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The ratio of energy of emitted radiation...

The ratio of energy of emitted radiation of a black body at `27^(@)C` and `927^(@)C` is

A

`1 : 4`

B

`1 : 16`

C

`1 : 64`

D

`1 : 256`

Text Solution

Verified by Experts

The correct Answer is:
D
As we use
`(P_(1))/(P_(2))=(27+273)^(4)/(927+273)^(4)=((300)/(1200))^(4)`
`implies (P_(1))/(P_(2))=(1)/(256)`
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