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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 241-270 of 323 questions

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241 Medium
241 of 323
Understanding +4 | -1
The null points are on the axial line of a bar
magnet, when it is placed such that its south
pole points:
View answer and explanation
Correct option: C
north
Option C (north) is correct because when the north pole of the bar magnet points geographically north, the null points are on the axial line.
242 Medium
242 of 323
Application +4 | -1
A steel wire of length 1 has a magnetic moment mm. It is bent into a semicircular arc. What is the new magnetic moment?
View answer and explanation
Correct option: A
2mπ\frac{2 m}{\pi}
Option A (2mπ\frac{2 m}{\pi}) is correct because the new magnetic moment is calculated by dividing the original magnetic moment by the original length and multiplying by the new separation between the poles.
243 Medium
243 of 323
Remembering +4 | -1
Which of the following materials is paramagnetic at 300 K300 {~K} ?
View answer and explanation
Correct option: D
Platinum
Option D (Platinum) is correct because Platinum is known to be paramagnetic at room temperature (300 K).
244 Medium
244 of 323
Understanding +4 | -1
Choose the correct option
View answer and explanation
Correct option: B
True dip is less than apparent dip.
Option B (True dip is less than apparent dip) is correct because the apparent dip is the angle in a plane not aligned with the magnetic meridian, and is always greater than or equal to the true dip.
245 Medium
245 of 323
Analyzing +4 | -1
Assertion (A): Magnetic poles are only mathematical assumptions having no real existence.

Reason (R): A charged particle is moved along a magnetic field line. The magnetic force on the particle is perpendicular to its velocity.
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 magnetic poles are mathematical constructs, but the force on a charged particle moving along a magnetic field line is zero, not perpendicular.
246 Medium
246 of 323
Understanding +4 | -1
Assertion: Ferromagnetic substance exhibits spontaneous magnetisation.
Reason: In the absence of external magnetic field, atomic dipoles of each domain are aligned in particular direction.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because ferromagnetic materials exhibit spontaneous magnetization due to aligned atomic dipoles within domains, even without an external field.
247 Medium
247 of 323
Application +4 | -1
A magnet performs 10 oscillations per minute in a horizontal plane at a place where the angle o dip is 45∘45^\circ and the total intensity is 0.707 units. The number of oscillations per minute at a place where dip angle is 60∘60^\circ and total intensity is 0.5 CGS units will be
View answer and explanation
Correct option: B
7
Option B (7) is correct because the number of oscillations depends on the horizontal component of the Earth's magnetic field, which is related to the total intensity and the angle of dip.
248 Medium
248 of 323
Application +4 | -1
A magnet performs 10 oscillations per minute in a horizontal plane at a place where the angle of dip is 45∘^ \circ and the total intensity is 0.707 CGS units. The number of oscillations per minute at a place where dip angle is 60∘^ \circ and total intensity is 0.5 CGS units will be
View answer and explanation
Correct option: B
7
Option C (400) is correct because the number of turns in the secondary coil can be calculated using the transformer equation, relating the number of turns to the voltage in the primary and secondary coils.
249 Medium
249 of 323
Application +4 | -1
The true value of angle of dip at a place is 60∘60^\circ, the apparent dip in a plane inclined at an angle of 30∘30^\circ with magnetic meridian is
View answer and explanation
Correct option: B
tan⁡−1(2){\tan ^{ - 1}}\left( 2 \right)
Option B (tan⁡−1(2){\tan ^{ - 1}}\left( 2 \right)) is correct because it is calculated using the formula relating apparent and true angles of dip.
250 Medium
250 of 323
Application +4 | -1
A rod of ferromagnetic material with dimensions 10  cm×0.5  cm×0.2  cm10\mathrm{\;cm} \times 0.5\mathrm{\;cm} \times 0.2\mathrm{\;cm} is placed in a magnetic field of strength 0.5×104A−m−10.5 \times {10^4}\mathrm{A} - {\mathrm{m}^{ - 1}} as a result of which of which a magnetic moment of 0.5 A−m−2\mathrm{A} - {\mathrm{m}^{ - 2}} is produced in rod. The value of magnetic induction will be
View answer and explanation
Correct option: B
6.28 T
Option B (6.28 T) is correct because it correctly calculates the magnetic induction using the given magnetic field strength and magnetic moment.
