JEE Main 2022 · 29 Jun · Shift 1 · Physics Q53

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PhysicsInteger+4 / -1Question 53 of 90

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 $\left(1+\dfrac5\alpha\right)$ K, then the value of $\alpha$ will be ______.

Compound slab: plate A (conductivity K, 4.0 cm thick) joined to plate B (2K, 2.5 cm), with the free face of A at 100°C and of B at 0°C.

Your answer

Answer and worked solution

Official answer: 21

Series slabs. $K_{eq}=\dfrac{4+2.5}{\frac4K+\frac{2.5}{2K}}=\dfrac{6.5}{5.25}K=\left(1+\dfrac{5}{21}\right)K$ .

Answer: 21

Topics:Conduction

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JEE Main 2022 · 29 Jun · Shift 1, Physics question 53