There are two spherical balls $A$ and $B$ of the same material with same surface, but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool. then

In the figure shown $AB$ is rod of length $30 \mathrm{cm}$ and area of cross-section $1 {\mathrm{cm}}^2$ and thermal conductivity of $336$ S.I. units. The ends $A$ and $B$ are maintained at temperature $20°\mathrm{C}$ and $40°\mathrm{C}$ respectively. A point $C$ of this rod is connected to a box $D$, containing ice at $0°\mathrm{C}$, through a highly conducting wire of negligible heat capacity. The rate at which ice melts in the box is $x$. The rate of heat flow through $AC$ is $y$ and the rate of heat flow through $BC$ is z. Then,

[Assume latent heat of fusion for ice $L_{ice}=80 \mathrm{cal} {\mathrm{g}}^{-1}$]

The top of insulated cylindrical container is covered by a disc having emissivity $0.6$ and thickness $1 \mathrm{cm}$. The temperature is maintained by circulating oil as shown in figure. If temperature of upper surface of disc is $127°\mathrm{C}$ and temperature of surrounding is $27°\mathrm{C}$, then the radiation lost to the surroundings will be (Take  $\sigma =\frac{17}{3}\times {10}^{-8} \mathrm{W}/{\mathrm{m}}^2{\mathrm{K}}^4$ )