Q 11-10-095JEE MainJEE Main 2021 (27 Aug, Shift 1)Medium
A rod $CD$ of thermal resistance $10.0$ K W$^{-1}$ is joined at the middle of an identical rod $AB$ as shown in figure. The ends $A$, $B$ and $D$ are maintained at $200^\circ$C, $100^\circ$C and $125^\circ$C respectively. The heat current in $CD$ is $P$ W. The value of $P$ is
Numerical value type. Enter your answer.
Answer: 2
Each half of $AB$ ($AC$ and $CB$) has thermal resistance $5$ K W$^{-1}$. Let the junction be at $T$. Net heat flow into $C$ is zero:
$$\frac{200 - T}{5} + \frac{100 - T}{5} + \frac{125 - T}{10} = 0$$
$$2(200 - T) + 2(100 - T) + (125 - T) = 0 \Rightarrow T = 145^\circ\text{C}$$
Heat current in $CD$: $P = \dfrac{145 - 125}{10} = 2$ W.
Solution by Sreeraj P, M.Sc Physics