JEE Main 2022PhysicsThermal Properties of MatterMediumMCQ

JEE Main 2022 — Thermal Properties of Matter Question with Solution

JEE Main 2022 (27 Jul Shift 1)

Question

If K1 and K2 are the thermal conductivities L1 and L2 are the lengths and A1 and A2 are the cross sectional areas of steel and copper rods respectively such that K2K1=9,A1A2=2,L1L2=2. Then, for the arrangement as shown in the figure. The value of temperature T of the steel - copper junction in the steady state will be

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Show full solutionCorrect option: C
Correct answer
C45°C

Step-by-step explanation

At steady state, the heat conducted by both the rods will be the same.

The rate of heat conduction is given by, dQdt=KAl∆T. Therefore, if the temperature of the junction is T,

K1 A1l1T1-T=K2 A2l2T-T2

⇒450-TT-0=K2 A2l1 K1 A1l2=9×12×2

⇒450-T=9 T⇒T=45°C

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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.