Question

Two rods of same length and same material; having area of cross section $A$ and $4A$ respectively are joined as shown in the figure. If $T_A$ and $T_B$ are temperatures of free ends in steady state. The temperature of junction point $C$ is :

(Assume no heat loss will take place through lateral surface).

• A. $\frac{T_A-T_B}{2}$
• B. $\frac{T_A+T_B}{2}$
• C. $\frac{T_A+4T_B}{5}$
• D. $\frac{2T_A+T_B}{5}$

Answer 1

The correct option is C $\frac{T_A+4T_B}{5}$

We know,

$\frac{dQ}{dt}=\frac{kA(T_2-T_1)}{l}$

Where,

k=conductivity, A=area of cross-section, l=length of rod between the terminal

Let the temperature at junction is $T_C{}^{0}C$

As both rods are in series, same rate of heat will be transferred,

$\frac{dQ}{dt}=\frac{kA(T_C-T_A)}{l}=\frac{k4A(T_B-T_C)}{l}$

$T_C-T_A=4(T_B-T_C)$

$5T_C=T_A+4T_B$

$T_C=\frac{T_A+4T_B}{5}$

Hence option $\text{C}$ is correct answer.