$A$ square is formed by four identical rods. If the temperature difference across one diagonal is $100^{\circ}C$,what will be the temperature difference across the other diagonal?

Two slabs $A$ and $B$ of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of $A$ and $B$ are $k_1$ and $k_2$ respectively. $A$ steady temperature difference of $12^{\circ} C$ is maintained across the composite slab. If $k_1=\frac{k_2}{2}$, the temperature difference across slab $A$ is (in $^{\circ} C$)

$A$ wall consists of two layers $A$ and $B$ of equal thickness. Their thermal resistances are $R_1$ and $R_2$. Find the temperature of the interface.

The three rods shown in the figure have identical dimensions. Heat flows from the hot end at a rate of $40 \, W$ in the arrangement $(a)$. Find the rate of heat flow when the rods are joined as in arrangement $(b)$ (Assume $K_{Al} = 200 \, W/m^{\circ}C$ and $K_{Cu} = 400 \, W/m^{\circ}C$). (in $, W$)

Three identical rods $A, B$ and $C$ are placed end to end. $A$ temperature difference is maintained between the free ends of $A$ and $C$. The thermal conductivity of $B$ is thrice that of $C$ and half of that of $A$. The effective thermal conductivity of the system will be ($K_{A}$ is the thermal conductivity of rod $A$).

Assertion: The equivalent thermal conductivity of two plates of the same thickness in contact is less than the smaller value of thermal conductivity.
Reason: For two plates of equal thickness in contact,the equivalent thermal conductivity is given by: $\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2}$