MAGNETISM AND MATTER Quiz-20
Dear Readers,
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. An inductor of 10 mH shows 50 mH when operate with a core mad e of ferrite. The susceptibility of ferrite is
Solution
μr=L'/L=50/10=5
μr=L'/L=50/10=5
Xm=μr-1=5-1=4
Q2. A magnet 20 cm long with its poles concentrated at its ends is placed vertically with its north pole on the table. At a point due 20 cm south (magnetic) of the pole, a neutral point is obtained. If H=0.3 G, then the pole strength of the magnet is approximately
Solution
NS is a magnet held vertically with its north pole on the table. P is neutral point, where NP=20 cm, figure. Clearly,
NS is a magnet held vertically with its north pole on the table. P is neutral point, where NP=20 cm, figure. Clearly,
Q3. The time period of a freely suspended magnet is 2 sec. If it is broken in length into two equal parts and one parts is suspended in the same way, then its time period will be
Solution
T'=T/n⇒T'=2/2=1sec
T'=T/n⇒T'=2/2=1sec
Q4. A short bar magnet of magnetic moment 255 JT-1 is placed with its axis perpendicular to earth’s field direction. At what distance from the center of the magnet, the resultant field is inclined at 45° with earth’s field, H=0.4×10-4T ?
Solution
Since, B and H are perpendicular to each other and the resultant field is inclined at an angle 45° with. So, B=H μ0/4Ï€ 2M/r3 =H ∴r3=μ0/4Ï€ 2M/H=0.5 m
Since, B and H are perpendicular to each other and the resultant field is inclined at an angle 45° with. So, B=H μ0/4Ï€ 2M/r3 =H ∴r3=μ0/4Ï€ 2M/H=0.5 m
Q5.A bar-magnet of moment of inertia 49×10-2 kg-m^2 vibrate in a magnetic field of induction 0.5×10-4 T. The time period of vibration is 8.8 s. The magnetic moment of the bar magnet is
Solution
Time period of magnet is T=2Ï€√(I/(MBH)) Or M=(4Ï€2 I)/(T2 BH ) 0r M=(4 ×(3.14)2 × 49 × 10-2)/((8.8)2× 0.5 ×10-4) Or M=5000 A-m2
Time period of magnet is T=2Ï€√(I/(MBH)) Or M=(4Ï€2 I)/(T2 BH ) 0r M=(4 ×(3.14)2 × 49 × 10-2)/((8.8)2× 0.5 ×10-4) Or M=5000 A-m2
Q6. A magnet performs 10 oscillations per minute in a horizontal plane at a place where the angle of dip is 45° and the total intensity is 0.707 CGS units. The number of oscillations per minute at a place where dip angle is 60° and total intensity is 0.5 CGS units will be
Solution
Q7. There is no couple acting when two bar magnets are placed coaxially separated by a distance because
Solution
It is a fact.
It is a fact.
Q8. The points A and B are situated perpendicular to the axis of 2 cm long bar magnet at large distances x and 3x from the centre on opposite sides. The ratio of magnetic fields at A and B will be approximately equal to
Solution
On equatorial line, magnetic field due to magnet varies inversely as cube of the distance, therefore, B1/B2=(3x/x)3=27∶1
On equatorial line, magnetic field due to magnet varies inversely as cube of the distance, therefore, B1/B2=(3x/x)3=27∶1
Q9.The correct relation is
[Where BH= Horizontal component of earth’s magnetic field; BV= Vertical component of earth’s magnetic field and B=Total intensity of earth’s magnetic field]
Solution
It is a fact.
It is a fact.
Q10. A short bar magnet with the north pole facing north forms a neutral point a P in the horizontal plane. If the magnet is rotated by 90° in the horizontal plane, the net magnetic induction at P is (Horizontal component of earth’s magnetic field=BH)
Solution
In Fig.(a), at neutral point P,
BH=μ0/4Ï€ (M/d3) In Fig. (b) Net magnetic induction at P = resultant of μ0/4Ï€ (2 M)/d3=2BH along horizontal and BH along vertical =√((2BH)2+(BH)2)=√5BH
In Fig.(a), at neutral point P,
BH=μ0/4Ï€ (M/d3) In Fig. (b) Net magnetic induction at P = resultant of μ0/4Ï€ (2 M)/d3=2BH along horizontal and BH along vertical =√((2BH)2+(BH)2)=√5BH
