Dimensional formula for thermal conductivity is (here $K$ denotes the temperature)

Two plates $A$ and $B$ have thermal conductivities $84 \, W m^{-1} K^{-1}$ and $126 \, W m^{-1} K^{-1}$ respectively. They have the same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of $A$ and $B$ are kept at $100^{\circ} C$ and $0^{\circ} C$ respectively,then the temperature of the surface of contact in steady state is $.......... \, ^{\circ} C$.

In a steady state, the graph between temperature $(\theta)$ and distance $(x)$ from the hot end is:

Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle $ABC$,right-angled at $B$. The points $A$ and $B$ are maintained at temperatures $T$ and $\sqrt{2}T$ respectively. In the steady state,the temperature of point $C$ is $T_C$. Assuming that only heat conduction takes place,$\frac{T_C}{T}$ is equal to

$A$ cylindrical rod has temperatures $T_1$ and $T_2$ at its ends. The rate of flow of heat is $Q_1 \text{ cal s}^{-1}$. If the length and radius of the rod are doubled while keeping the temperatures constant,then the new rate of flow of heat $Q_2$ will be:

An ice cube of dimensions $60\,cm \times 50\,cm \times 20\,cm$ is placed in an insulation box of wall thickness $1\,cm$. The box keeping the ice cube at $0^{\circ}C$ temperature is brought to a room of temperature $40^{\circ}C$. The rate of melting of ice is approximately. (Latent heat of fusion of ice is $3.4 \times 10^{5}\,J\,kg^{-1}$ and thermal conductivity of insulation wall is $0.05\,W\,m^{-1\circ}C^{-1}$)