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Physics Question on Thermodynamics

Three rods of identical cross-sectional area and made from the same metal from the sides of an isosceles triangle ABC, right angled at B. The points A and B are maintained at temperatures T and (2\sqrt 2)T respectively. In the steady state, the temperature of the point C is TcT_c. Assuming that only heat conduction takes place, Tc_c/T is

A

12(2−1)\frac{1}{2(\sqrt 2 -1)}

B

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

C

13(2−1)\frac{1}{\sqrt 3(\sqrt 2 -1)}

D

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

Answer

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

Explanation

Solution

The diagramatic representation of the given problem is shown in figure. Since, TB>TA_B > T_A the heat will flow from B to

A. Similarly, heat will also flow from B to C and C to A.
Applying the conduction formula
ΔQΔt=KAl(ΔT)\frac{\Delta Q}{\Delta t}=\frac{KA}{l}(\Delta T)
To the sides CA and BC, we get
(ΔT2a)CA=(ΔTa)BC    ⇒   TC−T2a=2T−TCa\bigg(\frac{\Delta T}{\sqrt 2a}\bigg)_{CA} =\bigg(\frac{\Delta T}{a}\bigg)_{BC} \, \, \, \, \Rightarrow \, \, \, \frac{T_C-T}{\sqrt 2 a}=\frac{\sqrt 2 T -T_C}{a}
3T=TC(2+1)   ⇒     TCT=3(2+1)3T =T_C(\sqrt 2 +1) \, \, \, \Rightarrow \, \, \, \, \, \frac{T_C}{T}=\frac{3}{(\sqrt 2 +1)}