If $K_1$ and $K_2$ are the thermal conductivities, $L_1$ and $L_2$ are the lengths and $A_1$ and $A_2$ are the cross sectional areas of steel and copper rods respectively such that $\dfrac{K_2}{K_1} = 9$, $\dfrac{A_1}{A_2} = 2$, $\dfrac{L_1}{L_2} = 2$. Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steel - copper junction in the steady state will be
Answer: (C) $45^\circ$C
The rods are in series; thermal resistance $R = \dfrac{L}{KA}$.
$$\frac{R_1}{R_2} = \frac{L_1}{L_2}\cdot\frac{K_2}{K_1}\cdot\frac{A_2}{A_1} = 2\times9\times\frac12 = 9$$
The same heat current flows through both, so the temperature drop divides as the resistances:
$$T - 0 = 450\times\frac{R_2}{R_1 + R_2} = 450\times\frac{1}{10} = 45^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics