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.

For the shown figure,calculate the equivalent thermal resistance if the bricks are made of the same material of thermal conductivity $K$.

The ratio of thermal conductivities of two different materials is $5:3$. If the thermal resistances of two rods of these materials having the same thickness are equal,then the ratio of the lengths of the rods will be:

Three conductors of the same length having thermal conductivities $k_1, k_2$ and $k_3$ are connected as shown in the figure. The areas of cross-section of the $1^{\text{st}}$ and $2^{\text{nd}}$ conductors are the same,and for the $3^{\text{rd}}$ conductor,it is double that of the $1^{\text{st}}$ conductor. The temperatures are given in the figure. In the steady-state condition,the value of $\theta$ is . . . . . . $^{\circ}C$. (Given: $k_1 = 60 \ J s^{-1} m^{-1} K^{-1}, k_2 = 120 \ J s^{-1} m^{-1} K^{-1}, k_3 = 135 \ J s^{-1} m^{-1} K^{-1}$)

Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at $0^{\circ}C$ and $90^{\circ}C$ respectively. The temperature of the junction of the three rods will be ...... $^{\circ}C$.

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 $f$.