Correct Answer - A::B

`T_2gtT_1`

`A=pi(R_2^2-R_1^2)`

`so, (dQ)/(dt)=(KA(T_2-T_1))/l`

`=(K pi(R_2^2-R_1^2)(T_2-T_1))/l`

considering a concentric cylindtical cell of radius R and thickness DR.The radial heat flow through the cell.

H=(dQ)/(dt)=-KA(d theta)/(dr)`

`[-ve because as r increases Q decreases]`

`A=2 pi rl,`

`H=-2 pi rl K(d theta)/(dr)`

`or, int_(r_1)^(r_2)(dr)/r=-(2 pi rlK)/H int_(T_1)^(T_2)d theta`

Integrating and simplifying we get

`H=(dQ)/(dt)=(2 pi Kl(T_2-T_1))/(In(R_2//R_1))`.