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 cm 2 for each plate. The equivalent thermal conductivity of the compound plate is 1 + 5 α K , then the value of α will be _____ .

Step-by-step solution

The thermal resistance of rod 1 will be R 1 = l 1 K A = 4 K A and rod 2 will be R 2 = l 2 2 K A = 2 . 5 2 K A . Since they are connected in series, R e q = R 1 + R 2 = 10 . 5 2 K A . If equivalent thermal conductivity is K e q , then R e q = l 1 + l 2 K e q A = 6 . 5 K e q A . Therefore, 10 . 5 2 K A = 6 . 5 K e q A ⇒ K e q = 26 21 K = 1 + 5 21 K Hence, α = 21 .