i) Thermal resitance of aluminium cube, `R_(1) = l/KA`
or `R_(1)=(3.0 xx 10^(-2))/(237)(3.0 xx 10^(-2))^(2)`
`0.14 KW^(-1)`
and thermal resistance of copper cube, `R_(2) = l/(KA)`
or `R_(2) = (3.0 xx 10^(-2))/(401)(3.0 xx 10^(-2))^(2)`
Aw these two resistance are in paralle, their equivalent resistance will be
`R=(R_(1)R_(2))/(R_(1)+R_(2))= ((0.14)(0.08))/(0.14) + (0.08) = 0.05 KW^(-1)`
`therefore` Thermal current, H= ("Temperature difference")/("Thermal resistance")`
`(100-20)/(0.05) = 1.6 xx 10^(3)`W
ii) Parallel, thermal current distributing in the inverse ratio of resistance. Hence ,
`(H_(Cu)/H_(Al) = R_(Al)/R_(Cu) = R_(1)/R_(2)=0.14/0.08 = 1.75`