Question

Seven identical rods of thermal conductivity $k$ are connected as shown in figure. All the rods are of identical length $l$ and cross-sectional area $A$. If one end A is kept at ${100}^{\circ }\text{C}$ and the other end B is kept at $0^{\circ }\text{C}$, which of the following statement(s) is/are true regarding the temperatures of the junctions C, D and E (${\theta }_C,{\theta }_D$ and ${\theta }_E)$ in steady state?

• A. ${\theta }_C>{\theta }_E>{\theta }_D$
• B. ${\theta }_E={62.5}^{\circ }\text{C}$ and ${\theta }_D={37.5}^{\circ }\text{C}$
• C. ${\theta }_E={50}^{\circ }\text{C},{\theta }_C={62.5}^{\circ }\text{C}$ and ${\theta }_D={37.5}^{\circ }\text{C}$
• D. ${\theta }_E={50}^{\circ }\text{C},{\theta }_C={60}^{\circ }\text{C}$ and ${\theta }_D={40}^{\circ }\text{C}$

Answer 1

Answer: (A), (C)

${\theta }_E={50}^{\circ }\text{C},{\theta }_C={62.5}^{\circ }\text{C}$ and ${\theta }_D={37.5}^{\circ }\text{C}$

This problem can be solved like an electric circuit problem.
Let $R_1,R_2,R_3,R_4,R_5,R_6$ and $R_7$ be the rates of heat flow through AE, EB, AC, CD, CE, ED and DB, respectively.


From symmetry, we can deduce
$R_1=R_2$ ; $R_5=R_6$ .....(i)
$R_3=R_4+R_5=R_7$.....(ii)
$R_4+R_6=R_7$.....(iii)
[using junction rule at C and D)

From (i), ${\theta }_E={50}^{\circ }\text{C}$
Also, $R_5=R_6$
$\Rightarrow \frac{KA({\theta }_C-50)}{l}=\frac{KA}{l}(50-{\theta }_D)$
$\Rightarrow {\theta }_C+{\theta }_D=100$ .... (iv)

From (ii),
$\frac{KA}{l}(100-{\theta }_C)=\frac{KA}{l}({\theta }_C-50)+\frac{KA}{l}({\theta }_C-{\theta }_D)=\frac{KA}{l}({\theta }_D-0)$

$\Rightarrow 2{\theta }_C-2{\theta }_D=50$ .... (v)

Solving (iv) and (v).
${\theta }_C={62.5}^{\circ }\text{C}$
& ${\theta }_D={37.5}^{\circ }\text{C}$
$\therefore {\theta }_C>{\theta }_E>{\theta }_D$

Hence, options (a) and (c) are correct.