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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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Magnets A and B are geometrically similar but the magnetic moment of A is twice that of B. If T₁and T₂ be the time periods of the oscillation when their like poles and unlike poles are kept together respectively, then $\frac{T_{1}}{T_{2}}$ will be

1/3

1/2

$\frac{1}{\sqrt{3}}$

$\sqrt{3}$

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Explanation

The time period of oscillation is inversely proportional to the square root of the magnetic moment. Since A's moment is twice B's, the ratio of their periods for like poles is 1:âˆš2. For unlike poles, it's âˆš2:1. Hence, T1/T2 = (âˆš2/1)*(1/âˆš2) = 1/âˆš3.

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A magnetic needle suspended by a silk thread is vibrating in the earth's magnetic field. If the temperature of the needle is increased by 500°C, then

The time period decreases

The time period remains unchanged

The time period increases

The needle stops vibrating

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Explanation

$T \propto \frac{1}{\sqrt{M}}$ . Since magnetic moment decreases with increase in temperature hence time period T increases.

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The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is T₀. If this magnet is replaced by another magnet of the same size and pole strength but with double the mass, the new time period will be

$\frac{T_{0}}{2}$

$\frac{T_{0}}{\sqrt{2}}$

$\sqrt{2} T_{0}$

2T₀

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Explanation

The time period of oscillation of a bar magnet is directly proportional to the square root of its mass. If the mass is doubled while keeping the magnetic moment constant, the new time period will be âˆš2 times the original period T_0.

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A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T', then ratio T'/T is

1/4

$\frac{1}{2 \sqrt{2}}$

1/2

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Explanation

$T = 2 π \sqrt{\frac{I}{MB}} \frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1} M_{2}}{I_{2} M_{1}}} = \sqrt{\frac{I}{\frac{I}{8}} \times \frac{\frac{M}{2}}{M}} = 2 \frac{T_{2}}{T_{1}} = \frac{1}{2}$

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