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JEE • PHYSICS • MAGNETISM AND MATTER

Magnetism and matter practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 271-300 of 323 questions

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271 Medium
271 of 323
Analyzing +4 | -1
Which of the following statements are correct?
A. Electric monopoles do not exist, whereas magnetic monopoles exist.
B. Magnetic field lines due to a solenoid at its ends and outside cannot be completely straight and confined.
C. Magnetic field lines are completely confined within a toroid.
D. Magnetic field lines inside a bar magnet are not parallel.
E. χ=−1\chi = - 1 is the condition for a perfect diamagnetic material, where χ\chi is its magnetic susceptibility.
Choose the correct answer from the options given below.
View answer and explanation
Correct option: A
C and E
Option A is correct because magnetic field lines are confined within a toroid (C) and χ=−1\chi = -1 is the condition for a perfect diamagnetic material (E).
272 Medium
272 of 323
Understanding +4 | -1
For a vibration magnetometer, the time period of suspended bar magnet can be reduced by
View answer and explanation
Correct option: C
moving it towards equator
Option C (moving it towards equator) is correct because the horizontal component of Earth's magnetic field is higher at the equator, decreasing the time period.
273 Medium
273 of 323
Application +4 | -1
A bar magnet is 0.10 m0.10 \mathrm{~m} long and its pole strength is 12 Am12\,\mathrm{Am}. Find the magnitude of the magnetic field at a point on its axis at a distance of 20 cm20 \mathrm{~cm} from it.
View answer and explanation
Correct option: A
3.4×10−5 T3.4 \times 10^{-5} \mathrm{~T}
Option A (3.4×10−5 T3.4 \times 10^{-5} \mathrm{~T}) is correct because it is calculated using the formula for the magnetic field on the axis of a bar magnet.
274 Medium
274 of 323
Application +4 | -1
A magnetic field strength (H)3×103Am−1(\mathrm{H}) 3 \times 10^3 \mathrm{Am}^{-1} produces a magnetic field of induction (B) of 12πT12 \pi \mathrm{T} in an iron rod. Find the relative permeability of iron?
View answer and explanation
Correct option: A
104{10^4}
Option A (104{10^4}) is correct because it's calculated using the formula for relative permeability, considering the magnetic field strength and magnetic field of induction.
275 Medium
275 of 323
Understanding +4 | -1
The resultant magnetic moment of neon atom will be
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because neon is a diamagnetic material with all electron spins paired, resulting in no net magnetic moment.
276 Medium
276 of 323
Analyzing +4 | -1
Assertion (A): The properties of paramagnetic and ferromagnetic substance are not effected by heating.

Reason (R): As temperature rises, the alignment of molecular magnets gradually increases.
View answer and explanation
Correct option: D
If both (A) and (R) are false
Option D is correct. The properties of paramagnetic and ferromagnetic substances are affected by heating, and as temperature rises, the alignment of molecular magnets gradually decreases, not increases.
277 Medium
277 of 323
Remembering +4 | -1
Permanent magnet has properties retentivity and coercivity respectively
View answer and explanation
Correct option: A
High-high
Option A (High-high) is correct because permanent magnets require both high retentivity (to maintain strong magnetism) and high coercivity (to resist demagnetization).
278 Medium
278 of 323
Application +4 | -1
Susceptibility of MgMg at 300  K300\;K is 1.2×10−51.2 \times {10^{ - 5}}. The temperature at which susceptibility will be 1.8×10−51.8 \times {10^{ - 5}}is
View answer and explanation
Correct option: B
200 KK
Option B (200 KK) is correct because the susceptibility is inversely proportional to the temperature, and the calculation follows from this relationship.
279 Medium
279 of 323
Understanding +4 | -1
A magnet of magnetic moment MM is rotated through 360∘360^\circ in a magnetic field HH. The work done will be
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the work done in rotating a magnet through a full circle in a magnetic field is zero, as the initial and final potential energies are the same.
280 Medium
280 of 323
Application +4 | -1
An electron in an atom revolves around the nucleus in an orbit of radius 0.5 Å. Calculate the equivalent magnetic moment if the frequency of revolution of the electron is 1010 MHz10^{10}\,\mathrm{MHz}.
View answer and explanation
Correct option: A
1.256×10−23Am21.256 \times 10^{-23} \mathrm{Am}^2
Option A (1.256×10−23Am21.256 \times 10^{-23} \mathrm{Am}^2) is correct because the magnetic moment is calculated using the current due to the electron's revolution and the area of the orbit.
281 Medium
281 of 323
Understanding +4 | -1
Assertion: Magnetic susceptibility of diamagnetic substance is small negative.
Reason: Diamagnetic substance have unpaired electron.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because diamagnetic substances have a small negative magnetic susceptibility, but this is due to paired electrons, not unpaired electrons.
282 Medium
282 of 323
Application +4 | -1
A magnet of magnetic moment 50i^  A50\hat i\;A-m2{m^2} is placed along the xx-axis in a magnetic field B⃗=(0.5  i^+3.0j^)T\vec B = \left( {0.5\;\hat i + 3.0\hat j} \right)T. The torque acting on the magnet is
View answer and explanation
Correct option: B
150  k^  N150\;\hat k\;N-mm
Option B (150  k^  N150\;\hat k\;N-mm) is correct because the torque is calculated using the cross product of the magnetic moment and the magnetic field.
283 Medium
283 of 323
Remembering +4 | -1
Identify the paramagnetic substance:
View answer and explanation
Correct option: B
Aluminium
Option B (Aluminium) is correct because aluminium is a paramagnetic substance, meaning it is weakly attracted to magnetic fields.
284 Medium
284 of 323
Remembering +4 | -1
The relationship between the magnetic susceptibility (χ)(\chi ) and the magnetic permeability (μ)(\mu) is given by:

