Question

Three rods of material $x$ and three rods of material $y$ are connected as shown in figure. All are identical in length and cross sectional area. If end $A$ is maintained at $60^{\circ }C$, end $E$ at $10^{\circ }C$, thermal conductivity of $x$ is $0.92cal{s}^{-1}c{m}^{-1\circ }{C}^{-1}$ and that of $y$ is $0.46cal{s}^{-1}c{m}^{-1}{}^{\circ }{C}^{-1}$ then find the temperatures of junctions $B,C,D.$

• A. $20^{\circ }C,30^{\circ }C,20^{\circ }C$
• B. $30^{\circ }C,20^{\circ }C,20^{\circ }C$
• C. $20^{\circ }C,20^{\circ }C,30^{\circ }C$
• D. $20^{\circ }C,20^{\circ }C,20^{\circ }C$

Answer 1

Answer: (B)

Let $L$ and $A$ be the length and area of cross-section of each rod respectively. Temperature of $A=60^{\circ }C$,
Temperature of $E=10^{\circ }C$
Let $T_1,T_2,T_3$ be respective temperatures of $B,C,D$.

Let $H_1,H_2,H_3,H_4,H_5$ and $H_6$ be the amounts of heat flowing per second respectively from $A$ to $B$; $B$ to $C;B$ to $D;C$ to $D;D$ to $E$ and $C$ to $E$. Then
$H_1=\frac{0.46A\left(60-T_1\right)}{L},H_2=\frac{0.92A\left(T_1-T_2\right)}{L}$
$H_3=\frac{0.46A\left(T_1-T_3\right)}{L},H_4=\frac{0.92A\left(T_2-T_3\right)}{L}$
$H_5=\frac{0.46A\left(T_3-10\right)}{L},H_6=\frac{0.92A\left(T_2-10\right)}{L}$
As $H_1=H_2+H_3$
$\therefore \frac{0.46A\left(60-T_1\right)}{L}$
$=\frac{0.92A\left(T_1-T_2\right)}{L}+\frac{0.46A\left(T_1-T_3\right)}{L}$
or $60-T_1=2\left(T_1-T_2\right)+T_1-T_3$
or $4T_1-2T_2-T_3=60\dots$(i)
Again, $H_2=H_4+H_6$ gives
$T_1-3T_2+T_3=-10\dots$(ii)
Again, $H_5=H_3+H_4$ gives
$T_1+2T_2-4T_3=-10$
Solving (i), (ii) and (iii), we get
$T_1=30^{\circ }C,T_2=20^{\circ }C,T_3=20^{\circ }C$