$A$ 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.

$A$ composite slab consists of two materials having coefficients of thermal conductivity $K$ and $2K$,and thicknesses $x$ and $4x$ respectively. The temperatures of the two outer surfaces of the composite slab are $T_2$ and $T_1$ $(T_2 > T_1)$. The rate of heat transfer through the slab in a steady state is $\left[\frac{A(T_2 - T_1)K}{x}\right] \cdot f$,where '$f$' is equal to:

The temperatures of the two outer surfaces of a composite slab,consisting of two materials having coefficients of thermal conductivity $K$ and $2K$ and thickness $x$ and $4x$,respectively,are $T_2$ and $T_1$ $(T_2 > T_1)$. The rate of heat transfer through the slab in a steady state is $\left( \frac{A(T_2 - T_1)K}{x} \right)f$,where $f$ is equal to:

In a composite rod,when two rods of different lengths $l_1$ and $l_2$ and of the same cross-sectional area are joined end to end,if $K$ is the effective coefficient of thermal conductivity,then the value of $(l_1 + l_2)/K$ is

Twelve identical conducting rods form the edges of a uniform cube of side length $l$. In the steady state,the junctions $B$ and $H$ are maintained at $100^{\circ}C$ and $0^{\circ}C$ respectively. Find the temperature of the junction $A$ in $^{\circ}C$.

$A$ wall has two layers $A$ and $B$,each made of different materials. Both layers are of the same thickness. The thermal conductivity of material $A$ is twice that of $B$. If in the steady state,the temperature difference across the wall is $24^{\circ} C$,then the temperature difference across the layer $B$ is (in $^{\circ} C$)