Two metal rods $1$ and $2$ of same lengths have same temperature difference between their ends. Their thermal conductivities are $K_1$ and $K_2$ and cross-sectional areas are $A_1$ and $A_2$,respectively. If the rate of heat conduction in rod $1$ is four times that in rod $2$,then:

Three rods of same dimensions have thermal conductivities $3K, 2K$ and $K$. They are arranged as shown with their ends at $100^{\circ}C, 50^{\circ}C$ and $0^{\circ}C$. The temperature of the junction is

$A$ slab consists of two identical plates of copper and brass. The free face of the brass is at $0^{\circ} C$ and that of copper at $100^{\circ} C$. If the thermal conductivities of brass and copper are in the ratio $1: 4$,then the temperature of the interface is (in $^{\circ} C$)

$A$ conducting rod of length $1 \,m$ has an area of cross-section $10^{-3} \,m^2$. One end is immersed in boiling water $(100^{\circ} C)$ and the other end in ice $(0^{\circ} C)$. If the coefficient of thermal conductivity of the rod is $96 \,cal/(s \cdot m \cdot ^{\circ}C)$ and the latent heat of fusion for ice is $8 \times 10^4 \,cal/kg$,then the amount of ice that will melt in one minute is:

$A$ rod $A$ of length $40\, cm$ has a temperature difference of $80^\circ C$ at its two ends. Another rod $B$ of length $60\, cm$ has a temperature difference of $90^\circ C$ at its ends. Both rods have the same area of cross-section. If the rate of flow of heat is the same for both,then the ratio of their thermal conductivities $(K_A : K_B)$ will be:

Rate of flow of heat through a cylindrical rod is $H_1$. The temperatures at the ends of the rod are $T_1$ and $T_2$. If all the dimensions of the rod become double and the temperature difference remains the same, and if the rate of flow of heat becomes $H_2$, then $H_2 =$