A composite slab consists of two materials having coefficient of thermal conductivity \(K\) and \(2 \mathrm{~K}\), thickness \(\mathrm{x}\) and \(4 \mathrm{x}\) respectively. The temperature of the two outer surfaces of a composite slab are \(\mathrm{T}_2\) and \(\mathrm{T}_1\left(\mathrm{~T}_2>\mathrm{T}_1\right)\). The rate of heat transfer through the slab in a steady state is \(\left[\frac{\mathrm{A}\left(\mathrm{T}_2-\mathrm{T}_1\right) \mathrm{K}}{\mathrm{x}}\right] \cdot \mathrm{f}\) where ' \(\mathrm{f}\) ' is equal to

\(
\begin{array}{ll}
& \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\
\because & \mathrm{R}_1=\frac{\mathrm{x}}{\mathrm{KA}}, \mathrm{R}_2=\frac{4 \mathrm{x}}{2 \mathrm{KA}} \\
\therefore & \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{x}}{\mathrm{KA}}+\frac{2 \mathrm{x}}{\mathrm{KA}}=\frac{3 \mathrm{x}}{\mathrm{KA}}
\end{array}
\)
Rate of heat transfer of composite slab is given by,
\(
\begin{aligned}
& \quad \frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{R}_{e \mathrm{q}}}=\frac{\mathrm{KA}\left(\mathrm{T}_2-\mathrm{T}_1\right)}{3 \mathrm{x}} \\
& \therefore \quad \mathrm{f}=\frac{1}{3}
\end{aligned}
\)