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 R₁  and  R₂  respectively then
 $R_1=\frac{{\ell }_1}{k_1{A}_1}=\frac{\ell }{k\pi r^2}$
And  $R_2=\frac{{\ell }_2}{k_2{A}_2}=\frac{\ell }{\left(2k\right)\left(\pi r_1{r}_2\right)}=\frac{\ell }{4k\pi r^2}$
 Equivalent thermal resistance  $R_{eq}=R_1+R_2$
  $\Rightarrow R_{eq}=\frac{5\ell }{4k\pi r^2}$  ……….. (1)
Now if we consider the same lamina with equivalent thermal conductivity $K_{eq}$ , then
 
$R_{eq}=R_1+R_2=\frac{\ell }{k_{eq}\pi r^2}+\frac{\ell }{k_{eq}\left(2\pi r^2\right)}=\frac{3\ell }{2k_{eq}\pi r^2}$  ……….. (2)
By equating the terms of $R_{eq}$  from eqn. (1) & (2), we get

$$\begin{gathered} \frac{5\ell }{4k\pi r^2}=\frac{3\ell }{2k_{eq}\pi r^2} \\ {k}_{eq}=\frac{6k}{5} \\ \therefore N=6 \end{gathered}$$