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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a
P/(Aeσ)
b
[P/(Aeσ)]2
c
[P/(Aeσ)]1/2
d
[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