JEE Main 2022PhysicsThermal Properties of MatterHardNumerical

JEE Main 2022Thermal Properties of Matter Question with Solution

JEE Main 2022 (29 Jun Shift 1)

Question

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+5αK, then the value of α will be _____ .

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Show full solutionCorrect answer: 21
Correct answer
21

Step-by-step explanation

The thermal resistance of rod 1 will be R1=l1KA=4KA and rod 2 will be R2=l22KA=2.52KA.

Since they are connected in series, Req=R1+R2=10.52KA.

If equivalent thermal conductivity is Keq, then Req=l1+l2KeqA=6.5KeqA. Therefore,

10.52KA=6.5KeqAKeq=2621K=1+521K

Hence, α=21.

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About this question

This is a previous-year question from JEE Main 2022, covering the Thermal Properties of Matter chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.