How much `.^(235)U` is consumed in a day in an atomic house operating at 400 MW, provided the whole of mass `.^(235)U` is converted into energy ?

Explanation

`P = 400 MW = 400xx10^(6)W`,
`c=3xx10^(8)m//s`
t = 24 hours `= 24xx60xx60` sec
`E = mc^(2)`
`(Pt)/(c^(2))=m [because P= (E )/(t)]`
`m = (400xx10^(-6)xx24xx60xx60)/(9xx10^(6))`
`= 384xx10^(-6)kg`
`therefore` Mass required `= 384xx10^(-6)xx10^(3)g`
`= 0.384 g`