Consider a compound lab consisting of two different materials that have equal lengths, thickness and thermal conductivities \[K\] and \[2K\] respectively. The equivalent thermal conductivity of the slab is:
(A) \[\sqrt 2 K\]
(B) \[3K\]
(C) \[\dfrac{4}{3}K\]
(D) \[\dfrac{2}{3}K\]

Hint:The equivalent thermal conductivity of the two slabs of different material in series, is obtained by adding them.
It is similar to that of the equivalent resistances connected in series.

Complete step-by-step solution:
We know, in case of resistors, we can find the equivalent resistance if we connect them in series or parallel.
In series connection:
\[{R_{eq}} = {R_1} + {R_2}\]
Where, \[{R_{eq}}\]is the equivalent resistance.
In case of Parallel connection:
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\]
\[{R_{eq}}\]is again the equivalent resistance.
We know, resistance can be expressed as:
\[R = \dfrac{{\rho L}}{A}\]
Here,
\[R = \]Resistance
\[\rho = \]Resistivity of the material
\[L = \]Length of the conductor
\[A = \]Area of the conductor.
We know that resistivity of the material is nothing but the reciprocal of thermal conductivity.
This means:
\[\rho = \dfrac{1}{K}\]
Putting this value, in the expression above, we get:
\[R = \dfrac{L}{{KA}}\]
Where, \[K\] is the thermal conductivity.
Let us consider, the slabs are connected in series.
They follow the formula as in case of resistances connected in series:
\[\dfrac{{{L_S}}}{{{K_S}{A_S}}} = \dfrac{{{L_1}}}{{{K_1}{A_1}}} + \dfrac{{{L_2}}}{{{K_2}{A_2}}}\]
Here,
\[\dfrac{{{L_S}}}{{{K_S}{A_S}}}\]indicates the equivalent quantities.
We know, in case of equivalent resistance, the length becomes twice that of the initial, and the area of slab remains the same.
Applying this concept:
\[\dfrac{{2{L_{}}}}{{{K_S}A}} = \dfrac{{{L_{}}}}{{{K_1}{A_{}}}} + \dfrac{L}{{{K_2}A}}\]
Since it is given that slabs have the same length, we have considered it to be \[L\].
Now, putting \[{K_1} = K\]and \[{K_2} = 2K\], we obtain
\[\dfrac{{2{L_{}}}}{{{K_S}A}} = \dfrac{{{L_{}}}}{{K{A_{}}}} + \dfrac{L}{{2KA}}\]
Cancelling the equal terms, we get;
\[\dfrac{{{2_{}}}}{{{K_S}}} = \dfrac{{{1_{}}}}{K} + \dfrac{1}{{2K}}\]
On solving the above equation, we obtain:
\[{K_s} = \dfrac{{4K}}{3}\]
This is the required thermal conductivity of the compound slab.

Therefore, option (C) is correct.

Note: Thermal conductivity can be defined as the rate at which heat is transferred by conduction through a cross sectional area of a material. Resistivity is defined as the resistance of a wire having unit length and unit cross sectional area.