In Searle's method for finding the thermal conductivity of metals,the temperature gradient along the bar:

Two vessels of different materials are similar in size in every respect. The same quantity of ice filled in them gets melted in $20 \text{ minutes}$ and $30 \text{ minutes}$ respectively. The ratio of their thermal conductivities will be

$A$ heat source at $T_1 = 10^3\, K$ is connected to another heat reservoir at $T_2 = 10^2\, K$ by a copper slab which is $1\, m$ thick. Given that the thermal conductivity of copper is $0.1\, W\, K^{-1}\, m^{-1}$,the energy flux through it in the steady state is ........... $W\, m^{-2}$.

The ends of two rods of different materials with their thermal conductivities,radii of cross-sections,and lengths all in the ratio $1:2$ are maintained at the same temperature difference. If the rate of flow of heat in the larger rod is $4 \; cal/sec$,the rate of flow of heat in the shorter rod in $cal/sec$ will be:

$A$ large cylindrical rod of length $L$ is made by joining two identical rods of copper and steel of length $(\frac{L}{2})$ each. The rods are completely insulated from the surroundings. If the free end of the copper rod is maintained at $100\,^oC$ and that of the steel rod at $0\,^oC$,then the temperature of the junction is........$^oC$ (Thermal conductivity of copper is $9$ times that of steel).

$Assertion :$ $A$ brass tumbler feels much colder than a wooden tray on a chilly day.
$Reason :$ The thermal conductivity of brass is more than the thermal conductivity of wood.