Question

Three rods of material $x$ and three rods of material $y$ of identical length and cross sectional area are connected as shown in the figure. If the end $A$ is maintained at $60^{\circ }C$ and the junction $E$ at $10^{\circ }C$, then the temperature of junction $C$ is [Given the thermal conductivity of $x$ is $0.92W/m.K$ and that of $y$ is $0.46W/m.K$ ]

• A. $10^{\circ }C$
• B. $20^{\circ }C$
• C. $30^{\circ }C$
• D. $40^{\circ }C$

Answer 1

Answer: (B)

Treating the given network of rods in terms of thermal resistance $R_x$ and $R_y$ with
$R_x=\frac{L}{A\times 0.92}$ and $R_y=\frac{L}{A\times 0.46}[$ as $R=\frac{L}{AK}]$
so that if $R_x=R,R_y=2R_x=2R$
Now as in this bridge $[(P/Q)=(R/S)]$, so the bridge is balanced, i.e., the temperature at junctions $C$ and $D$ is equal and the rod CD becomes ineffective as no heat will flow through it.
Now as the thermal resistance of the bridge between junctions $B$ and $E$ is $\frac{1}{(R{)}_{nE}}=\frac{1}{(R+R)}+\frac{1}{(2R+2R)}$, i.e., $(R{)}_{BE}=\frac{4}{3}R$
The total resistance of bridge between $A$ and $E$ will be
$R_{\text{eq }}=R_{AB}+R_{BE}=2R+(4/3)R=(10/3)R$
So the net rate of flow of heat through the bridge will be
Now if $T_B$ is the temperature at $B$,
But ${\left[\frac{dQ}{dt}\right]}_{AB}=\frac{dQ}{dt}$,i.e., $\frac{60-T_B}{2R}=\frac{15}{R}$,i.e., $T_B=30^{\circ }C$
Also at $B$
${\left[\frac{dQ}{dt}\right]}_{AB}={\left[\frac{dQ}{dt}\right]}_{BC}+{\left[\frac{dQ}{dt}\right]}_{BD}$
i.e. $\frac{15}{R}=\frac{30-T_C}{R}+\frac{30-T_D}{2R}$
and as $T_C=T_D=T,30=3(30-T)$,
i.e., $T_C=T_D=T=20^{\circ }C$