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Magnetism Mcq
Quiz 7
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Q.1
A bar magnet of length 10cm and having the pole strength equal to 10- weber is kept in the magnetic field having magnetic induction (B) equal to 4π × 10-3 tesla. It makes an angle of 30° with the direction of magnetic induction (B). The value of the torque acting on the magnet is
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a) 2π × 10-7 newton × metre
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b)2π × 10-5 newton × metre
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c)0.5 newton × metre
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d)0.5 × 102 newton × metre
Explanation
τ=m×l×Bsinθ Answer:(a)
Q.2
The angle of dip at place i 40.6° and the intensity of the vertical component of the earth's magnetic field V=6 × 10-5 tesla. The total intensity of earth's magnetic field at this place is
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a) 7 × 10-5 tesla
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b) 6 × 10-5 tesla
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c) 5 × 10-5 tesla
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d) 9.2 × 10-5 tesla
Explanation
Answer: (d)
Q.3
Due to earth's magnetic field, charged cosmic ray particles [ CBSE 1997]
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a)requires greater kinetic energy to reach equator than the pole
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b) require less kinetic energy to reach the equator than the pole
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c)can never reach the pole
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d)can never reach the equator
Explanation
Answer: (d)
Q.4
The needle of dip circle when placed at a geometric pole stays long
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a) south north direction only
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b) east west direction only
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c)vertical direction
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d)none of above
Explanation
Answer: (c)
Q.5
The magnetic susceptibility for diamagnetic material is
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a) small and negative
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b) small and positive
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c)large and positive
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d)large and negative
Explanation
Answer:(a)
Q.6
The correct value of dip angle at a place is 45°. If the dip circle is rotated 45° out of the meridian, then the tangent of the angle of apparent dip is
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a) 1
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b) 1/√2
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c) √2
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d) 1/2
Explanation
New plane is at angle of α=45&dge with magnetic meridian thus H'=Hcosα H'=Hcos45=H/√2 Vertical component is same for both the planes In magnetic meridian tanδ=V/H In new plane tanδ'=V/H'cosα=tanα/cosα=tan45/cos45=√2 Answer: (c)
Q.7
At certain place, the angle of dip is 30° and the horizontal component of earth's magnetic field is 0.5 oersted. the earth's total magnetic field ( in oersted) is
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a)√3
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b) 1
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c)1 / √3
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d)1/2
Explanation
Use tan30=V/H find value of V B2=V2 + H2 Answer: (c)
Q.8
Two small magnets each of magnetic moment 10 Am2 are placed in end on position 0.1m apart from their centres. The force acting between them is
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a) 0.6 × 107 N
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b) 0.006 × 107 N
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c)0.6 N
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d)0.06 N
Explanation
Force between two magnetic dipoles placed in end on position is given by Answer:(c)
Q.9
At a certain place, horizontal component is √3 times the vertical component. the angle of dip at this place is
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a) 0
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b) π/3
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c)π/6
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d)none of the above
Explanation
Answer: (c)
Q.10
A steel wire of length l has a magnetic moment M. It is then bent into a semicircular arc. the new magnetic moment is
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a) M
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b) 2M/π
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c) M/l
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d) M × l
Explanation
Answer: (b)
Q.11
Two identical magnetic dipoles of magnetic moment 1.0 Am2 each, placed at sepration of 2m with their axis perpendicular to each other. The resultant magnetic field at a point midway between the dipole is
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a)5 × 10-7 T
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b) √5 × 10-7 T
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c)10-7 T
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d)2 × 10-7 T
Explanation
As the axis are perpendicular, mid point lies on axial line of one magnet and on the equatorial line of their magnetMagnetic field at axial line MAgnetic field at equatorial line AS B1 is perpendicular to B2 resultant magnetic field is Answer: (b)
Q.12
The force between two short bar magnets with magnetic moments M1 and M2 whose centres are r metre apart is 8N, when their axes are in same line. If the sepration is increased to 2r, the force between them is reduced to
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a) 4N
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b) 2N
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c)1N
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d)0.5N
Explanation
Answer: (d)
Q.13
The force experienced by a pole of strength 100 A-m at a distance of 0.2m from a short magnet of length 5cm and pole strength of 200A-m on its axial line will be
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a) 2.5 × 10-2 N
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b) 2.5 × 10-3 N
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c)5.0 × 10-2 N
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d)5.0 × 10-3 N
Explanation
Using formula for axial point find magnetic field Force=F × m Answer:(a)
Q.14
A magnet 10 cm long has a pole strength of 12 A-m. Find the magnitude of magnetic field strength B at a point on its axis at a distance of 20 cm from it. What would be the value of B, if the point were to lie at the same distance on equatorial of magnet
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a) 3.4 × 10-5 T
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b) 1.4 × 10-5 T
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c) 1.7 × 10-5 T
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d) 0.85 × 10-5 T
Explanation
since d is not very large compared to l we should use following formula for B at axial point We know that Magnetic field at equatorial point is approximately half of axial point Answer: (c)
Q.15
A magnet of moment M is lying in a magnetic field of induction B. W1 is the work done in turning it from 0° to 60 ° and W2 is the work done in turning it from 30° to 90 ° . Then
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a)W2=W1
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b) W2=W1/2
