Q: Three rods of identical cross-sectional area 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 √2 T , respectively, in the steady state. Assuming that only heat conduction takes palace, temperature of point C is

(a) $ \displaystyle \frac{3T}{\sqrt{T} + 1 } $

(b) $ \displaystyle \frac{T}{\sqrt{T} + 1 } $

(c) $ \displaystyle \frac{T}{3(\sqrt{T} + 1 )} $

(d) $ \displaystyle \frac{T}{\sqrt{T} – 1 } $

Ans: (a)

Sol:

R_{AB} = R_{BC} = R

R_{AC} = √2 R

Since T_{B} > T_{A} , heat will flow from B to A through two paths BA and BCA.

In path BCA, current will be same in BC and CA

Let temp. of point C = T_{C}

$ \displaystyle I = \frac{\sqrt 2 T – T_C}{R_{BC}} = \frac{T_C – T}{R_{AC}}$

$ \displaystyle \frac{\sqrt 2 T – T_C}{R} = \frac{T_C – T}{\sqrt 2 R}$

2T -√2 T_{C} = T_{C} – T

(√2 + 1)T_{C} = 3T

$ \displaystyle T_C = \frac{3T}{\sqrt 2 + 1} $