(μ0(\mu_0 is the permeability of free space and μr\mu_r is relative permeability)
View answer and explanation
Correct option: A
χ=μμ0−1\chi=\frac{\mu}{\mu_0}-1
Option A (χ=μμ0−1\chi=\frac{\mu}{\mu_0}-1) is correct because it represents the correct relationship between magnetic susceptibility, magnetic permeability, and permeability of free space.
285 Medium
285 of 323
Remembering +4 | -1
Which of the following material is not
ferromagnetic in nature?
View answer and explanation
Correct option: D
Al
Option D (Al) is correct because Aluminum is paramagnetic, not ferromagnetic.
286 Medium
286 of 323
Understanding +4 | -1
Assertion (A): Magnetic moment of an atom is due to both, the orbital motion and spin motion of every electron.

Reason (R): A charged particle at rest produces a magnetic field.
View answer and explanation
Correct option: C
If (A) is true but the (R) is false
Option C (If (A) is true but the (R) is false) is correct because the magnetic moment of an atom is due to both orbital and spin motion of electrons, but a charged particle at rest does not produce a magnetic field.
287 Medium
287 of 323
Understanding +4 | -1
Which one of the following characteristics is
not associated with a paramagnetic material?
View answer and explanation
Correct option: D
Its magnetic susceptibility increases with rise in temperature
Option D (Its magnetic susceptibility increases with rise in temperature) is correct because the magnetic susceptibility of paramagnetic materials decreases with increasing temperature, following Curie's law.
288 Medium
288 of 323
Application +4 | -1
At any place on earth, the horizontal component of earth's magnetic field is 3\sqrt{3} times the vertical component. The angle of dip at that place will be
View answer and explanation
Correct option: D
30∘30^{\circ}
Option D (30∘30^{\circ}) is correct because the angle of dip is related to the horizontal and vertical components of the Earth's magnetic field by tan⁡θ=BVBH\tan \theta = \frac{{B_V}}{{B_H}}.
289 Medium
289 of 323
Analyzing +4 | -1
Statement I Time period of oscillation of two magnets, when like poles are in same direction (in a vibration magnetometer) is smaller, than the period of vibration when like poles are in opposite direction.

