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 Multiple Choice QuestionsMultiple Choice Questions

101.

A place where the vertical component of the earth's magnetic field is zero has angle of dip is

  • 45°

  • 90°

  • 60°


102.

The magnetic moment of a short bar magnet is 4 Am2 . The magnetic induction at a point 10 cm away from its  mid point on axial line is B1 , and on the equatorial line is B2 , then

  • B1 = 8 × 10-4 T , B2 = 4 × 10-4 T

  • B1 = 2 × 10-4 T , B2 = 1 × 10-4 T

  • B1 = 10 × 10-4 T , B2 = 5 × 10-4 T

  • B1 = 6 × 10-4 T , B2 = 3 × 10-4 T


103.

The value of angle of dip is zero at the magnetic equator because on it

  • V and H are equal

  • the value of V is zero

  • the value of H is zero

  • the value of V is infinity


104.

A bar magnet has a coercivity of 4 x 103 Am-1. It is placed inside a solenoid of length 12 cm having 60 turns in order to demagnetise it. The amount of current that should be passed through the solenoid is

  • 16 A

  • 8 A

  • 4 A

  • 2 A


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105.

A bar magnet of length l and magnetic dipole moment M is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be

      

  • M2

  • M

  • 3Mπ

  • 2Mπ


106.

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

  • OQ should be large, OR should be small

  • both OQ and OR should be large

  • OQ should be small and OR should be large

  • both OQ and OR should be small


107.

A frog can be levitated in magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog behaves as

  • paramagnetic

  • diamagnetic

  • ferromagnetic

  • anti-ferromagnetic


108.

The magnetic susceptibility of a material of a rod is 499. The absolute permeability of vacuum is 4π x 10-7 Hm-1 .  The absolute permeability of the material of rod is

  • π × 10-4 Hm-1

  • 2π × 10-4 Hm-1

  • 3π × 10-4 Hm-1

  • 4π × 10-4 Hm-1


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109.

At a certain place, the horizontal component of earth's magnetic field is 3 times the vertical component. The angle of dip at that place is

  • 30°

  • 60°

  • 45°

  • 90°


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110.

Magnetic field intensity H at the centre of a circular loop of radius r carrying current I emu is 

  • rI  oersted

  • 2πIr oersted

  • I2πr oersted

  • 2πrI oersted


B.

2πIr oersted

H = μ0I2 r = μ04 π × 2 πlrIn emu system μ04π = 1So, H = 2 πlr oersted


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