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 $0.92 cal s ^{-1} cm ^{-1 \circ} C ^{-1}$ and that of $y$ is $0.46 cal s ^{-1} cm ^{-1}{ }^{\circ} C ^{-1}$ 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{0.46 A\left(60-T_{1}\right)}{L}, H_{2}=\frac{0.92 A\left(T_{1}-T_{2}\right)}{L} $
$H_{3}=\frac{0.46 A\left(T_{1}-T_{3}\right)}{L}, H_{4}=\frac{0.92 A\left(T_{2}-T_{3}\right)}{L} $
$H_{5}=\frac{0.46 A\left(T_{3}-10\right)}{L}, H_{6}=\frac{0.92 A\left(T_{2}-10\right)}{L}$
As $H_{1}=H_{2}+H_{3}$
$\therefore \frac{0.46 A\left(60-T_{1}\right)}{L}$
$=\frac{0.92 A\left(T_{1}-T_{2}\right)}{L}+\frac{0.46 A\left(T_{1}-T_{3}\right)}{L}$
or $ 60-T_{1}=2\left(T_{1}-T_{2}\right)+T_{1}-T_{3}$
or $ 4 T_{1}-2 T_{2}-T_{3}=60 \dots$(i)
Again, $ H_{2}=H_{4}+H_{6} $ gives
$T_{1}-3 T_{2}+T_{3}=-10\dots$(ii)
Again, $ H_{5}=H_{3}+H_{4} $ gives
$T_{1}+2 T_{2}-4 T_{3}=-10$
Solving (i), (ii) and (iii), we get
$T_{1}=30^{\circ} C , T_{2}=20^{\circ} C , T_{3}=20^{\circ} C$