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Physics MCQs for NEET — Practice Questions with Answers

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Two lines of force due to a bar magnet

Intersect at the neutral point

Intersect near the poles of the magnet

Intersect on the equatorial axis of the magnet

Do not intersect at all

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Explanation

(d)

Two lines of force due to a bar magnet do not intersect at all.

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A dip needle in a plane perpendicular to magnetic meridian will remain

Vertical

Horizontal

In any direction

At an angle of dip to the horizontal

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Explanation

(a) Dip needle can not rotate horizontally. When it is placed perpendicular to the magnetic meridian, it experiences a force due to vertical component of the earth's magnetic field only. SO it remains vertical.

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If the angles of dip at two places are 30^o and 45^orespectively, then the ratio of horizontal components of earth's magnetic field at the two places will be

$\sqrt{3} : \sqrt{2}$

$1 : \sqrt{2}$

$1 : \sqrt{3}$

1:2

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A line passing through places having zero value of magnetic dip is called

Isoclinic line

Agonic line

Isogonic line

Aclinic line

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Explanation

(d)

It is called a magnetic equator or aclinic line.

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The earth's magnetic field at a certain place has a horizontal component 0.3 Gauss and the total strength 0.5 Gauss. The angle of dip is

$\tan^{- 1} (\frac{3}{4})$

$\sin^{- 1} (\frac{3}{4})$

$\tan^{- 1} (\frac{4}{3})$

$\sin^{- 1} (\frac{3}{5})$

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Two bar magnets with magnetic moments 2 M and M are fastened together at right angles to each other at their centres to form a crossed system, which can rotate freely about a vertical axis through the centre. The crossed system sets in earth’s magnetic field with magnet having magnetic moment 2M making an angle $θ$ with the magnetic meridian such that

$θ = \tan^{- 1} (\frac{1}{\sqrt{3}})$

$θ = \tan^{- 1} (\sqrt{3})$

$θ = \tan^{- 1} (\frac{1}{2})$

$θ = \tan^{- 1} (\frac{3}{4})$

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The time period of oscillation of a freely suspended bar magnet with usual notations is given by

$T = 2 π \sqrt{\frac{I}{{MB}_{H}}}$

$T = 2 π \sqrt{\frac{M B_{H}}{I}}$

$T = 2 \sqrt{\frac{I}{{MB}_{H}}}$

$T = 2 π \sqrt{\frac{B_{H}}{M I}}$

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Explanation

(a)

$T = 2 π \sqrt{\frac{I}{M B_{H}}}$

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The period of oscillation of a magnet in vibration magnetometer is 2 sec. The period of oscillation of a magnet whose magnetic moment is four times that of the first magnet is

1 sec

4 sec

8 sec

0.5 sec

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A magnetic needle is made to vibrate in uniform field H, then its time period is T. If it vibrates in the field of intensity 4H, its time period will be

T/2

2/T

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Explanation

The time period of oscillation of a magnetic needle in a vibration magnetometer is inversely proportional to the square root of the magnetic field intensity. If the field intensity is increased by 4 times, the time period will decrease by a factor of 1/sqrt(4) = 1/2. Hence, the new time period will be T/2.

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A bar magnet A of magnetic moment M_A is found to oscillate at a frequency twice that of magnet B having identical size as magnet A, of magnetic moment M_B when placed in a vibrating magneto-meter. We may say that

$M_{A} = 2 M_{B}$

M_A = 8M_B

M_A = 4M_B

M_B = 8M_A

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Explanation

The frequency of oscillation of a magnet in a vibrating magnetometer is inversely proportional to the square root of its magnetic moment. If the frequency of A is twice that of B, then M_A = (2)^2 * M_B = 4M_B.

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