JEE Main 2020PhysicsMagnetic Properties of MatterEasyMCQ

JEE Main 2020 — Magnetic Properties of Matter Question with Solution

JEE Main 2020 (04 Sep Shift 2)

Question

A paramagnetic sample shows a net magnetisation of 6 A/m when it is placed in an external magnetic field of 0.4 T at a temperature of 4 K. When the sample is placed in an external magnetic field of 0.3 T at a temperature of 24 K, then the magnetisation will be :

Choose an option

Show full solutionCorrect option: D
Correct answer
D0.75 A/m

Step-by-step explanation

According to Curie's Law,
χ∝1T
Therefore, χ1T1=χ2T2
Also, χ=MB
So, 
M1T1B1=M2T2B2
On substituting the values,
6×40.4=M2×240.3

Therefore, M2=0.75 A m-1

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About this question

This is a previous-year question from JEE Main 2020, covering the Magnetic 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.