Three rods of material x and three rods of material y of identical...

Treating the given network of rods in terms of thermal resistance R_x and R_y with

${\text{R}}_{\text{x}} = \dfrac{\text{L}}{\text{A} \times \text{0.92}} \text{and} {\text{R}}_{\text{y}} = \dfrac{\text{L}}{\text{A} \times \text{0.46}} \left[\text{as R} = \dfrac{\text{L}}{\text{AK}}\right]$

so that if R_x = R, R_y = 2R_x = 2R

Now as in this bridge [(P/Q) = (R/S)], so the bridge is balanced, i.e., the temperature at junctions C and D is equal and the rod CD becomes ineffective as no heat will flow through it.

Now as the thermal resistance of the bridge between junctions B and E is

$\dfrac{1}{{\text{R}}_{\text{BE}}} = \dfrac{1}{\left(\text{R} + \text{R}\right)} + \dfrac{1}{\left(\text{2R} + \text{2R}\right)} \text{,} \text{i.e.,} {\text{R}}_{\text{BE}} = \dfrac{4}{3} \text{R}$

The total resistance of bridge between A and E will be

R_eq = R_AB + R_BE = 2R + (4/3) R = (10/3) R

So the net rate of flow of heat through the bridge will be

$\dfrac{\text{dQ}}{\text{dt}} = \dfrac{\text{\Delta} \text{\theta}}{{\text{R}}_{\text{eq}}} = \dfrac{\left(\text{60} - \text{10}\right)}{\left(\text{10/3}\right) \text{R}} = \dfrac{\text{15}}{\text{R}}$

Now if T_B is the temperature at B,

$[ \dfrac{\text{dQ}}{\text{dt}} {]}_{\text{AB}} = \dfrac{\text{\Delta} \text{\theta}}{{\text{R}}_{\text{AB}}} = \dfrac{\text{60} - {\text{T}}_{\text{B}}}{2 \text{R}}$

$\text{But} [ \dfrac{\text{dQ}}{\text{dt}} {]}_{\text{AB}} = \dfrac{\text{dQ}}{\text{dt}} \text{,} \text{i.e.,} \dfrac{\text{60} - {\text{T}}_{\text{B}}}{2 \text{R}} = \dfrac{\text{15}}{\text{R}} \text{,} \text{i.e.,} {\text{T}}_{\text{B}} = {\text{30}}^{\text{o}} \text{C}$

Also at B

$[ \dfrac{\text{dQ}}{\text{dt}} {]}_{\text{AB}} = [ \dfrac{\text{dQ}}{\text{dt}} {]}_{\text{BC}} + [ \dfrac{\text{dQ}}{\text{dt}} {]}_{\text{BD}}$

$\text{i.e.,} \dfrac{\text{15}}{\text{R}} = \dfrac{\text{30} - {\text{T}}_{\text{C}}}{\text{R}} + \dfrac{\text{30} - {\text{T}}_{\text{D}}}{\text{2R}}$

and as T_C = T_D = T, 30 = 3 (30 - T),

i.e., T_C = T_D = T = 20^o C