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A 60 w bulb has a filament temperature of 2000 K. what will be the wattage of another bulbs of same filament area and material at a temperature of 4000 K?A. 8 wB. 32 wC. 64 wD. 960 w

Answer» Correct Answer - D
`(P_(2))/(P_(1))=(T_(2)^(4))/(T_(1)^(4))=((4000)/(2000))^(4)=16`
`P_(2)=16xxP_(1)=16xx60=960w`
`P=sigma A rhoT^(4)`


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