251 Medium
251 of 323
Understanding +4 | -1
A sphere of gold when brought towards a
powerful magnet experiences
View answer and explanation
Correct option: D
repulsive force
Option D (repulsive force) is correct because gold is a diamagnetic material, which is repelled by strong magnetic fields.
252 Medium
252 of 323
Understanding +4 | -1
The materials suitable for making electromagnets should have
View answer and explanation
Correct option: B
low retentivity and low coercivity
Option B (low retentivity and low coercivity) is correct because electromagnets need to be easily magnetized and demagnetized, which requires low retentivity and coercivity.
253 Medium
253 of 323
Understanding +4 | -1
A thin light weight rod of diamagnetic substance such as silver is suspended in uniform external magnetic field. It will align itself with its length
View answer and explanation
Correct option: C
perpendicular to magnetic field
The correct answer is C (perpendicular to magnetic field) because diamagnetic substances tend to align themselves perpendicular to the magnetic field.
254 Medium
254 of 323
Application +4 | -1
The coil in a tangent galvanometer is 16 cm in radius. If a current of 20 mA is to produce a deflection of 45∘45^\circ then the number of turns wound on it, is (Take horizontal component of earth’s magnetic field=0.36×10−4  T  and  μ0=4π×10−7  Wb  A−1  m−1= 0.36 \times {10^{ - 4}}\mathrm{\;T\;and\;}{\mathrm{\mu }_0} = 4\pi \times {10^{ - 7}}\;\mathrm{Wb\;}{\mathrm{A}^{ - 1}}\mathrm{\;}{\mathrm{m}^{ - 1}}
View answer and explanation
Correct option: B
458
Option B (458) is correct because the number of turns is calculated using the tangent galvanometer formula, relating current, radius, magnetic field, and deflection angle.
255 Medium
255 of 323
Application +4 | -1
A bar magnet of magnetic moment 2.5 Am22.5\,\mathrm{Am}^2 is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east-west direction. Find the kinetic energy of the magnet as it aligns itself in the north-south direction. The horizontal component of earth's magnetic field is 0.3 G0.3\,\mathrm{G}.
View answer and explanation
Correct option: A
75μJ75 \mu \mathrm{J}
Option A (75μJ75 \mu \mathrm{J}) is correct because the kinetic energy equals the loss in potential energy as the magnet aligns.
256 Medium
256 of 323
Remembering +4 | -1
The strength of earth's magnetic field on the earth's surface is of the order
View answer and explanation
Correct option: A
10−5 T10^{-5} \mathrm{~T}
Option A (10−5 T10^{-5} \mathrm{~T}) is correct because the Earth's magnetic field strength on the surface is approximately on the order of 10−510^{-5} Tesla.
257 Medium
257 of 323
Application +4 | -1
At a certain place, the angle of dip is 30∘^ \circ and the horizontal component of earth’s magnetic field is 0.50 oersted. The earth’s total magnetic filed (in oersted) is
View answer and explanation
Correct option: C
13\frac{1}{{\sqrt 3 }}
Option C (13\frac{1}{{\sqrt 3 }}) is correct because the earth's total magnetic field is related to the horizontal component and the angle of dip by the formula B=BHcos⁡δB = \frac{B_H}{\cos \delta}.
258 Medium
258 of 323
Application +4 | -1
A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through 181∘181^{\circ} to deflect magnet by 37∘37^{\circ} from magnetic meridian. Now this magnet is replaced by another magnet and upper end of wire has to be rotated by 273∘273^{\circ} to deflect magnet by 53∘53^{\circ} from magnetic meridian. The ratio of magnetic moment of the two magnets respectively is
View answer and explanation
Correct option: B
4855\frac{48}{55}
Option B (48/55) is correct because it is the ratio of magnetic moments calculated using the torsion constant and deflection angles.
259 Medium
259 of 323
Remembering +4 | -1
For a paramagnetic material, the dependence of the magnetic susceptibility XX on the absolute temperature is given as
View answer and explanation
Correct option: C
X∝1/TX \propto 1/T
Option C (X∝1/TX \propto 1/T) is correct because the magnetic susceptibility of a paramagnetic material is inversely proportional to the absolute temperature, as described by Curie's Law.
