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 ?

Options

A. `75.25`
B. `80.25`
C. `85.25`
D. None of the above

Answer

B

Explanation

`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).