Magnetism and Matter NEET PYQReal NEET & AIPMT 2005–2020 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Magnetism and Matter question from the NEET & AIPMT papers on this site (2005–2020), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Magnetism and Matter previous year questions
18 questions. Tap an option to check your answer.
Q1NEET 2020 · Q171
An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m⁻¹. The permeability of the material of the rod is : (µ₀ = 4π×10⁻⁷ T m A⁻¹)
Not quite — the answer is A.
µ = µ₀(1 + χ) = 4π×10⁻⁷ × 600 = 2400π×10⁻⁷ = 2.4π×10⁻⁴ T m A⁻¹. The magnetising field of 1200 A m⁻¹ is not needed for permeability.
Q2NEET 2017 (Gujarati paper) · Q161
Two reasons for using soft iron as the material for electromagnets.
Not quite — the answer is A.
An electromagnet should become strongly magnetised by a small current (high permeability) and lose its magnetism when the current is switched off (low retentivity and coercivity). High retentivity suits permanent magnets such as steel, not electromagnets.
Q3NEET 2016 Phase 2 · Q28
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is
Not quite — the answer is B.
Work to rotate from 0° to 60° is W = MB(1 − cos60°) = MB/2, so MB = 2W. Torque at 60° is MB sin60° = 2W × √3/2 = √3W. Using sin60° with MB = W gives √3W/2.
Q4NEET 2016 Phase 1 · Q3
The magnetic susceptibility is negative for:
Not quite — the answer is A.
A diamagnetic material develops magnetisation opposite to the applied field, so its susceptibility is small and negative. Paramagnetic and ferromagnetic materials are magnetised along the field and have positive susceptibility.
Q5NEET 2013
A bar magnet of length l and magnetic dipole moment M is bent in the form of an arc of a circle of radius r subtending an angle of 60° at the centre. The new magnetic dipole moment will be
Not quite — the answer is B.
Pole strength stays m with M = ml. The arc length l = r(π/3), so r = 3l/π. The poles are now joined by a chord subtending 60°, whose length equals r, so M′ = m × 3l/π = 3M/π.
Q6AIPMT 2012 Mains
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be
Not quite — the answer is B.
Work to turn from 0° to 60° is W = MB(1 − cos 60°) = MB/2 = √3 J, so MB = 2√3 J. The torque to hold it there is τ = MB sin 60° = 2√3 × √3/2 = 3 (in N m).
Q7AIPMT 2011 Mains
A short bar magnet of magnetic moment 0.4 J T⁻¹ is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is
Not quite — the answer is C.
U = −MB cos θ; stable equilibrium is θ = 0, so U = −MB = −0.4 × 0.16 = −0.064 J. +0.064 J is the unstable (antiparallel) equilibrium.
Q8AIPMT 2011 Prelims
There are four light-weight-rod samples, A, B, C, D separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted (i) A is feebly repelled (ii) B is feebly attracted (iii) C is strongly attracted (iv) D remains unaffected Which one of the following is true?
Not quite — the answer is B.
Feeble attraction towards a magnet is the signature of a paramagnetic material. Feeble repulsion indicates diamagnetic (so the first sample is diamagnetic, not non-magnetic), strong attraction indicates ferromagnetic, and the unaffected sample is non-magnetic, not ferromagnetic.
Q9AIPMT 2010 Mains · Q75
A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10⁻⁴ m² carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10⁻² tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be
Not quite — the answer is C.
Magnetic moment m = NIA = 2000 × 2.0 × 1.5 × 10⁻⁴ = 0.6 A m². Torque τ = mB sin 30° = 0.6 × 5 × 10⁻² × 0.5 = 1.5 × 10⁻² N m. Using cos 30° or dropping the factor N leads to the other values.
Q10AIPMT 2010 Mains · Q86
The magnetic moment of a diamagnetic atom is
Not quite — the answer is D.
In a diamagnetic atom the orbital and spin magnetic moments of the electrons cancel completely, so the net (permanent) magnetic moment is zero. A moment between zero and one or greater than one would describe paramagnetic or ferromagnetic atoms.
Q11AIPMT 2010 Prelims · Q34
Electromagnets are made of soft iron because soft iron has
Not quite — the answer is D.
An electromagnet must lose its magnetism as soon as the current is switched off, so it needs low retentivity, and it must be easy to demagnetise, so low coercivity. High retentivity and high coercivity suit permanent magnets.
Q12AIPMT 2010 Prelims · Q35
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be
Not quite — the answer is A.
T ∝ 1/√B. The opposing field reduces the net horizontal field to 24 − 18 = 6 μT. T′ = 2 × √(24/6) = 4 s. Adding the fields (42 μT) would give a shorter period.
Q13AIPMT 2009 Prelims · Q28
A bar magnet having a magnetic moment of 2 × 10⁴ J T⁻¹ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 × 10⁻⁴ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is:
Not quite — the answer is B.
Work = MB(cos 0° − cos 60°) = 2 × 10⁴ × 6 × 10⁻⁴ × ½ = 6 J. The value 12 J is MB itself, which would be the work only for a 90° turn.
Q14AIPMT 2009 Prelims · Q34
If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is:
Not quite — the answer is D.
Diamagnetic materials get a weak induced moment opposite to the applied field, so they are pushed towards weaker field regions. This happens regardless of which pole is used, so they are repelled by both poles.
Q15AIPMT 2008 Prelims · Q37
Curie temperature is the temperature above which
Not quite — the answer is B.
Above the Curie temperature, thermal agitation destroys the alignment of the domains, and a ferromagnet behaves as a paramagnet. It does not become diamagnetic, because the atoms still carry permanent magnetic moments.
Q16AIPMT 2007 Prelims · Q45
Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show
Not quite — the answer is D.
Above the Curie temperature thermal agitation destroys the domain alignment and a ferromagnet becomes paramagnetic. It does not become diamagnetic or lose all magnetic response.
Q17AIPMT 2006 Prelims · Q41
Above Curie temperature :
Not quite — the answer is A.
Above its Curie temperature a ferromagnet loses its domain alignment because thermal agitation wins, and it behaves as a paramagnet. Curie temperature is a property of ferromagnets; paramagnets do not turn ferromagnetic or diamagnetic on heating.
Q18AIPMT 2005 Prelims · Q33
If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material are denoted by μd, μp and μf respectively, then :
Not quite — the answer is C.
Atoms of diamagnetic substances have no net permanent dipole moment, while atoms of paramagnetic and ferromagnetic substances do. So μd = 0 and μp ≠ 0; any option with μd ≠ 0 or μp = 0 is wrong.
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