$A$ wall is made of two layers of thickness $1\ cm$ and $4\ cm$ and thermal conductivities $K$ and $3K$ respectively. If the temperature difference across the wall is $50\ ^oC$,then the temperature difference across the thin layer will be

$A$ hollow sphere of inner radius $R$ and outer radius $2R$ is made of a material of thermal conductivity $K$. It is surrounded by another hollow sphere of inner radius $2R$ and outer radius $3R$ made of the same material of thermal conductivity $K$. The inside of the smaller sphere is maintained at $0^oC$ and the outside of the bigger sphere at $100^oC$. The system is in steady state. The temperature of the interface will be ........ $^oC$.

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 $f$.

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:

$A$ wall consists of two layers $A$ and $B$. Both layers have the same thickness but are made of different materials. The thermal conductivity of $A$ is twice that of $B$. In the steady state,the temperature difference across the two ends of the wall is $36^{\circ}C$. What is the temperature difference across the two ends of layer $A$ in $^{\circ}C$?

In a composite rod,when two rods of different lengths $l_1$ and $l_2$ and of the same cross-sectional area are joined end to end,if $K$ is the effective coefficient of thermal conductivity,then the value of $(l_1 + l_2)/K$ is