L^2][T]} = [M T^{-3}] \] So emissive power has dimensional formula \([M T^{-3}]\).
Step 3: Dimensional formula of black body emissive power.
A black body is an ideal emitter, but its emissive power still represents the same physical quantity as energy emitted per unit area per unit time. Therefore, its dimensional formula is also: \[ [E_b] = [M T^{-3}] \] This shows both numerator and denominator have identical dimensions.
Step 4: Finding the dimensional formula of emissivity.
Now substituting the dimensions into the formula: \[ e = \frac{[M T^{-3}]}{[M T^{-3}]} \] On dividing, all fundamental dimensions cancel completely: \[ e = [M^0 L^0 T^0] \] Thus emissivity is a dimensionless quantity.
Step 5: Analysis of options.
- Option (1) \([M^1 L^0 T^{-3}]\): This corresponds to emissive power, not emissivity.
- Option (2) \([M^1 L^2 T^{-3}]\): Dimensionally incorrect combination.
- Option (3) \([M^0 L^0 T^0]\): Correct, as emissivity is dimensionless.
- Option (4) \([M^1 L^2 T^{-2}]\): Represents energy, not emissivity.
Step 6: Final conclusion.
Since emissivity is a ratio of two identical physical quantities (emissive powers), all dimensions cancel out completely, making it a pure number. Therefore, the correct answer is: \[ \boxed{[M^0 L^0 T^0]} \]
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