260 Medium
260 of 323
Understanding +4 | -1
For a diamagnetic material:
View answer and explanation
Correct option: C
μr<1,χ<0\mu_r < 1, \chi < 0
Option C (μr<1,χ<0\mu_r < 1, \chi < 0) is correct because diamagnetic materials have a relative permeability less than 1 and a negative magnetic susceptibility.
261 Medium
261 of 323
Remembering +4 | -1
The susceptibility and permeability of a perfectly diamagnetic substance is
View answer and explanation
Correct option: C
-1 and 0
Option C (-1 and 0) is correct because perfectly diamagnetic substances have a susceptibility of -1 and a relative permeability of 0.
262 Medium
262 of 323
Understanding +4 | -1
For which among the following house
appliances, magnet is an essential part?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because magnets are essential components in the motors of calling bells, fans, and washing machines.
263 Medium
263 of 323
Application +4 | -1
A short magnet produces a deflection of 37∘37^{\circ} in a deflection magnetometer in Tan- AA position when placed at a separation of 10 cm10 \mathrm{~cm} from the needle. Find the ratio of the magnetic moment of the magnet to the earth's horizontal magnetic field (in  A−m2 T−1\mathrm{~A-m}^2 \mathrm{~T}^{-1}).
View answer and explanation
Correct option: A
3.75×1033.75 \times 10^3
Option A (3.75×1033.75 \times 10^3) is correct because it is calculated using the formula for the magnetic field in the Tan-A position of a deflection magnetometer.
264 Medium
264 of 323
Understanding +4 | -1
Permanent magnets are those substances that ________.
View answer and explanation
Correct option: A
retain their ferromagnetic property for a long time at room temperature
Option A (retain their ferromagnetic property for a long time at room temperature) is correct because permanent magnets maintain their magnetism at room temperature.
265 Medium
265 of 323
Understanding +4 | -1
If the magnetizing field on a ferromagnetic material is increased, its permeability
View answer and explanation
Correct option: A
Decreased
Option A is correct because as the magnetizing field increases in a ferromagnetic material, the permeability decreases due to saturation.
266 Medium
266 of 323
Application +4 | -1
Two short bar magnets of moments ' MM ' and 3M′\sqrt{3} M^{\prime} are joined like a cross (+) This cross suspended from its centre with its plane horizontal in the earth's magnetic field. When the cross comes to equilibrium, angle made by the weaker magnet with the earth's magnetic field BHB_H is
View answer and explanation
Correct option: D
60∘60^{\circ}
Option D (60∘60^{\circ}) is correct because the angle made by the weaker magnet with the earth's magnetic field is 60 degrees when the torques are balanced.
267 Medium
267 of 323
Application +4 | -1
A magnet is parallel to a uniform magnetic field. If it is rotated by 60∘60^\circ, the work done is 0.8  J0.8\;J. How much work is done in moving it 30∘30^\circ further
View answer and explanation
Correct option: A
0.8×107erg0.8 \times {10^7}erg
Option A (0.8×107erg0.8 \times {10^7}erg) is correct because it calculates the work done using the formula W=MB(cos⁡θ1−cos⁡θ2)W = MB(\cos{\theta _1} - \cos{\theta _2}) and converts Joules to ergs.
268 Medium
268 of 323
Remembering +4 | -1
The intensity of magnetic field due to an isolated pole of strength mm at a point distance rr from it will be
View answer and explanation
Correct option: A
mr2\frac{m}{{{r^2}}}
Option A (mr2\frac{m}{{{{r^2}}}}) is correct because the magnetic field intensity due to an isolated pole is inversely proportional to the square of the distance.
269 Medium
269 of 323
Remembering +4 | -1
Which of the following is responsible for the earth's magnetic field?
View answer and explanation
Correct option: A
Convective currents in earth's core.
Option A (Convective currents in earth's core) is correct because the Earth's magnetic field is generated by the movement of molten iron in the Earth's outer core, which creates electric currents and thus a magnetic field.
270 Medium
270 of 323
Remembering +4 | -1
Curie law χT=\chi T= constant, relating magnetic susceptibility (χ)(\chi) and absolute temperature (T) of magnetic substances, is obeyed by
View answer and explanation
Correct option: B
paramagnetic substances
Option B (paramagnetic substances) is correct because Curie's law, which states that magnetic susceptibility is inversely proportional to absolute temperature (χT=constant\chi T = constant), is obeyed by paramagnetic substances.
Revision notes and key takeaways

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