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 cm² 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

$\frac{ΔQ}{Δt} = (\frac{1}{R}) ΔT$
R : Thermal resistance
$\begin{array}{l} ∴ R_{1} = \frac{L_{1}}{K_{1} A} = \frac{L_{1}}{K (120)} \\ L_{1} = 4 cm \\ A = 120 {cm}^{2} \\ R_{2} = \frac{2.5}{(2 K) (120)} \end{array}$
Now, R_eqof this series combination
$\begin{array}{l} R_{eq} = R_{1} + R_{2} \\ where L_{eq} = 4 + 2 .5 = 6 .5 \\ \frac{L_{eq}}{K_{eq} (A)} = \frac{4}{K (120)} + \frac{2.5}{2 K (120)} \\ \frac{6 .5}{K_{eq} (120)} = \frac{4}{K (120)} + \frac{5}{4 K (120)} \\ \frac{6 .5}{K_{eq}} = \frac{21}{4 K} \end{array}$
$\begin{array}{l} K_{eq} = \frac{26}{21} K = (1 + \frac{5}{21}) K \\ ∴ α = 21 \end{array}$