As per the given figure, two plates A and B of thermal conductivity K and 2 K 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 ___.
The correct answer is: 21
\(\frac{L_1}{K_1A_1}+\frac{L_1}{K_1A_1}=\frac{L_1+L_2}{K_{eff}A_eff}\)
\(⇒K_{eff}=\frac{13K}{10.5}=(1+\frac{5}{21})K\)
⇒ α = 21
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\(F(\frac{dy}{dt},y,t) = 0\)
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Read More: Differential Equations