$A$ slab consists of two parallel layers of two different materials of same thickness having thermal conductivities $K_1$ and $K_2$. The equivalent conductivity of the combination is

Six identical rods are arranged as shown in the figure. The temperature of the junction $B$ will be......... $^oC$

The ratio of thermal conductivities of two materials is $1 : 2$. If the ratio of the lengths of the rods of these materials is $1 : 2$ and the ratio of their cross-sectional areas is $2 : 1$,then the ratio of their thermal resistances is:

Two cylindrical rods $A$ and $B$ made of different materials are joined in a straight line. The ratios of lengths,radii,and thermal conductivities of these rods are $\frac{L_A}{L_B} = \frac{1}{2}$,$\frac{r_A}{r_B} = 2$,and $\frac{K_A}{K_B} = \frac{1}{2}$. The free ends of rods $A$ and $B$ are maintained at $400 \ K$ and $200 \ K$,respectively. The temperature of the rods' interface is . . . . . . $K$,when equilibrium is established.

$A$ cylinder of thermal conductivity $k_1$ and radius $r$ is surrounded by a cylindrical shell of inner radius $r$,outer radius $2r$,and thermal conductivity $k_2$. Both cylinders have the same length and the temperature difference across their ends is the same. The equivalent thermal conductivity of the system is:

The temperatures of the two outer surfaces of a composite slab are $T_2$ and $T_1$ $(T_2 > T_1)$. The thermal conductivities of the materials are $K$ and $2K$,and their thicknesses are $x$ and $4x$,respectively. In the steady state,the rate of heat flow through the slab is given by $f \left( \frac{A(T_2 - T_1)K}{x} \right)$. Find the value of $f$.