JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Heat & Thermodynamics

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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 + \frac{5}{α}) K

, then the value of

α

will be ______.

Answer: 21

Solution

For plates in series, thermal resistances add:

R_{e q} = R_{1} + R_{2}

where

R = \frac{L}{K A}

\frac{L _{e q}}{K _{e q} A} = \frac{L _{1}}{K _{1} A} + \frac{L _{2}}{K _{2} A}

With

L_{1} = 4

cm,

K_{1} = K

L_{2} = 2.5

cm,

K_{2} = 2 K

, and

L_{e q} = 6.5

cm:

\frac{6.5}{K _{e q}} = \frac{4}{K} + \frac{2.5}{2 K} = \frac{4}{K} + \frac{1.25}{K} = \frac{5.25}{K} = \frac{21}{4 K}

K_{e q} = \frac{6.5 \times 4 K}{21} = \frac{26 K}{21} = (1 + \frac{5}{21}) K

Therefore,

α = 21

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Heat & Thermodynamics

Solution

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