A black rectangular surface of area A emits energy E per second at `27^circC`. If length and breadth are reduced to one third of initial value and temperature is raised to `327^circC`, then energy emitted per second becomes

Options

A. `(4E)/(9)`
B. `(7E)/(9)`
C. `(10E)/(9)`
D. `(16E)/(9)`

Answer

D

Explanation

According to question
`E=esigmaA(T^(4)-T_(0)^(4))`
[where symbols have their usual meanings]
When l and b change to `(l)/(3)and (b)/(3)` , respectively .
Area becomes `(l)/(3)xx (b)/(3)=(lb)/(3)=(A)/(3)" " (becauseA=lb)`
Now for two different cases
`(E')/(E)=(A')/(A)((227+373)^(4))/((27+273)^(4))=(1)/(9)((600)/(300))^(4)`
`therefore" " E'=(1)/(9)xx(2)^(4)xxE`
`rArr" " E'=(16E)/(9)`