Condider two rods of same length and different specific heats `(s_(1), s_(2))`, thermal conductivities `(K_(1), K_(2))` and areas of cross-section `(A_(1), A_(2))` and both having temperatures `(T_(1), T_(2))` at their ends. If their rate of loss of heat due to conduction are equal, then
A. `K_(1)A_(1) = K_(2)A_(2)`
B. `(K_(1)A_(1))/(s_(1)) = (K_(2)A_(2))/(s_(2))`
C. `K_(2)A_(1) = K_(1)A_(2)`
D. `(K_(2)A_(1))/(s_(2)) = (K_(1)A_(2))/(s_(1))`
