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 $^{\circ} C$)

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 . . . . . .

The ratio of thermal conductivities of two different materials is $5:3$. If the thermal resistances of two rods of these materials having the same thickness are equal,then the ratio of the lengths of the rods will be:

Four identical conducting rods are joined as shown in the figure. Points $A$ and $D$ are maintained at temperatures of $200^{\circ}C$ and $20^{\circ}C$ respectively. The temperature of junction $B$ will be ....... $^{\circ}C$.

$ABCDE$ is a regular pentagon made of uniform wire. Heat enters at $A$ and leaves at $C$. $T_B$ and $T_D$ are the temperatures of points $B$ and $D$ respectively. Find the temperature $T_C$.

Two metal slabs of same cross-sectional area have thicknesses $d_1$ and $d_2$,and thermal conductivities $K_1$ and $K_2$ respectively,are connected in series. The free ends of the two slabs are kept at temperatures $T_1$ and $T_2$ $(T_1 > T_2)$. The temperature $T$ of their common junction is