The three rods depicted in the figure have identical geometrical...

Let the length and cross-sectional area of each rod be $L$ and $A$ respectively.

The thermal resistance of a rod is given by $R = \dfrac{L}{KA}$.

Let the thermal resistance of the aluminium rod be $R_{Al} = R$. Since $K_{Cu} = 2K_{Al}$, the thermal resistance of the copper rod is $R_{Cu} = \dfrac{R}{2}$.

In arrangement (a), the three rods are connected in series. The equivalent thermal resistance is:

$R_{a} = R_{Al} + R_{Cu} + R_{Al} = R + \dfrac{R}{2} + R = \dfrac{5R}{2}$

The rate of heat flow in arrangement (a) is:

$H_a = \dfrac{\Delta T}{R_a} = \dfrac{100}{\dfrac{5R}{2}} = \dfrac{40}{R}$

Given $H_a = 40 \text{ W}$, we have \dfrac{40}{R} = 40 \Rightarrow R = 1 ^{}^{{}^{\circ}}\text{C W}^{-1}.

In arrangement (c), the three rods are connected in parallel. The equivalent thermal resistance $R_c$ is given by:

$\dfrac{1}{R_c} = \dfrac{1}{R_{Al}} + \dfrac{1}{R_{Cu}} + \dfrac{1}{R_{Al}} = \dfrac{1}{R} + \dfrac{2}{R} + \dfrac{1}{R} = \dfrac{4}{R}$

$\Rightarrow R_c = \dfrac{R}{4}$

The rate of heat flow in arrangement (c) is:

$H_c = \dfrac{\Delta T}{R_c} = \dfrac{100}{\dfrac{R}{4}} = \dfrac{400}{R}$

Substituting $R = 1$, we get $H_c = 400 \text{ W}$.