In this situation, a rod of length L and area of cross-section `piR^(2)` and another of same length L and area of cross-section `pi[(3R)^(2)-R^(2)]=8pi(R^(2)` will coduct heat simultaneously, so total heat flowing per second will be
`(dQ)/(dt) = (dQ_(1)/(dt) + (dQ_(2))/(dt)`
`=K_(1)pir^(2)(T_(1)-T_(2))/(L) +(K_(2)8piR^(2)(T_(1)-T_(2))/(L)`..............(i)
Now , if the equivalent conductivity is K.
Then, `(dQ)/(dt) = K(9piR^(2)(T-(1)-T_(2))/(L)` [As A `=pi(3R)^(2)]`.......(ii)
So, from Eqs, (i) amnd (ii), we have
9K = `K_(1) + 8K_(2)`
i.e., `K = (K_(1)+8K_(2))/9 = (K+8xx2K)/(9) = (17K)/(9)`