Correct Answer - A

a) We know that, relation between the temperature gradient (TG) and thermal conductivity (K),

`(dQ)/(dt) = -KA(d(theta))/(dx)=-KAxx(TG)`

i.e., `TG alpha 1/K` `(dQ)/(dt)` = constant

or `Kalpha1/t rArrK_(1)alpha1/(t_(1))`...........(i)

`K_(2)alpha1/t_(2)`..............(ii)

From the Eqs. (i) and (ii), we get

`K_(1)/K_(2)=(1//t_(1))/(1//t_(2)) rArrK_(1):K_(2)=t_(2):t_(1)`