The temperatures of the two outer surfaces of a composite slab,consisting of two materials having coefficients of thermal conductivity $K$ and $2K$ and thicknesses $x$ 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:

Two rectangular blocks,having identical dimensions,can be arranged either in configuration $I$ or in configuration $II$ as shown in the figure. One of the blocks has thermal conductivity $k$ and the other $2k$. The temperature difference between the ends along the $x$-axis is the same in both the configurations. It takes $9 \ s$ to transport a certain amount of heat from the hot end to the cold end in the configuration $I$. The time to transport the same amount of heat in the configuration $II$ is: (in $s$)

Three conducting rods $AB, BC$ and $BD$ made of the same material and same cross-section are arranged as shown in the figure. Temperatures at the points $A, D$ and $C$ are maintained at $20^{\circ} C, 90^{\circ} C$ and $0^{\circ} C$ respectively. The ratio of lengths of $BD$ to $BC$ when there is no heat flow in $AB$ is:

Six identical conducting rods are joined as shown in the figure. Points $A$ and $D$ are maintained at $100^{\circ} C$ and $10^{\circ} C$ respectively. The temperature of junction $C$ will be $.... ^{\circ} C$.

Two rods $A$ and $B$ of the same cross-sectional area $A$ and length $l$ are connected in series between a source $(T_1 = 100^{\circ}C)$ and a sink $(T_2 = 0^{\circ}C)$ as shown in the figure. The rods are laterally insulated. The thermal conductivities of rods $A$ and $B$ are $3K$ and $K$ respectively. The ratio of the thermal resistance of rod $A$ to that of rod $B$ is:

In a composite slab, there are 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 $T_1$ $(T_2 > T_1)$. $T_2$ is on the side with conductivity $K$ and $T_1$ is on the side with conductivity $2K$. 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: