A sphere at temperature `600 K` is placed in an enviroment to temperature is `200 K`. Its cooling rate is `H`. If its temperature reduced to `400 K` then cooling rate in same enviorment will become

Options

A. `(3)/(16)R`
B. `(16)/(3)R`
C. `(9)/(27)R`
D. None

Answer

A

Explanation

According to stefan's law -
`(R')/(R)=((400)^4-(200)^4)/((600)^4-(200)^4)=(4^4-2^4)/(6^4-2^4)`
`=((4^2+2^2)(4^2-2^2))/((6^2+2^2)(6^2-2^2))=(20xx12)/(40xx32)`
`R'=3/16R`