The ratio of thermal conductivity of two rods of different materials is $4:3$. The two rods have same area of cross-section and same thermal resistance. The lengths of rods are in the ratio

Mud houses are cooler in summer and warmer in winter because

Two sheets of thickness $d$ and $3d$ are touching each other. The temperature just outside the thinner sheet side is $A$,and on the side of the thicker sheet is $C$. The interface temperature is $B$. If $A, B$,and $C$ are in arithmetic progression,the ratio of the thermal conductivity of the thinner sheet to the thicker sheet is:

Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures $T_1=300 \ K$ and $T_2=100 \ K$,as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are $K_1$ and $K_2$ respectively. If the temperature at the junction of the two cylinders in the steady state is $200 \ K$,then $K_1 / K_2 = . . . . . .$

One end of a copper rod of length $1.0 \; m$ and area of cross-section $10^{-3} \; m^2$ is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is $92 \; cal/(m \cdot s \cdot ^\circ C)$ and the latent heat of ice is $8 \times 10^4 \; cal/kg$,then the amount of ice which will melt in one minute is:

The length of a metal bar is $20 \ cm$ and the area of cross-section is $4 \times 10^{-4} \ m^2$. If one end of the rod is kept in ice at $0^{\circ} C$ and the other end is kept in steam at $100^{\circ} C$,the mass of ice melted in one minute is $5 \ g$. The thermal conductivity of the metal in $W \ m^{-1} \ K^{-1}$ is (Latent heat of fusion $= 80 \ cal/g$):