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

Let the thermal resistance of the aluminium rod be $R_{Al} = R$.

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

In arrangement (a), the three rods are 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 is $H_a = \dfrac{\Delta T}{R_a} = \dfrac{100}{\dfrac{5R}{2}} = \dfrac{40}{R}$.

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

In arrangement (b), one aluminium rod is in series with a parallel combination of a copper rod and an aluminium rod. The equivalent resistance of the parallel part $R_p$ is:

$\dfrac{1}{R_p} = \dfrac{1}{R_{Cu}} + \dfrac{1}{R_{Al}} = \dfrac{2}{R} + \dfrac{1}{R} = \dfrac{3}{R} \Rightarrow R_p = \dfrac{R}{3}$

The total equivalent resistance for arrangement (b) is:

$R_b = R_{Al} + R_p = R + \dfrac{R}{3} = \dfrac{4R}{3}$

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