Three rods of identical area of cross-section and made from same metal form sides of an isosceles tr

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10-2.Transmission of Heat

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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

A

$\frac{1}{{(\sqrt 2 + 1)}}$

B

$\frac{3}{{(\sqrt 2 + 1)}}$

C

$\frac{1}{{2(\sqrt 2 - 1)}}$

D

$\frac{1}{{\sqrt 3 (\sqrt 2 - 1)}}$

(IIT-1995)

Solution

(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 }}$

Standard 11

Physics

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