Correct Answer - (a) `1:4`
(b) `2.9 :1`
`C `-cooling rate `H-` heat loss
`(H_(Al))/(H_(Cu)) = (sigmaA_(AI)(T_(AI^(4)-T_(0)^(4))))/(sigmaeA_(Cu)(T_(Cu^(4)-T_(0)^(4))))`
at `t = 0 T_(Al) = T_(Cu) =T =(A_(Al))/(A_(Cu)) = (R^(2))/((2R)^(2)) = (1)/(4)`
(b) `C-` cooling rate
`(C_(Al))/(C_(Cu)) = ((A_(Al))/(m_(Al)S_(Al)))((m_(Cu)S_(Cu))/(A_(Cu)))`
`= ((A_(Al))/(A_(Cu)))((m_(Cu))/(m_(Al)))((S_(Cu))/(S_(Al)))`
`= ((R )/(2R))^(2) xx {((2R)/(R))^(3)xx3.4}xx(390)/(900)=2.9`