Three rods of identical area of cross-section and made from the same metal form the sides of an isosceles triangle $ABC$, right angled at $B$. The points $A$ and $B$ are maintained at temperatures $T$ and $\sqrt 2 T$ respectively. In the steady state the temperature of the point C is ${T_C}$. Assuming that only heat conduction takes place, $\frac{{{T_C}}}{T}$ is equal to
IIT 1995, Diffcult
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(b) $\because {T_B} > {T_A}$ ==> Heat will flow $B$ to $A$ via two paths $(i)$ $B$ to $A$ $(ii)$ and along $BCA$ as shown.
Rate of flow of heat in path $BCA$ will be same
i.e. ${\left( {\frac{Q}{t}} \right)_{BC}} = {\left( {\frac{Q}{t}} \right)_{CA}}$
$ \Rightarrow \frac{{k(\sqrt 2 T - {T_C})A}}{a} = \frac{{k({T_C} - T)A}}{{\sqrt 2 a}}$
$ \Rightarrow \frac{{{T_C}}}{T} = \frac{3}{{1 + \sqrt 2 }}$
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