Question

Three cylindrical rods $A,B,C$ of equal lengths and equal diameters are joined in series as shown in the figure. Their thermal conductivities are $2k,k,0.5k$ respectively. In the steady state, the free ends of rods A and C are at 100${}^o$C and 0${}^o$C respectively. Neglect loss of heat from the curved surfaces of rods.

The temperature of the junction between rods B and C is :

• A. ${57.1}^{o}C$
• B. ${47.1}^{o}C$
• C. ${37.1}^{o}C$
• D. ${85.7}^{o}C$

Answer 1

The correct option is A ${57.1}^{o}C$

Let temperature between $A$ and $B=T_1$ and temperature between $B$ and $C=T_2$
$\Rightarrow \frac{Q}{t}=\frac{A(T_1-T_2)}{L}$

$\Rightarrow \frac{K_{A}A(100-T_1)}{l}=\frac{K_{B}A(T_1-T_2)}{l}=\frac{K_{C}A(T_2-0)}{l}$

And it is given that:

$K_A=2K$, $K_B=K$, $K_C=0.5K$
$200-2T_1=T_1-T_2=0.5T_2$
So, we find $T_2=57.1℃$