Two vessels of different materials are similar in size in every respect. The same quantity of ice filled in them gets melted in $20$ minutes and $40$ minutes respectively. The ratio of thermal conductivities of the materials is

$A$ cylindrical rod has temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q$. All the linear dimensions of the rod are doubled while keeping the temperatures constant. The new rate of flow of heat is

The temperature difference across two cylindrical rods $A$ and $B$ of the same material and same mass are $40^{\circ} C$ and $60^{\circ} C$ respectively. In steady state,if the rates of flow of heat through the rods $A$ and $B$ are in the ratio $3: 8$,the ratio of the lengths of the rods $A$ and $B$ is (in $: 3$)

Three rods made of the same material and having the same cross-sectional area but different lengths $10 \ cm$,$20 \ cm$,and $30 \ cm$ are joined as shown. The temperature of the joint is ....... $^oC$.

$A$ thermocol box has a total wall area (including the lid) of $1.0 \,m^2$ and wall thickness of $3 \,cm$. It is filled with ice at $0^{\circ} C$. If the average temperature outside the box is $30^{\circ} C$ throughout the day, the amount of ice that melts in one day is (Given: $K_{\text{thermocol}} = 0.03 \,W/mK$, $L_{\text{fusion(ice)}} = 3.00 \times 10^5 \,J/kg$) (in $\,kg$)

$A$ 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 $kg$)