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

Let the thermal resistance of the aluminium rod be $R_{Al} = \dfrac{L}{K_{Al}A} = R$.

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

For arrangement (a), 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 \implies R = 1 ^{}^{{}^{\circ}}\text{C W}^{-1}.

For arrangement (b), the equivalent thermal resistance is:

$R_b = R_{Al} + \dfrac{R_{Cu} R_{Al}}{R_{Cu} + R_{Al}} = R + \dfrac{\dfrac{R}{2} \cdot R}{\dfrac{R}{2} + R} = R + \dfrac{R}{3} = \dfrac{4R}{3}$

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

$H_b = \dfrac{\Delta T}{R_b} = \dfrac{100}{\dfrac{4R}{3}} = \dfrac{75}{R}$

Substituting R = 1 ^{}^{{}^{\circ}}\text{C W}^{-1}, we get $H_b = 75 \text{ W}$.

Answer: $75 \text{ W}$