Statement II Moment of inertia increases in same position.
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: A
Statement I is true but Statement II is false.
Option A (Statement I is true but Statement II is false) is correct because the time period is smaller when like poles are in the same direction, and the moment of inertia does not increase in the same position.
290 Medium
290 of 323
Remembering +4 | -1
The susceptibility of certain magnetic material is 5000 . What is the class of magnetic material?
View answer and explanation
Correct option: C
Ferromagnetic
Option C (Ferromagnetic) is correct because ferromagnetic materials have very high susceptibility values (much greater than 1).
291 Medium
291 of 323
Remembering +4 | -1
Lines passing through points of zero declination is
View answer and explanation
Correct option: C
Agonic lines
Option C (Agonic lines) is correct because agonic lines are defined as imaginary lines on the Earth's surface connecting points where the magnetic declination is zero, meaning the magnetic north and true north align.
292 Medium
292 of 323
Application +4 | -1
The period of oscillation of a magnet at a place is 2 seconds. When it is remagnetised, so that the pole strength becomes 4 times the initial value. The period of oscillation in second is
View answer and explanation
Correct option: B
11
Option B (1) is correct because the time period is inversely proportional to the square root of the magnetic moment, which is proportional to the pole strength.
293 Medium
293 of 323
Understanding +4 | -1
For an isotropic medium B,  μ,  HB,\;\mu ,\;H and MM are related as (where B,  μ0,  HB,\;{\mu _0},\;H and MM have their usual meanings in the context of magnetic material)
View answer and explanation
Correct option: D
B=μ0(H+M)B = {\mu _0}\left( {H + M} \right)
Option D (B=μ0(H+M)B = {\mu _0}(H + M)) is correct because it represents the relationship between magnetic induction (B), magnetic field intensity (H), and magnetization (M) in an isotropic medium.
294 Medium
294 of 323
Understanding +4 | -1
Assertion: Super conductors can be easily levitated in strong magnetic field.
Reason: Magnetic susceptibility of super conductor is extremely positive.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because superconductors can be levitated, but their magnetic susceptibility is extremely negative (not positive).
295 Medium
295 of 323
Remembering +4 | -1
Substances, when placed in a magnetic field
acquire feeble magnetization in the direction
opposite to that of the applied field are called
View answer and explanation
Correct option: A
Diamagnetic substances
Option A (Diamagnetic substances) is correct because diamagnetic materials are repelled by a magnetic field, acquiring feeble magnetization in the opposite direction.
296 Medium
296 of 323
Application +4 | -1
The needle of a dip circle shows an apparent dip of 45∘45^{\circ} in a particular position and 53∘53^{\circ} when the circle is rotated through 90∘90^{\circ}. Find the true dip.
View answer and explanation
Correct option: C
39∘39^{\circ}
Option C (39∘39^{\circ}) is correct because the true dip is calculated using the formula involving the apparent dips in two perpendicular positions.
297 Medium
297 of 323
Application +4 | -1
The magnetic susceptibility of a material of a rod is 499. The absolute permeability of vacuum is 4π×10−7  Hm−14\pi \times {10^{ - 7}}\;\mathrm{H}{\mathrm{m}^{ - 1}}. The absolute permeability of the material of a rod is
View answer and explanation
Correct option: B
2π×10−4  Hm−12\pi \times {10^{ - 4}}\;\mathrm{H}{\mathrm{m}^{ - 1}}
Option B (2π×10−4  Hm−12\pi \times {10^{ - 4}}\;\mathrm{H}{\mathrm{m}^{ - 1}}) is correct because it is calculated using the formula for absolute permeability, considering the magnetic susceptibility and permeability of vacuum.
298 Medium
298 of 323
Remembering +4 | -1
The magnetic moment of a diamagnetic atoms is:
View answer and explanation
Correct option: D
Equal to zero
Option D (Equal to zero) is correct because diamagnetic atoms have no permanent magnetic dipole moment.
299 Medium
299 of 323
Application +4 | -1
A magnet of distance moment 2 J T−1{\mathrm{T}^{ - 1}} is aligned in the direction of magnetic field of 0.1 T. What is the net work done to bring, the magnet normal to the magnetic field?
View answer and explanation
Correct option: B
0.2 J
Option B (0.2 J) is correct because it is calculated by finding the difference in potential energy of the magnet in the initial and final positions.
300 Medium
300 of 323
Application +4 | -1
The magnetic susceptibility of a paramagnetic material is 1.0×10−51.0 \times 10^{-5} at 270C27^0 \mathrm{C} Temperature. Then, at what temperature its magnetic susceptibility would be 1.5×10−51.5 \times 10^{-5} ?
View answer and explanation
Correct option: C
−73∘C-73^{\circ} \mathrm{C}
Option C (−73∘C-73^{\circ} \mathrm{C}) is correct because the magnetic susceptibility of a paramagnetic material is inversely proportional to its absolute temperature.
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