The diagram shows four slabs of different materials with equal thickness, placed side by side. Heat flows from left to right and the steady-state temperatures of the interfaces are given. Rank the materials according to their thermal conductivities, smallest to largest.

The figure shows the face and interface temperatures of a composite slab consisting of four materials, of identical thickness, through which the heat transfer is steady. Rank the materials according to their thermal conductivities, greatest first.

A partition wall has two layers A and B, in contact, each made of a different material. They have the same thickness but the thermal conductivity of layer A is twice that of layer B. If the steady state temperature difference across the wall is 60 K, then the corresponding difference across the layer A is

The figure here shows four rectangular metal plates, with sides of , L , 2L or 3L. They are all made of the same material , and their temperature is to be increased by the same amount , Rank the plates according to the expected increase in (a) their vertical heights and (b) their areas , greatest first.

Two identical rods are made of different materials whose thermal conductivities are K_(1) and K_(2) . They are placed end to end between two heat reservoirs at temperatures theta_(1) and theta_(2) . The temperature of the junction of the rod is

The end of two rods of different materials with their thermal conductivities, area of cross-section and lengths all in the ratio 1:2 are maintained at the same temperature difference. If the rate of flow of heat in the first rod is 4 cal//s . Then, in the second rod rate of heat flow in cal//s will be

Consider a compound slab consisting of two different material having equal thickness and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is

A composite slab consists of two slabs A and B of different materials but of the same thickness placed one on top of the other. The thermal conductivities of A and B are k_(1) and K_(2) respectively. A steady temperature difference of 12^(@)C is maintained across the composite slab. If k_(1) = K_(2)//2 , the temperature difference across slab A will be:

A wall has two layers A and B, each made of different materials. Both layers are of same thickness. But, the thermal conductivity of material A is twice that of B. If, in the steady state, the temperature difference across the wall is 24^(@)C , then the temperature difference across the layer B is :

If two conducting slabs of thickness d_1 and d_2 , and thermal conductivity K_1 and K_2 are placed together face to face as shown in figure in the steady state temperature of outer surfaces are theta_1 and theta_2 . The temperature of common surface is-