$A$ slab consists of two parallel layers of copper and brass of equal thickness. The ratio of their thermal conductivities is $1:4$. If the temperature of the free side of the brass is $100^{\circ}C$ and that of the copper is $0^{\circ}C$,find the temperature of the interface in $^{\circ}C$.

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: $k_1 = 60 \ J s^{-1} m^{-1} K^{-1}, k_2 = 120 \ J s^{-1} m^{-1} K^{-1}, k_3 = 135 \ J s^{-1} m^{-1} K^{-1}$)

Two conducting rods $A$ and $B$ of same length and cross-sectional area are connected $(i)$ In series $(ii)$ In parallel as shown. In both combinations,a temperature difference of $100^{\circ}C$ is maintained. If the thermal conductivity of $A$ is $3K$ and that of $B$ is $K$,then the ratio of heat current flowing in the parallel combination to that flowing in the series combination is:

$A$ 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:

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

Two slabs $A$ and $B$ of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of $A$ and $B$ are $k_1$ and $k_2$ respectively. $A$ steady temperature difference of $12^{\circ} C$ is maintained across the composite slab. If $k_1=\frac{k_2}{2}$,the temperature difference across slab $A$ is (in $^{\circ} C$)