$Q=\frac{KA\,(\Delta \ta )t}{l}$                    $\because $ Q and $\Delta \ta $ are same for both spheres hence                    $K\propto \frac{l}{At}\propto \frac{l}{{{r}^{2}}t}$Or, $\frac{{{K}_{\text{larger}}}}{{{K}_{\text{smaller}}}}=\frac{{{l}_{l}}}{{{l}_{s}}}\times {{\left( \frac{{{r}_{s}}}{{{r}_{l}}} \right)}^{2}}\times \frac{{{t}_{s}}}{{{t}_{l}}}$. It is given that ${{r}_{l}}=2{{r}_{s}},$ ${{l}_{l}}=\frac{1}{4}{{l}_{s}}$ and ${{t}_{1}}=25$min,  ${{t}_{s}}=16$min.            Or, $\frac{{{k}_{\text{larger}}}}{{{k}_{\text{smaller}}}}=\left( \frac{1}{4} \right){{\left( \frac{1}{2} \right)}^{2}}\times \frac{16}{25}=\frac{1}{25}$

$Q\propto \frac{A}{l}\propto \frac{{{r}^{2}}}{l}$ Or, $\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\times \frac{{{l}_{1}}}{{{l}_{2}}}$            Or, $\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{4}{1}\times \frac{1}{2}$ Or, ${{Q}_{2}}=2{{Q}_{1}}$

 $\frac{dQ}{dt}=-KA\frac{d\ta }{dx}$; when $K=\infty ,\ \ \frac{d\ta }{dx}=0$ i.e. q  is independent of x i.e. constant or uniform.

Temperature of interface $\ta =\frac{{{K}_{1}}{{\ta }_{1}}{{l}_{2}}+{{K}_{2}}{{\ta }_{2}}{{l}_{1}}}{{{K}_{1}}{{l}_{2}}+{{K}_{2}}{{l}_{1}}}$ It is given that ${{K}_{Cu}}=9{{K}_{S}}.$ So if ${{K}_{S}}={{K}_{1}}=K$ n ${{K}_{Cu}}={{K}_{2}}=9K$            Or, q $=\frac{9K\times 100\times 6+K\times 0\times 18}{9K\times 6+K\times 18}$=$\frac{5400K}{72K}=75{}^\circ C$

$\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{l_{1}^{2}}{l_{2}^{2}}$ $\refore $${{K}_{2}}=\frac{{{K}_{1}}l_{2}^{2}}{l_{1}^{2}}=\frac{0.92\times {{(4.2)}^{2}}}{{{(8.4)}^{2}}}=0.23$

Given \[{{A}_{1}}={{A}_{2}}\] and \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{5}{4}\] Q \[{{R}_{1}}={{R}_{2}}\Rightarrow \] \[\frac{{{l}_{1}}}{{{K}_{1}}A}=\frac{{{l}_{2}}}{{{K}_{2}}A}\]Þ \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{5}{4}.\]

Thermal conductivity of composite plate            ${{K}_{eq}}=\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}=\frac{2\times 2\times 3}{2+3}$$=\frac{12}{5}$=2.4