$A$ 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

As per the given figure,two plates $A$ and $B$ of thermal conductivity $K$ and $2K$ are joined together to form a compound plate. The thickness of the plates are $4.0 \,cm$ and $2.5 \,cm$ respectively,and the area of cross-section is $120 \,cm^{2}$ for each plate. If the equivalent thermal conductivity of the compound plate is $\left(1+\frac{5}{\alpha}\right) K$,then the value of $\alpha$ will be . . . . . .

Six identical conducting rods are joined as shown in the figure. Points $A$ and $D$ are maintained at temperatures of $200^oC$ and $20^oC$ respectively. The temperature of junction $B$ 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 $f$.

$A$ cylinder of radius $R$ is made of a material with thermal conductivity $K_1$. It is surrounded by a cylindrical shell of a material with thermal conductivity $K_2$,having an inner radius $R$ and an outer radius $2R$. The two ends of this composite system are maintained at two different temperatures. There is no heat loss from the surface of the cylinder in the steady state. Find the equivalent thermal conductivity of the system.

Two plates of thermal conductivities $k_1$ and $k_2$,cross-sectional areas $A_1$ and $A_2$,and the same thickness $l$ are joined as shown in the figure. The equivalent thermal conductivity $k$ of the combination is: