As per the given figure, two plates A and B of thermal conductivity K and 2K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross section is $120 c m^{2}$ for each plate. The equivalent thermal conductivity of the compound plate is $(1 + \frac{5}{α}) K,$ then the value of $α$ will be ______

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answer is 21.

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Detailed Solution

$\frac{L_{1}}{K_{1} A_{1}} + \frac{L_{2}}{K_{2} A_{2}} = \frac{L_{1} + L_{2}}{K_{e f f} A_{e f f}}$ $\frac{4}{K} + \frac{2.5}{2 K} = \frac{6.5}{K_{e f f}}$ $\frac{10.5}{2 K} = \frac{6.5}{K_{e f f}}$ $K_{e f f} = \frac{13 K}{10.5} = (1 + \frac{5}{21}) K$ $\Rightarrow α = 21$