Two identical conducting rods are first connected independently to two vessels,one containing water at $100^{\circ}C$ and the other containing ice at $0^{\circ}C$. In the second case,the rods are joined end to end and connected to the same vessels. Let $q_1$ and $q_2$ $g/s$ be the rate of melting of ice in the two cases respectively. The ratio $q_2/q_1$ is

$A$ composite slab is made of two different metals of the same thickness,having thermal conductivities $K$ and $2K$. The equivalent thermal conductivity of the composite slab is:

$A$,$B$ 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 $^{\circ} C$)

Five identical rods are joined as shown in the figure. Points $A$ and $C$ are maintained at temperatures $120^\circ C$ and $20^\circ C$ respectively. The temperature of junction $B$ will be....... $^\circ C$

$A$ 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 $^{\circ}C$?

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 . . . . . .