JEE Main 2019PhysicsThermal Properties of MatterMediumMCQ

JEE Main 2019 — Thermal Properties of Matter Question with Solution

JEE Main 2019 (09 Apr Shift 2)

Question

Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are θ2 and θ1 respectively, θ2>θ1. The temperature at the interface is

Choose an option

Show full solutionCorrect option: D
Correct answer
Dθ110+9θ210

Step-by-step explanation



Let the temperature of the junction T °C .

Rate of heat flow in Rod 1 = rate of heat flow in Rod 2

3kAdθ2-T=kA3dT-θ1

⇒9θ2-T=T-θ1

⇒10T=9θ2+θ1

⇒T=9θ2+θ110=θ110+9θ210

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About this question

This is a previous-year question from JEE Main 2019, 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.