Consider a shell of thickness $$(dr)$$  and of radius $$(r)$$  and the temperature of inner and outer surfaces of this shell be $$T,\left( {T - dT} \right)$$
$$\eqalign{ & H = \frac{{KA\left[ {\left( {T - dT} \right) - T} \right]}}{{dr}} = \frac{{ - KAdT}}{{dr}} \cr & H = - 4\,\pi K{r^2}\frac{{dT}}{{dr}}\,\,\left( {\because \,\,A = 4\,\pi {r^2}} \right) \cr & {\text{Then, }}\left( H \right)\int\limits_{{r_1}}^{{r_2}} {\frac{1}{{{r^2}}}dr = - 4\,\pi K\int\limits_{{T_1}}^{{T_2}} {dT} } \cr & H\left[ {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right] = - 4\,\pi K\left[ {{T_2} - {T_1}} \right] \cr & {\text{or, }}H = \frac{{ - 4\,\pi K{r_1}{r_2}\left( {{T_2} - {T_1}} \right)}}{{\left( {{r_2} - {r_1}} \right)}} \cr} $$

As from law of heat transfer through conduction
$$\eqalign{ & H = \frac{{\Delta Q}}{{\Delta t}} = KA\left( {\frac{{{T_1} - {T_2}}}{1}} \right) \cr & \Rightarrow H \propto \frac{{{r^2}}}{l}\,......\left( {\text{i}} \right) \cr} $$
(A) When $$r = 2{r_0};l = 2{l_0}$$
$$H \propto \frac{{{{\left( {2{r_0}} \right)}^2}}}{{2{l_0}}} \Rightarrow H \propto \frac{{2r_0^2}}{{{l_0}}}$$
(B) When $$r = 2{r_0};l = {l_0}$$
$$H \propto \frac{{{{\left( {2{r_0}} \right)}^2}}}{{{l_0}}} \Rightarrow H \propto \frac{{4r_0^2}}{{{l_0}}}$$
(C) When $$r = {r_0};l = {l_0} \Rightarrow H \propto \frac{{r_0^2}}{{{l_0}}}$$
(D) When $$r = {r_0};l = 2{l_0} \Rightarrow H \propto \frac{{r_0^2}}{{2{l_0}}}$$
It is obvious that heat conduction will be more in case (B).

For a rod of length $$L$$ and area of cross-section $$A$$ whose faces are maintained at temperatures $${T_1}$$ and $${T_2}$$ respectively.
Then in steady state the rate of heat flowing from one face to the other face in time $$t$$ is given by $$\frac{{dQ}}{{dt}} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$$

The heat flow rate is given by
$$\eqalign{ & \frac{{dQ}}{{dt}} = \frac{{kA\left( {{\theta _1} - \theta } \right)}}{x} \cr & \Rightarrow \,\,{\theta _1} - \theta = \frac{x}{{kA}}\frac{{dQ}}{{dt}} \cr & \Rightarrow \,\,\theta = {\theta _1} - \frac{x}{{kA}}\frac{{dQ}}{{dt}} \cr} $$
where $${\theta _1}$$ is the temperature of hot end and $${\theta}$$ is temperature at a distance $$x$$ from hot end.
The above equation can be graphically represented by option (A) .

Heat current $$H = {H_1} + {H_2}$$
$$\eqalign{ & = \frac{{{K_1}\;A\left( {\;{T_1} - {T_2}} \right)}}{d} + \frac{{{K_2}\;A\left( {\;{T_1} - {T_2}} \right)}}{d} \cr & \frac{{{K_{EQ}}2A\left( {\;{T_1} - {T_2}} \right)}}{d} = \frac{{A\left( {{T_1} - {T_2}} \right)}}{d}\left[ {{K_1} + {K_2}} \right] \cr} $$
Hence equivalent thermal conductivities for two rods of equal area is given by
$${K_{EQ}} = \frac{{{k_1} + {k_2}}}{2}$$

The thermal resistance
$$\eqalign{ & \frac{x}{{KA}} + \frac{{4\,x}}{{2\,KA}} = \frac{{3\,x}}{{KA}} \cr & \therefore \,\,\frac{{dQ}}{{dt}} = \frac{{\Delta T}}{{\frac{{3\,x}}{{KA}}}} \cr & = \frac{{\left( {{T_2} - {T_1}} \right)KA}}{{3\,x}} \cr & = \frac{1}{3}\left\{ {\frac{{A\left( {{T_2} - {T_1}} \right)K}}{x}} \right\} \cr & \therefore \,\,f = \frac{1}{3} \cr} $$

Rate of loss of heat by conduction is, $$H = \frac{{\Delta Q}}{{\Delta t}} = KA\left( {\frac{{{T_1} - {T_2}}}{l}} \right)$$
All the symbols have their usual meaning.
For first rod, $${H_1} = {K_1}{A_1}\left( {\frac{{{T_1} - {T_2}}}{{{l_1}}}} \right)$$
For second rod, $${H_2} = {K_2}{A_2}\left( {\frac{{{T_1} - {T_2}}}{{{l_2}}}} \right)$$
but $${l_1} = {l_2}$$  i.e. of same length
and $${H_1} = {H_2}$$   i.e. same rate of loss of heat through conduction
So, we have
$${K_1}{A_1}\left( {{T_1} - {T_2}} \right) = {K_2}{A_2}\left( {{T_1} - {T_2}} \right)$$
or $${K_1}{A_1} = {K_2}{A_2}$$
NOTE
Heat transfer occurs only between regions that are at different temperatures, and the rate of heat flow is $$\frac{{dQ}}{{dt}}.$$

Heat flow from $$B$$ to $$A,A$$  to $$C$$ and $$C$$ to $$B$$ (for steady state condition, $$\frac{{\Delta Q}}{{\Delta t}}$$ is same)
Where $$\frac{{\Delta Q}}{{\Delta t}} = \frac{{kA\Delta T}}{\ell }$$
For sides $$AC$$ and $$CB{\left( {\frac{{\Delta T}}{{\sqrt 2 a}}} \right)_{AC}} = {\left( {\frac{{\Delta T}}{a}} \right)_{CB}}$$
$$\eqalign{ & \Rightarrow \frac{{T - {T_c}}}{{\sqrt 2 a}} = \frac{{{T_c} - \sqrt 2 T}}{a} \cr & \Rightarrow T - {T_c} = \sqrt 2 {T_c} - 2T \cr & \Rightarrow 3T = {T_c}\left( {\sqrt 2 + 1} \right) \Rightarrow \frac{{{T_c}}}{T} = \frac{3}{{\sqrt 2 + 1}} \cr} $$

$$\eqalign{ & \frac{{dQ}}{{dt}} = \frac{{KA\left( {100 - 0} \right)}}{1} = m\frac{{dL}}{{dt}}, \cr & \frac{{K \times 100 \times 0.36}}{{0.1}} = \frac{{4.8 \times 3.36 \times {{10}^5}}}{{60 \times 60}} \cr & K = 1.24\,J/m/s{/^ \circ }C \cr} $$

(A) Conduction is the process of transmission of heat in a body from the hotter part to the colder part without any bodily movement of constituent atoms or molecules of the body.
(B) In convection, the heated lighter particles move upward and colder heavier particles move downward to their place. This depends on weight and hence, on gravity.
(C) Radiation is the process of transmission of heat from one body to another body through electromagnetic waves even through vacuum, irrespective of their temperatures. Hence, choice (B) is correct.