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## MAGNETISM AND MATTER Quiz-12

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. Due to the earth’s magnetic field, charged cosmic ray particles
•  Require greater kinetic energy to reach the equator than the poles
•  Require less kinetic energy to reach the equator than the poles
•  Can never reach the equator
•  Can never reach the poles
Solution
As they enter the magnetic field of the earth, they are deflected away from the equator

Q2. Magnetic lines of force due to a bar magnet do not intersect because
•  A point always has a single net magnetic field
•  The lines have similar charges and so repel each other
•  The lines always diverge from a single point
•  The lines need magnetic lenses to be made to intersect
Solution
A point always has a single net magnetic field

Q3. Tangent galvanometer is used to measure
•  Current impulses
•  Magnetic moments of bar magnets
•  Earth’s magnetic field
Solution

Q4. The figure illustrates how B, the flux density inside a sample of unmagnetised ferromagnetic material, varies with B0, the magnetic flux density in which the sample is kept. For the sample to be suitable for making a permanent magnet

•  OQ should be large, OR should be small
•  OQ and OR should both be large
•  OQ should be small and OR should be large
•  OQ and OR should both be small
Solution
In the given figure OQ refers to retentivity while OR refers to coercivity. For permanents both retentivity and coercivity should be high

Q5.A bar magnet of magnetic moment 104 J/T is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of 4×10-5 T to a direction 60° from the field will be
•  0.2 J
•  2.0 J
•  4.18 J
•  2×102 J
Solution
Magnetic moment of bar M=104 J/T B=4×10-5 T Hence work done W=M ⃗.B ⃗ =104×4×10-5×cos⁡60°=0.2 J

Q6. Two like magnetic poles of strength 10 and 45 SI units are separated by a distance 30 cm. The intensity of magnetic field is zero on the line joining them
•  At a point 10 cm from the stronger pole
•  At a point 20 cm from the stronger pole
• At the mid-point
•  At infinity
Solution
Suppose magnetic field is zero at point P which lies at a distance x from 10 unit pole. Hence at P
So from stronger pole distance is 20 cm.

Q7.Domain formation is the necessary feature of
•  Ferromagnetism
•  Paramagnetism
•  Diamagnetism
•  All of these
Solution
Domain formation is the necessary feature of ferromagnetism

Q8.A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.16 tesla experiences a torque of magnitude 0.032 J. The magnetic moment of bar magnet will be
•  0.23 J/T
•  0.40 J/T
•  0.80 J/T
•  Zero
Solution
Torque, Ï„=MBH sin⁡Î¸ ⇒ 0.032=M×0.16sin⁡30° ⇒ M=0.4 J/T

Q9.Some equipotential surfaces of the magnetic scalar potential are shown in the figure. Magnetic field at a point in the region is

•  10-4 T
•  2×10-4 T
•  0.5×10-4 T
•  None of these
Solution
T=2Ï€√(1/MH= ) ∞

Q10. A bar magnet when placed at an angle of 30° to the direction of magnetic field induction of 5×10-2 T, experiences a moment of couple 25×10-6 N-m. If the length of the magnet is 5 cm, its pole strength is
•  2×10-2A-m
•  5×10-2A-m
•  2 A-m
• 5 A-m
Solution
Torque, Ï„=MB sin⁡Î¸ ⇒ Ï„=(mL)B sin⁡Î¸ ⇒ 25×10-6=(m×5×10-2 )×510-2 × sin⁡30°) ⇒ m=2×10-2 A-m

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