One number sorts all three classes, and most lost marks here come from reading only half of it. That number is the magnetic susceptibility χ, defined by M = χH — magnetisation per unit magnetising field (NCERT Class 12 Physics, Chapter 5, page 146). A stem gives you χ and expects the class back; or gives the class and expects the sign back. The half-read version remembers "χ is small for dia and para" and forgets that the sign is what separates them.
Diamagnetic: χ is small and negative, roughly −10⁻⁵ in magnitude, so μ_r = 1 + χ sits just below 1 and the field inside the specimen is slightly weaker than outside. The atoms have no permanent magnetic moment; the field induces one that opposes it. Consequence: a diamagnetic specimen is repelled, drifting from stronger field towards weaker, and a freely suspended diamagnetic rod sets itself perpendicular to the field (page 146).
Paramagnetic: χ is small and positive, again of order 10⁻⁵, so μ_r is just above 1. Here the atoms do carry permanent moments, randomly oriented until an applied field partially aligns them. The specimen is weakly attracted towards stronger field, and a suspended rod turns parallel to it (page 147).
Ferromagnetic: χ is positive and large — 10³ and beyond — with μ_r of the same order. Permanent atomic moments are already aligned in domains; an applied field grows the favourably-oriented domains. Strong attraction, and magnetisation that survives after the field is removed (page 148).
For NEET, the three facts that carry the marks are sign of χ, order of magnitude of χ, and direction of drift in a non-uniform field. They are not independent — fix the sign and the other two follow.
Watch out for the μ_r conversion. μ_r = 1 + χ, not μ_r = χ. For a diamagnetic substance that difference is the whole answer. (How χ changes with temperature is handled in the separate temperature-effects lesson.)
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The relative permeability μ_r of a diamagnetic substance satisfies which of the following?
Answer: C. For a diamagnetic substance χ is small and negative, and μ_r = 1 + χ, so μ_r falls just below 1 — the field inside is slightly weaker than the applied field (NCERT Class 12 Physics, Chapter 5, page 146).
Why A is wrong: A is wrong because a relative permeability of order 10³ is the signature of a ferromagnetic substance, not a diamagnetic one (Chapter 5, page 148).
Why B is wrong: B is wrong because μ_r slightly above 1 describes a paramagnetic substance, whose χ is small and positive (Chapter 5, page 147).
Why D is wrong: D is wrong because μ_r equals exactly 1 only for vacuum; a diamagnetic specimen measurably reduces the field inside it (Chapter 5, page 146).
A small specimen placed in a non-uniform magnetic field drifts from the region of stronger field towards the region of weaker field. The specimen is
Answer: A. Only a diamagnetic specimen is repelled by a magnetic field, because its induced moment opposes the applied field; it therefore moves towards the weaker-field region (NCERT Class 12 Physics, Chapter 5, page 146).
Why B is wrong: B is wrong because a paramagnetic specimen is weakly attracted and drifts towards the stronger-field region (Chapter 5, page 147).
Why C is wrong: C is wrong because a ferromagnetic specimen is strongly attracted and drifts towards the stronger-field region (Chapter 5, page 148).
Why D is wrong: D is wrong because the drift direction is set by the sign of χ, not by the specimen's shape; both paramagnetic and ferromagnetic specimens move towards stronger field (Chapter 5, pages 147-148).
In which class of magnetic material do the constituent atoms carry zero net magnetic moment in the absence of an external field?
Answer: D. Diamagnetism arises in atoms whose electronic moments cancel to zero; an applied field induces a moment that opposes it, which is why χ comes out negative (NCERT Class 12 Physics, Chapter 5, page 146).
Why A is wrong: A is wrong because paramagnetic atoms carry a permanent net moment that is merely randomly oriented until a field partially aligns it (Chapter 5, page 147).
Why B is wrong: B is wrong because ferromagnetic atoms carry permanent moments that are already mutually aligned within domains (Chapter 5, page 148).
Why C is wrong: C is wrong because it names the two classes that do possess permanent atomic moments (Chapter 5, pages 147-148).
