Correct Answer - C
The energy radiated per second , `P`, by a body of surface area `A`, at temperature `T` Kelvin, is given by
`E = e sigma AT^(4)`
Hence, `E_(1) = e sigma 4 pi(1)^(2) xx (4000)^(4)`
`=e sigma pi xx 1024 xx 10^(12)J`
And `E_(2) = e sigma[4 pi (4)^(2)](2000)^(4)`
`=e sigma pi xx 1024 xx 10^(12)J`
so, `E_(1)=E_(2)`.