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

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P/(Aeσ)

[P/(Aeσ)]2

[P/(Aeσ)]1/2

[P/(Aeσ)]1/4

answer is D.

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Detailed Solution

We know that the energy radiated per second per unit area at temp. T is given by =σeT4The power radiated from the filament of area A is given by P=σeT4×A∴ T=PAeσ1/4

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