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:

The dimensions of thermal resistance are

$A$ composite block is made of slabs $A, B, C, D$ and $E$ of different thermal conductivities (given in terms of a constant $K$) and sizes (given in terms of length,$L$) as shown in the figure. All slabs are of same width. Heat $Q$ flows only from left to right through the blocks. Then in steady state:
$(A)$ Heat flow through $A$ and $E$ slabs are same.
$(B)$ Heat flow through slab $E$ is maximum.
$(C)$ Temperature difference across slab $E$ is smallest.
$(D)$ Heat flow through $C =$ Heat flow through $B +$ Heat flow through $D$.

In a composite rod,when two rods of different lengths and of the same area are joined end to end,then if $K$ is the effective coefficient of thermal conductivity,$\frac{{\ell _1} + {\ell _2}}{K}$ is equal to

Three identical rods $A, B$ and $C$ are placed end to end. $A$ temperature difference is maintained between the free ends of $A$ and $C$. The thermal conductivity of $B$ is thrice that of $C$ and half of that of $A$. The effective thermal conductivity of the system will be ($K_{A}$ is the thermal conductivity of rod $A$).

Two rods of the same length and material transfer a given amount of heat in $12 \, s$ when they are joined in parallel. But when they are joined in series,then they will transfer the same amount of heat under the same conditions in ........ $s$.