Key idea In a series combination of heat conductors, the rate of heat flow remains constant.
According to the question, the figure given below shows the conduction of heat through series combinations of rods.


where, \(T_1, T_2=\) temperature of junctions,
\(I=\text { heat current }=\frac{d Q}{d t}\)
As, \(\frac{d Q_1}{d t}=\frac{d Q_2}{d t}=\frac{d Q_3}{d t}=\frac{T_i-T_f}{k_{\text {eq }}}\)
So, \(\frac{k_1 A_1\left(T_i-T_1\right)}{l_1}=\frac{k_2 A_2\left(T_1-T_2\right)}{l_2}=\frac{k_3 A_3\left(T_2-T_f\right)}{l_3}\)
\(\begin{array}{rlrl}
& \ddots & A_1 & =A_2=A_3 \\
& \text {and } & l_1 & =l_2=l_3 \\
\Rightarrow & 2 k\left(100-T_1\right) & =k\left(T_1-T_2\right)=0.5\left(T_2-0\right) \ldots (i)
\end{array}\)
Equivalent coefficient of thermal conductivity,
\(\begin{aligned}
& \frac{1}{k_{\mathrm{eq}}}=\frac{1}{2 k}+\frac{1}{k}+\frac{2}{k} \\
& \Rightarrow k_{\text {eq }}=\frac{2 k}{7} \\
& \because \text { Heat current, } \frac{d Q}{d t}=\frac{d Q_1}{d t} \\
& \Rightarrow \frac{k_{e q} A\left(T_i-T_f\right)}{l}=\frac{k_1 A_1\left(T_i-T_1\right)}{l_1}
\end{aligned}\)
here, \(l=l_1+l_2+l_3=l_1+l_1+l_1=3 l_1\) and
\(A=A_1+A_2+A_3=A_1+A_1+A_1=3 A_1\)
So, \(\frac{2 k}{7} \times \frac{3 A_1}{3 l_1}(100-0)=\frac{2 k A_1\left(100-T_1\right)}{l_1}\)
or \(\frac{1}{7} \times 100=100-T_1\)
\(\begin{array}{ll}
\Rightarrow & T_1=100-\frac{100}{7} \\
\Rightarrow & T_1=\frac{600^{\circ} \mathrm{C}}{7}
\end{array}\)
Similarly,
From Eq. (i), we get
\(\begin{aligned}
2 k\left(100-T_1\right) & =0.5 k T_2 \\
\Rightarrow 200-2 \times \frac{600}{7} & =0.5 T_2 \\
\Rightarrow T_2 & =\frac{400^{\circ}}{7} \mathrm{C}
\end{aligned}\)
Hence, the correct option is (a).