Three specimens P, Q and R have magnetic susceptibilities −9.5 × 10⁻⁶, +2.3 × 10⁻⁵ and +1.4 × 10³ respectively. P, Q and R are, in that order,
Answer: B. A small negative χ marks a diamagnetic substance, a small positive χ a paramagnetic one, and a χ of order 10³ a ferromagnetic one (NCERT Class 12 Physics, Chapter 5, pages 146-148).
Why A is wrong: A is wrong because it swaps Q and R: +2.3 × 10⁻⁵ is a small positive χ (paramagnetic), while +1.4 × 10³ is enormous (ferromagnetic) (Chapter 5, pages 147-148).
Why C is wrong: C is wrong because the negative susceptibility belongs to the diamagnetic specimen; a paramagnetic χ is positive (Chapter 5, page 146).
Why D is wrong: D is wrong because it inverts the ordering entirely; the sign and order of magnitude of χ, not the specimen's position in the list, fix the class (Chapter 5, pages 146-148).
A material has magnetic susceptibility χ = −1.8 × 10⁻⁵. Its relative permeability is
Answer: D. Relative permeability follows from μ_r = 1 + χ = 1 + (−1.8 × 10⁻⁵) = 0.999982, just below 1 as required for a diamagnetic material (NCERT Class 12 Physics, Chapter 5, page 146).
Why A is wrong: A is wrong because it reports the magnitude of χ itself; χ is dimensionless and must be added to 1 to give μ_r (Chapter 5, page 146).
Why B is wrong: B is wrong because it adds the magnitude of χ instead of its signed value; χ is negative here, so μ_r must fall below 1 (Chapter 5, page 146).
Why C is wrong: C is wrong because μ_r is the ratio μ/μ₀ of two permeabilities and is always positive (Chapter 5, page 146).
A paramagnetic specimen of susceptibility 5.0 × 10⁻⁴ is placed in a magnetising field of magnitude 2.0 × 10³ A/m. The magnitude of the magnetisation induced in it is
Answer: A. Magnetisation is M = χH = (5.0 × 10⁻⁴)(2.0 × 10³ A/m) = 1.0 A/m; χ is dimensionless, so M carries the unit of H (NCERT Class 12 Physics, Chapter 5, page 146).
Why B is wrong: B is wrong because it reports H unchanged; magnetisation is χ times H, and χ here is far smaller than 1 (Chapter 5, page 145).
Why C is wrong: C is wrong because it reports χ alone; a dimensionless susceptibility cannot be a magnetisation in A/m (Chapter 5, page 145).
Why D is wrong: D is wrong because it divides H by χ instead of multiplying, which would make the magnetisation larger than the magnetising field (Chapter 5, page 145).
A thin rod of paramagnetic material and an identical rod of diamagnetic material are each freely suspended in a uniform magnetic field. At equilibrium, their long axes lie
Answer: C. The paramagnetic rod's permanent moments align with the field, turning its axis parallel; the diamagnetic rod's induced moment opposes the field, so it settles with its axis at right angles to it (NCERT Class 12 Physics, Chapter 5, pages 146-147).
Why A is wrong: A is wrong because only the paramagnetic rod seeks alignment along the field; the diamagnetic rod sets itself across it (Chapter 5, page 146).
Why B is wrong: B is wrong because a paramagnetic rod's moments align with, not against, the applied field and so turn its axis parallel (Chapter 5, page 147).
Why D is wrong: D is wrong because it reverses both cases; the sign of the moment relative to the field is what decides the orientation, and it is negative for the diamagnetic rod (Chapter 5, pages 146-147).
A specimen has magnetic permeability μ = 3.14 × 10⁻³ T·m/A. Taking μ₀ = 4π × 10⁻⁷ T·m/A, its susceptibility and class are
Answer: B. First μ_r = μ/μ₀ = (3.14 × 10⁻³)/(4π × 10⁻⁷) ≈ 2.50 × 10³, then χ = μ_r − 1 ≈ 2.50 × 10³; a positive susceptibility of this order identifies a ferromagnetic substance (NCERT Class 12 Physics, Chapter 5, pages 146 and 148).
