Transmission Of Heat Question 104
Question: 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]
Options:
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)} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \because $
$ T _{B}>T _{A} $
Therefore 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.
$ \frac{Q}{t}_BC=\frac{Q}{t}_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}} $
BETA
