Temperature of $$B$$ will be higher because, due to expansion centre of mass $$B$$ will comedown same heat is supplied but in $$B,$$ Potential energy is decreased therefore internal energy gain will be more.

$$t \propto \frac{\ell }{A},t' \propto \frac{{2\ell }}{{\frac{A}{2}}} \Rightarrow t' = 4t = 4 \times 12 = 48$$

From formula temperature of junction;
$$\eqalign{ & \theta = \frac{{{K_{{\text{copper}}}}{\theta _{{\text{copper}}}}{l_{{\text{steel}}}} + {K_{{\text{steel}}}}{\theta _{{\text{steel}}}}{l_{{\text{copper}}}}}}{{{K_{{\text{copper}}}}{l_{{\text{steel}}}} + {K_{{\text{steel}}}}{l_{{\text{copper}}}}}} \cr & = \frac{{9k \times 100 \times \frac{L}{2} + k \times 0 \times \frac{L}{2}}}{{9k \times \frac{L}{2} + k \times \frac{L}{2}}} = \frac{{\frac{{900}}{2}kL}}{{\frac{{10kL}}{2}}} = {90^ \circ }C \cr} $$

The heat flowrate is same

$$\eqalign{ & \therefore \,\,\frac{{KA\left( {400 - T} \right)}}{\ell } = \frac{{2\,KA\left( {T - 10} \right)}}{\ell } \cr & \therefore \,\,T = {140^ \circ }C \cr} $$
The temperature gradient access $$Pd$$  is
$$\eqalign{ & \frac{{dT}}{{dx}} = \frac{{140 - 10}}{1} \cr & \therefore \,\,dt = 130\,dx \cr} $$
Therefore change temperature at a cross - section $$M$$ distant $$'x'$$ from $$P$$ is
$$\eqalign{ & dl = dx\alpha \,\Delta T \cr & = dx\,\alpha \left( {130x} \right) \cr & \therefore \int {dl = 130\alpha \int\limits_0^1 {xdx} } \cr & \therefore \,\,\Delta l = 130 \times 1.2 \times {10^{ - 5}} \times \frac{1}{2} \cr & = 78 \times {10^{ - 5}}m \cr} $$

$$\eqalign{ & {\theta _A} - {\theta _B} = {36^ \circ }C\,\,\left( {{\text{Given}}} \right) \cr & {K_A} = 2\,{K_B}\,\,\left( {{\text{Given}}} \right) \cr & {\theta _C} = \frac{{\frac{{{K_A}}}{\ell }{\theta _A} + \frac{{{K_B}}}{\ell }{\theta _B}}}{{\frac{{{K_A}}}{\ell } + \frac{{{K_B}}}{\ell }}} \cr} $$

$$\eqalign{ & \therefore \,\,{\theta _C} = \frac{{2\,{\theta _A} + {\theta _B}}}{3} \cr & = \frac{{2\,{\theta _A} + {\theta _A} - 36}}{3} \cr & = \frac{{3\left( {{\theta _A} - 12} \right)}}{3} \cr & \therefore \,\,{\theta _A} - {\theta _C} = 12 \cr} $$

The quantity of heat flowing across a slab in time $$t,$$ $$Q = \frac{{KA\Delta \theta }}{l}$$
$$K =$$  thermal conductivity
$$\Delta Q =$$  change in temperature
$$A =$$  area of slab
$$l =$$  thickness
For same heat flow through each slab and (composite slab), we have
$$\eqalign{ & \frac{{{K_1}A\left( {\Delta {\theta _1}} \right)}}{l} = \frac{{{K_2}A\left( {\Delta {\theta _2}} \right)}}{l} = \frac{{K'A\left( {\Delta {\theta _1} + \Delta {\theta _2}} \right)}}{{2l}} \cr & {\text{or}}\,\,{K_1}\Delta {\theta _1} = {K_2}\Delta {\theta _2} = \frac{{K'}}{2}\left( {\Delta {\theta _1} + \Delta {\theta _2}} \right) = C\,\,\left( {{\text{say}}} \right) \cr & {\text{So,}}\,\,\Delta {\theta _1} = \frac{C}{{{K_1}}},\Delta {\theta _2} = \frac{C}{{{K_2}}} \cr & {\text{and}}\,\,\left( {\Delta {\theta _1} + \Delta {\theta _2}} \right) = \frac{{2C}}{{K'}} \cr & {\text{or}}\,\,\frac{C}{{{K_1}}} + \frac{C}{{{K_2}}} = \frac{{2C}}{{{K^\prime }}} \cr & {\text{or}}\,\,C\left( {\frac{{{K_1} + {K_2}}}{{{K_1}{K_2}}}} \right) = \frac{{2C}}{{K'}} \cr & \therefore K' = \frac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}} \cr & {\text{Given,}}\,\,{K_1} = K,{K_2} = 2K \cr & {\text{So,}}\,\,K' = \frac{{2K \times 2K}}{{K + 2K}} = \frac{4}{3}K \cr} $$

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

As the temperature difference $$\Delta T = {10^ \circ }C$$   as well as the thermal resistance is same for both the cases, so thermal current or rate of heat flow will also be same for both the cases.

Let $$T$$ be temperature of the junction
Here, $${K_A} = 2{K_B},T - {T_B} = 50\,K$$
At the steady state, $${H_A} = {H_B}$$
$$\eqalign{ & \therefore \frac{{{K_A}A\left( {{T_A} - T} \right)}}{L} = \frac{{{K_B}A\left( {T - {T_B}} \right)}}{L} \cr & 2{K_B}\left( {{T_A} - T} \right) = {K_B}\left( {T - {T_B}} \right) \cr & {T_A} - T = \frac{{T - {T_B}}}{2} = \frac{{50K}}{2} = 25K \cr} $$

Temperature difference between the end points $$A$$ and $$D = 200 - 20 = {180^ \circ }C$$
As the resistances for the three parts are equal, the temperature difference must be distributed equally in the three parts $$\left( {\frac{{180}}{3} = {{60}^ \circ }C} \right)$$
$$\therefore $$ Temperature of $$B = {200^ \circ }C - {60^ \circ } = {140^ \circ }C.$$