Why A is wrong: A is wrong because a paramagnetic susceptibility is of order 10⁻⁵, thousands of times smaller than the value computed here (Chapter 5, page 147).
Why C is wrong: C is wrong because it inverts the ratio, computing μ₀/μ instead of μ/μ₀; that also flips the apparent sign of the class (Chapter 5, page 146).
Why D is wrong: D is wrong because it quotes μ itself, in T·m/A, as though it were the dimensionless χ; χ must be obtained from μ_r = μ/μ₀ first (Chapter 5, page 146).
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Given
A paramagnetic specimen of volume V = 1.00 × 10⁻⁵ m³ is placed in a magnetising field of magnitude H = 1.20 × 10⁴ A/m. It acquires a net magnetic moment of magnitude m_net = 2.40 × 10⁻⁶ A·m².
Required
The magnetisation M, the susceptibility χ, the relative permeability μ_r, and confirmation of the material class.
Concept
Magnetisation is net magnetic moment per unit volume. Susceptibility is the constant of proportionality between magnetisation and magnetising field, and relative permeability is one more than the susceptibility (NCERT Class 12 Physics, Chapter 5, pages 145-146).
Formula
M = m_net / V; χ = M / H; μ_r = 1 + χ.
Substitution
M = (2.40 × 10⁻⁶ A·m²)/(1.00 × 10⁻⁵ m³); χ = M/(1.20 × 10⁴ A/m); μ_r = 1 + χ.
Calculation
M = 0.240 A/m. Then χ = (0.240 A/m)/(1.20 × 10⁴ A/m) = 2.00 × 10⁻⁵, which is dimensionless because A/m cancels A/m. Then μ_r = 1 + 2.00 × 10⁻⁵ = 1.0000200. The 1 in μ_r = 1 + χ is an exact number from the definition of relative permeability — it does not enter the significant-figure count, so the three significant figures of the data are preserved.
Final answer
M = 0.240 A/m, χ = 2.00 × 10⁻⁵, μ_r = 1.0000200. The susceptibility is positive and of order 10⁻⁵, confirming a paramagnetic specimen.
Common trap
Dividing m_net by V and stopping there, then calling 0.240 the susceptibility. Magnetisation carries the unit A/m; susceptibility is what you get after dividing by H, and it is dimensionless. A second version of the same slip is reporting μ_r as 2.00 × 10⁻⁵ — that is χ, not μ_r.
Similar NEET-style question
A specimen of volume 4.00 × 10⁻⁶ m³ develops a net magnetic moment of 1.60 × 10⁻⁶ A·m² in a magnetising field of 2.00 × 10⁴ A/m. Find its susceptibility and state whether the specimen is attracted towards or repelled from the stronger-field region of a non-uniform field.
χ = M/H, where M is magnetisation and H is magnetising field. Dimensionless. Distinguishes types: diamagnetic (χ < 0, small), paramagnetic (χ > 0, small), ferromagnetic (χ >> 0, large).
-- NCERT Class 12 Physics, Ch. 5, p. 147All materials weakly diamagnetic (induced opposing field). Bismuth, copper, water. χ < 0, ~−10⁻⁵. Independent of T.
-- NCERT Class 12 Physics, Ch. 5, p. 148Atoms have permanent dipoles aligning weakly with B. Aluminium, sodium, calcium, oxygen (at STP), copper chloride. χ > 0, small. χ and μr depend on the sample temperature as well as the material; as the field is increased or the temperature lowered, the magnetisation increases towards saturation.
-- NCERT Class 12 Physics, Ch. 5, p. 148Strong domain alignment. Iron, cobalt, nickel, gadolinium, etc.; relative magnetic permeability > 1000, chi very large. The ferromagnetic property depends on temperature: at high enough temperature a ferromagnet becomes a paramagnet, the domain structure disintegrating; the loss of magnetisation is gradual.
-- NCERT Class 12 Physics, Ch. 5, p. 149More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
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