\((a)\) \(\because\) The rods have identical dimensions.

Let their area of crossection be \(=A\) and length be \(=L\)

So each rod would have heat resistance of

\(R=\frac{L}{K A}\)

\(R_{\text {eff }}=R_1+R_2+R_3\)

\(=\frac{L}{K_{A l} A}+\frac{L}{K_{C u} A}+\frac{L}{K_{A l} A}\)

\(R_{\text {eff }}=\frac{L}{A} \times \frac{5}{400}\)   \(\left[\because K_N=200, K_{C u}=400\right]\)

\(H_1=\frac{\Delta T}{R_{\text {eff }}}=\frac{100}{\frac{L}{A} \times \frac{5}{400}} \quad \ldots (1)\)

\((b)\) When rods are connected in parallel

\(\frac{1}{R_{\text {eff }}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

\(R_{\text {eff }}=\frac{L}{A} \times 800\)

\(H_2=\frac{100}{\frac{L}{A} \times 800} \quad \ldots (2)\)

\((2)\) by \((1)\)

\(\frac{40}{H_2}=\frac{400}{800 \times 5} \Rightarrow H_2=400\,W\)