$A$,$B$ and $C$ are three identical conductors but made from different materials. They are kept in contact as shown. Their thermal conductivities are $K$,$2K$ and $K/2$. The free end of $A$ is at $100^{\circ} C$ and the free end of $C$ is at $0^{\circ} C$. During steady state,the temperature of the junction of $A$ and $B$ is nearly (in $^{\circ} C$)

The ratio of thermal conductivity of two rods of different materials is $5 : 4$. If the two rods have the same area of cross-section and the same thermal resistance,what is the ratio of their lengths?

$A$ copper rod of length $75 \ cm$ and an iron rod of length $125 \ cm$ are joined together end to end. Both are of circular cross-section with diameter $2 \ cm$. The free ends of the copper and iron are maintained at $100^{\circ} C$ and $0^{\circ} C$ respectively. The surfaces of the bars are insulated thermally. The temperature of the copper-iron junction is (Thermal conductivity of copper is $386.4 \ W/m-K$ and thermal conductivity of iron is $48.46 \ W/m-K$) (in $^{\circ} C$)

$A$ copper rod of length $18 \ cm$ and a steel rod of length $6 \ cm$ are joined together to form a composite rod of uniform cross-section. The temperature of the free end of the copper rod is $100 \ ^\circ C$ and the temperature of the free end of the steel rod is $0 \ ^\circ C$. What is the temperature at the junction in $^\circ C$? (Thermal conductivity of copper is $9$ times that of steel. The rod is in a steady state.)

$A$ cylinder of thermal conductivity $k_1$ and radius $r$ is surrounded by a cylindrical shell of inner radius $r$,outer radius $2r$,and thermal conductivity $k_2$. Both cylinders have the same length and the temperature difference across their ends is the same. The equivalent thermal conductivity of the system 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: