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Question

The tungsten filament of an electric lamp has a surface area A and a power rating P. If the emissivity of the filament is ϵ and σ is Stefan's constant, the steady temperature of the filament will be


A

T=(PAϵσ)2

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B

T=PAϵσ

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C

T=(PAϵσ)12

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D

T=(PAϵσ)14

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Solution

The correct option is D

T=(PAϵσ)14


The energy radiated per second per unit area at temperature T is σϵT4.
Thus, the energy radiated per second (or power radiated) from the filament of area A will be given by P=AσϵT4T=(PAσϵ)14
Hence, the correct choice is (d).


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