1.

Due to fall in main voltage, the temperature of filament of a bulb falls from 3000 K to 2000 K. What is the percentage fall in power consumption ?

Answer»

`75.25`
`80.25`
`85.25`
NONE of the above

Solution :`P_(1)=sigmaAT_(1)^(4)` and `P_(2)=AT_(2)^(4)`
`:.(P_(1))/(P_(2))=((T_(1))/(T_(2)))^(4)=((3000)/(2000))^(4)=(81)/(16)`
`(P_(2))/(P_(1))=(16)/(81)""rArr""1-(P_(2))/(P_(1))=1-(16)/(81)%` FALL in power `(P_(1)-P_(2))/(P_(1))xx100=(81-16)/(81)xx100`
`=80*25%`
Hence the CORRECT choice is (b).


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