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There rods of the same dimensions have thermal conductivities $3K, 2K$ and $K$. They are arranged as shown in fig. with their ends at $100\,^oC, 50\,^oC$ and $20\,^oC$. The temperature of their junction is....... $^oC$
$60$
$70$
$50$
$35$
Solution
$\frac{{dQ}}{{dt}} = KA\frac{{\Delta T}}{L}$
$For\,the\,first\,rod,\left( {\frac{{dQ}}{{dt}}} \right) = \frac{{3KA}}{L}\left( {100 – \theta } \right)$
$Similarly,{\left( {\frac{{dQ}}{{dt}}} \right)_2} = 2K\frac{A}{L}\left( {\theta – 50} \right)$
${\left( {\frac{{dQ}}{{dt}}} \right)_3} = K\frac{A}{L}\left( {\theta – 20} \right)$
$Now,{\left( {\frac{{dQ}}{{dt}}} \right)_1} = {\left( {\frac{{dQ}}{{dt}}} \right)_2} + {\left( {\frac{{dQ}}{{dt}}} \right)_3}$
$ \Rightarrow \,\,3\left( {100 – \theta } \right) = 2\left( {\theta – 50} \right) + \left( {\theta – 20} \right)$
$ \Rightarrow \theta = {70^ \circ }$
