JEE Main 2023PhysicsThermal Properties of MatterEasyNumerical

JEE Main 2023Thermal Properties of Matter Question with Solution

JEE Main 2023 (31 Jan Shift 1)

Question

A thin rod having a length of 1 m and area of cross-section 3×10-6 m2 is suspended vertically from one end. The rod is cooled from 210°C to 160°C. After cooling, a mass M is attached at the lower end of the rod such that the length of rod again becomes 1 m. Young's modulus and coefficient of linear expansion of the rod are 2×1011 N m-2 and 2×10-5 K-1, respectively. The value of M is ______ kg. (Take g=10 m s-2)

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Show full solutionCorrect answer: 60
Correct answer
60

Step-by-step explanation

Let Δl be the decease in length of rod due to decease in temperature

Then, Δl=lαΔT, here, coefficient of linear expansion is α=2×10-5 K-1 and ΔT=(483-433) K=50 K

Thus, Δl=1×2×10-5×50=10-3 m

Young's modulus is stated as  Y=FAΔll, here, F=Mg.

So, 2×1011=Mg×13×10-6×10-3

Mg=2×1011×3×10-9=6×10-2

Thus, M=60 kg

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About this question

This is a previous-year question from JEE Main 2023, 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.