A composite slab is made of two parallel layers of two different materials of thermal conductivities `K_1 and K_2` are of same thickness. Show that the equivalent thermal conductivity of the slab is `(2K_1 K_2)/( K_1 + K_2)`

A slab consists of two layers of different materials of the same thickness and having thermal conductivities K_(1) and K_(2) . The equivalent thermal conductivity of the slab is

Show that the SI units of thermal conductivity are W//m-K .

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

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

Three rods of same dimension are joined in series having thermal conductivity k_1 , k_2 and k_3 . The equivalent thermal conductivity is

IF two metallic plates of equal thickness and thermal conductivities K_(1) and K_(2) are put together face to face anda common plate is constructed, then the equivalent thermal conductivity of this plate will be

Statement I: Two solid cylindrical rods of identical size and different thermal conductivity K_1 and K_2 are connected in series. Then the equivalent thermal conductivity of two rods system is less than that value of thermal conductivity of either rod. Statement II: For two cylindrical rods of identical size and different thermal conductivity K_1 and K_2 connected in series, the equivalent thermal conductivity K is given by (2)/(K)=(1)/(K_1)+(1)/(K_2)

Statement I: Two solid cylindrical rods of identical size and different thermal conductivity K_1 and K_2 are connected in series. Then the equivalent thermal conductivity of two rods system is less than that value of thermal conductivity of either rod. Statement II: For two cylindrical rods of identical size and different thermal conductivity K_1 and K_2 connected in series, the equivalent thermal conductivity K is given by (2)/(K)=(1)/(K_1)+(1)/(K_2)

A composite slab is prepared by pasting two plates of thickness L_(1) and L_(2) and thermal conductivities K_(1) and K_(2) . The slab have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K_(1) and K_(2) . The thermal conductivity of the composite rod will be