A sphere and a cube of same material and same volume are heated up to same temperature and allowed to cool in the same sorroundings. The radio of the amounts of radiations emitted in equal time intervals will be

Options

A. `1:1`
B. `(4pi)/(3):(1)/(86400)//^@C 1`
C. `((pi)/(6))^(1//3):1`
D. `(1)/(2)((4pi)/(3))^(2//3):1`

Answer

C

Explanation

`Q=sigmaAt(T^4-T_0^4)`
If `T,T_0,sigma` and t are same for both bodies then
`(Q_("sphere"))/(Q_("cube"))=(A_("sphere"))/(A_("cube"))=(4pir^3)/(6a^2)` ..(i)
But according to problem, volume of sphere = volume of cube
`implies(4)/(3)pir^3=a^3`
`impliesa=((4)/(3)pi)^(1//3)r`
substituting the value of a in Eq. (i) we get
`(Q_("sphere"))/(Q_("cube"))=(4pir^2)/(6a^2)`
`=(4pir^2)/(6{((4)/(3)pi)^(1//3)r}^2)=(4pir^2)/(6{((4)/(3)pi)^(2//3)r^2))`
`=((pi)/(6))^(1//3):1`