A composite object is formed by combining a uniform rod of circular cross-section with  thermal conductivity K and a frustum of same length with thermal conductivity 2K as  shown in the figure. The equivalent thermal conductivity of the object is given as $\frac{NK}{5}$,  find ‘N’.

Consider the object as two portion A ‘uniform rod’ and ‘A frustum’ with thermal  resistance $R_{1}and{R}_2$  respectively then

$R_1=\frac{l_1}{k_1{A}_1}=\frac{l}{k\pi r^2}$  and  $R_2=\frac{l_2}{k_2{A}_2}=\frac{l}{\left(2k\right)\left(\pi r_1{r}_2\right)}=\frac{l}{4k\pi r^2}$.
$\therefore$ Equivalent resistance $R_{eq}=R_1+R_2$.
 $\Rightarrow R_{eq}=\frac{5l}{4k\pi r^2}\mathrm{.........}\left(1\right)$.
Now if we consider the same lamina with equivalent thermal conductivity $K_{eq}$ ,then
$R_{eq}=R_1+R_2=\frac{l}{k_{eq}\pi r^2}+\frac{l}{k_{eq}\left(2\pi r^2\right)}=\frac{3l}{2k_{eq}\pi r^2}\mathrm{........}\left(2\right)$ .
By equating the terms of $R_{eq}$ from eqn. (1) & (2), we get 
 $\frac{5l}{4k\pi r^2}=\frac{3l}{2k_{eq}\pi r^3}$
$k_{eq}=\frac{6k}{5}$