Seven rods A , B , C , D , E , F and G are connected as depicted in...

Let the thermal resistance of a rod be given by $R = \dfrac{l}{KA}$.

The thermal resistances of the given rods are:
$R_A = \dfrac{l}{K_0 A}$
$R_B = \dfrac{l}{2K_0 A} = \dfrac{R_A}{2}$
$R_C = \dfrac{l}{K_0 A} = R_A$
$R_D = \dfrac{l}{2K_0 A} = \dfrac{R_A}{2}$
$R_F = \dfrac{l}{4K_0 A} = \dfrac{R_A}{4}$

Since rod E is maintained at a steady temperature $T_1$ and rod G is maintained at a steady temperature $T_2$, they act as isothermal reservoirs. The entire rod E is at $T_1$ and the entire rod G is at $T_2$. Therefore, their thermal conductivities ($K_E$ and $K_G$) do not contribute to the thermal resistance of the heat flow path.

The rods A, B, C, D, and F form a Wheatstone bridge circuit between the temperature reservoirs at $T_2$ and $T_1$. Heat flows from rod G ($T_2$) to rod E ($T_1$).

Let's check the balance condition of the Wheatstone bridge:
$\dfrac{R_B}{R_A} = \dfrac{\dfrac{l}{2K_0 A}}{\dfrac{l}{K_0 A}} = \dfrac{1}{2}$
$\dfrac{R_D}{R_C} = \dfrac{\dfrac{l}{2K_0 A}}{\dfrac{l}{K_0 A}} = \dfrac{1}{2}$

Since $\dfrac{R_B}{R_A} = \dfrac{R_D}{R_C}$, the Wheatstone bridge is balanced. This implies that the junctions connected by rod F are at the same temperature, and no heat flows through rod F.

The equivalent thermal resistance of the network $R_{eq}$ is the parallel combination of the top branch (rods B and A in series) and the bottom branch (rods D and C in series).

Resistance of the top branch:
$R_{top} = R_B + R_A = \dfrac{l}{2K_0 A} + \dfrac{l}{K_0 A} = \dfrac{3l}{2K_0 A}$

Resistance of the bottom branch:
$R_{bottom} = R_D + R_C = \dfrac{l}{2K_0 A} + \dfrac{l}{K_0 A} = \dfrac{3l}{2K_0 A}$

The equivalent thermal resistance $R_{eq}$ is:
$\dfrac{1}{R_{eq}} = \dfrac{1}{R_{top}} + \dfrac{1}{R_{bottom}} = \dfrac{2K_0 A}{3l} + \dfrac{2K_0 A}{3l} = \dfrac{4K_0 A}{3l}$
$\Rightarrow R_{eq} = \dfrac{3l}{4K_0 A}$

The rate of heat flowing from the source is:
$H = \dfrac{T_2 - T_1}{R_{eq}} = \dfrac{T_2 - T_1}{\dfrac{3l}{4K_0 A}} = \dfrac{4K_0 A (T_2 - T_1)}{3l}$

Answer: $\dfrac{4 K_0 A (T_2 - T_1)}{3l}$