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c) W2=2W1
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d) W2=√3 W1
Explanation
Answer: (d)
Q.16
A bar magnet of magnetic moment 4.0 A-m2 is free to rotate about a vertices axis through its centre. The magnet is released from rest from east-west position. Kinetic energy of the magnet in north-south position will be [ H=25µT)
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a) 10-2 J
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b) 10-4 J
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c)10-6 J
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d)0
Explanation
Answer: (b)
Q.17
The length of bar magnet is 10cm and its pole strength is 10-3 Weber. It is placed in a magnetic field 4 π×10-3 T . in the direction making an angle 30° with electric field direction. The value of torque acting on the magnet will be
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a) 2π × 10-7 N-m
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b) 2π × 10-5 N-m
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c)0.5 × 102 N-m
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d)none of these
Explanation
Answer:(a)
Q.18
A magnet having a magnetic moment of 1.0 × 104 J/T is free to rotate in horizontal plane where a magnetic field 4 × 10-5 T exists. Find the work done in rotating the magnet slowly done in rotating the magnet slowly from a direction parallel to the field to a direction 60° from the field
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a) 0.4 J
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b) 2 J
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c) 0.2 J
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d) 1 J
Explanation
Answer: (c)
Q.19
A magnetic needle of magnetic moment 60 amp-m2 experience a torque of 1.2 ×10-3 N-m directed in geographical north. If the horizontal intensity of earth's magnetic field at that place is 40 Wb/m2, then the angle of direction will be
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a)30°
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b) 45°
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c)60°
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d)90°
Explanation
Answer: (a)
Q.20
A bar magnet with its poles 25 cm apart and pole strength 24.0 A-m rests with its centre on a frictionless pivot. A force F is applied on the magnet at a distance of 12cm from pivot so that it is held in equilibrium at an angle of 30° with respect to a magnetic field of induction 0.25T. The value of force F is
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a) 5.62 N
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b) 2.56 N
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c)6.52 N
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d)6.25 N
Explanation
USe formula F d=m×l×Bsinθ 0.12F=24×0.25×0.25sin30 F=0.75/0.12=6.25 NAnswer: (d)
Q.21
A current of 1 mp. is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment will be
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a) 0.314 A-m2
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b) 3140 A-m2
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c)100 A-m2
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d)µ0 A-m2
Explanation
Magnetic moment=INA where A is the cross-sectional area of coil Answer:(a)
Q.22
If the radius of circular coil is doubled and the current flowing through in it is halved then new magnetic moment will be if its initial magnetic moment is 4 units
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a) 8 units
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b) 4 units
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c) 2 units
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d) zero
Explanation
Use M=NIA for both the cases Answer: (a)
Q.23
A short bar magnet is placed with its north pole pointing south. the neutral point is 10 cm away from the centre of magnet. If H=0.4 gauss, calculate magnetic moment of the magnet
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a)2 Am2
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b) 1 Am2
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c)0.1 Am2
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d)0.2 Am2
Explanation
Answer: (d)
Q.24
At any place on earth, the horizontal component of earth's magnetic field is √3 times the vertical component. The angle of dip at that place will be
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a) 60 °
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b) 45°
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c)90°
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d)30°
Explanation
Answer: (d)
Q.25
The horizontal component of earth's magnetic field at any place is 0.36×10-4 Weber/mIf the angle of dip at that place is 60° then the value of vertical component of earth's magnetic field will be ( in Wb/m2)
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a) 0.12 × 10-4
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b) 0.24 ×10-4
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c)0.40 ×10-4
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d)0.62 ×10-4
Explanation
Answer:(d)
Q.26
The radius of the coil of tangent galvanometer is 16 cm. How many turns of the wire should be used if a current of 40 mA is to be produced a deflection of 45°, given horizontal component of earth's field is 0.36 ×10-4 T.
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a) 458
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b) 229
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c) 200
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d) 115
Explanation
Use formula Answer: (b)
Q.27
The tangent law applied when
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a) magnet is suspended in a uniform magnetic field
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b) horizontal component of earth's field is present
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c) there are two magnetic filed's
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d) there are two uniform magnetic field acting perpendicular to each other
Explanation
Answer: (d)
Q.28
The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into equal parts length wise then the time period of each part will be
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a)4 sec
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b) 2 sec
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c)0.5 sec
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d)0.25 sec
Explanation
Note Magnet is cut along the length, thus length has remained same. Moment of inertia reduced to half and magnetic moment is reduced to half will cancel each others effect Answer: (a)
Q.29
A thin magnetic needle oscillates in a horizontal plane with a period T. It is broken into n equal parts. the time period of each part will be
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a) T
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b) T/n2
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c)Tn2
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d)T/n
Explanation
Answer: (d)
Q.30
The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the same place in a similar manner. The size of two magnet is same but the magnetic moment of B is four times that of A. The time period of B will be
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a) T/4
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b) T/2
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c)2T
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d)4T
Explanation
Answer:(b)
0 h : 0 m : 1 s
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