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

Let the thermal resistance of rod A be $R_A = \dfrac{l}{K_0 A} = R_0$.

The thermal resistances of the other relevant rods are:
$R_C = \dfrac{l}{K_0 A} = R_0$
$R_B = \dfrac{l}{2K_0 A} = \dfrac{R_0}{2}$
$R_D = \dfrac{l}{2K_0 A} = \dfrac{R_0}{2}$

Since rod E is maintained at a steady temperature $T_1$ and rod G is maintained at a steady temperature $T_2$, the junctions of A and C with E are at temperature $T_1$, and the junctions of B and D with G are at temperature $T_2$.

Let the temperature of the upper junction (between A, B, and F) be $T_x$ and the lower junction (between C, D, and F) be $T_y$.

Applying the principle of continuity of heat flow (Kirchhoff's junction law) at the upper junction:
$\dfrac{T_1 - T_x}{R_A} + \dfrac{T_2 - T_x}{R_B} + \dfrac{T_y - T_x}{R_F} = 0$
$\Rightarrow \dfrac{T_1 - T_x}{R_0} + \dfrac{2(T_2 - T_x)}{R_0} + \dfrac{T_y - T_x}{R_F} = 0$ --- (1)

Similarly, at the lower junction:
$\dfrac{T_1 - T_y}{R_C} + \dfrac{T_2 - T_y}{R_D} + \dfrac{T_x - T_y}{R_F} = 0$
$\Rightarrow \dfrac{T_1 - T_y}{R_0} + \dfrac{2(T_2 - T_y)}{R_0} + \dfrac{T_x - T_y}{R_F} = 0$ --- (2)

Due to the symmetry of the upper and lower branches ($R_A = R_C$ and $R_B = R_D$), the temperature of the upper and lower junctions must be equal, i.e., $T_x = T_y$.

Since $T_x = T_y$, no heat flows through rod F, and it possesses a uniform temperature $T_F = T_x = T_y$.

Substituting $T_y = T_x$ into equation (1):
$\dfrac{T_1 - T_x}{R_0} + \dfrac{2(T_2 - T_x)}{R_0} + 0 = 0$
$\Rightarrow T_1 - T_x + 2T_2 - 2T_x = 0$
$\Rightarrow 3T_x = T_1 + 2T_2$
$\Rightarrow T_x = \dfrac{T_1 + 2T_2}{3}$

Thus, the uniform temperature of rod F is $\dfrac{T_1 + 2T_2}{3}$.

Answer: $\dfrac{T_1 + 2T_2}{3}$