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:

$A$ composite slab is prepared with two different materials $A$ and $B$. The relation between their coefficients of thermal conductivity and thickness is given as $K_A = \frac{K_B}{2}$ and $X_A = 2 X_B$,respectively. If the temperatures of the outer faces of $A$ and $B$ are $75^{\circ} C$ and $50^{\circ} C$ respectively,what will be the temperature of the common surface (in $^{\circ} C$)?

$A$ rod of length $L$ with sides fully insulated is made of a material whose thermal conductivity varies with temperature $T$ as $K = \frac{\alpha}{T}$,where $\alpha$ is a constant. The ends of the rod are kept at temperatures $T_1$ and $T_2$. The temperature $T$ at a distance $x$ from the end kept at $T_1$ is:

$A$ rectangular ice box of total surface area of $1000 \,cm^2$ initially contains $1.5 \,kg$ of ice at $0^{\circ}C$. If the thickness of the walls of the box is $2 \,mm$ and the temperature outside the box is $42^{\circ}C$, then the mass of the ice remaining in the box after $160 \,minutes$ is (Thermal conductivity of the material of the box $= 10^{-2} \,W m^{-1} K^{-1}$ and latent heat of the fusion of ice $= 336 \times 10^3 \,J kg^{-1}$) (in $kg$)

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

$A$ heat flux of $4000 \, J/s$ passes through a rod of length $10 \, cm$ and cross-sectional area $100 \, cm^2$. The thermal conductivity of copper is $400 \, W/m^{\circ}C$. The ends of the rod must be maintained at a temperature difference of ....... $^{\circ}C$.