$A$ composite slab is made of two different materials of equal thickness,having thermal conductivities $K$ and $2K$ respectively. The equivalent thermal conductivity of the slab is ........

Five rods of identical dimensions are arranged as shown in the figure. Their thermal conductivities are $k_1, k_2, k_5, k_4$ and $k_3$. When $A$ and $B$ are maintained at different temperatures,no heat flows through the central rod if:

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 slabs $A$ and $B$ of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab $A$ is twice that of $B$. The coefficient of thermal conductivity of slab $A$ is twice that of $B$. The first surface of slab $A$ is maintained at $100^{\circ} C$, while the second surface of slab $B$ is maintained at $25^{\circ} C$. The temperature at the contact of their surfaces is (in $^{\circ} C$)

$A$ composite slab consists of two materials having coefficients of thermal conductivity $K$ and $2K$,thickness $x$ and $4x$ respectively. The temperatures of two outer surfaces of the composite slab are $T_2$ and $T_1$ respectively $(T_2 > T_1)$. 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:

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.