Four identical rods AB , CD , CF and DE are connected as shown in...

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

The resistances of the given rods are:
$R_{AB} = R$
$R_{CF} = R$
$R_{DE} = R$

Since CD is a single rod of length $l$ and B is its midpoint, the lengths of sections CB and BD are $l/2$. Therefore, their thermal resistances are:
$R_{CB} = \dfrac{R}{2}$
$R_{BD} = \dfrac{R}{2}$

Let the temperatures at junctions B, C, and D be $T_B$, $T_C$, and $T_D$ respectively. Applying the principle of continuity of heat flow (Kirchhoff's junction law) at junction C:
$\dfrac{T_3 - T_C}{R} + \dfrac{T_B - T_C}{R/2} = 0$
$\Rightarrow T_3 - T_C + 2(T_B - T_C) = 0$
$\Rightarrow 3T_C = T_3 + 2T_B$
$\Rightarrow T_C = \dfrac{T_3 + 2T_B}{3}$

Applying the junction law at D:
$\dfrac{T_2 - T_D}{R} + \dfrac{T_B - T_D}{R/2} = 0$
$\Rightarrow T_2 - T_D + 2(T_B - T_D) = 0$
$\Rightarrow 3T_D = T_2 + 2T_B$
$\Rightarrow T_D = \dfrac{T_2 + 2T_B}{3}$

Applying the junction law at B:
$\dfrac{T_1 - T_B}{R} + \dfrac{T_C - T_B}{R/2} + \dfrac{T_D - T_B}{R/2} = 0$
$\Rightarrow T_1 - T_B + 2(T_C - T_B) + 2(T_D - T_B) = 0$
$\Rightarrow T_1 + 2T_C + 2T_D - 5T_B = 0$

Substituting the expressions for $T_C$ and $T_D$ into the equation for junction B:
$T_1 + 2\left(\dfrac{T_3 + 2T_B}{3}\right) + 2\left(\dfrac{T_2 + 2T_B}{3}\right) - 5T_B = 0$

Multiplying the entire equation by $3$ to clear the denominator:
$3T_1 + 2T_3 + 4T_B + 2T_2 + 4T_B - 15T_B = 0$
$\Rightarrow 3T_1 + 2(T_2 + T_3) - 7T_B = 0$
$\Rightarrow 7T_B = 3T_1 + 2(T_2 + T_3)$
$\Rightarrow T_B = \dfrac{3T_1 + 2(T_2 + T_3)}{7}$

Answer: $\dfrac{3 T_1 + 2(T_2 + T_3)}{7}$