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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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Two identical short bar magnets, each having magnetic moment M, are placed a distance of 2d apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is

$\frac{μ_{0}}{4 π} (\sqrt{2}) \frac{M}{d^{3}}$

$\frac{μ_{0}}{4 π} (\sqrt{3}) \frac{M}{d^{3}}$

$(\frac{2 μ_{0}}{4 π}) \frac{M}{d^{3}}$

$\frac{μ_{0}}{4 π} (\sqrt{5}) \frac{M}{d^{3}}$

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A magnet oscillating in a horizontal plane has a time period of 2 second at a place where the angle of dip is 30^o and 3 seconds at another place where the angle of dip is 60^o. The ratio of resultant magnetic fields at the two places is

$\frac{4 \sqrt{3}}{7}$

$\frac{4}{9 \sqrt{3}}$

$\frac{9}{4 \sqrt{3}}$

$\frac{9}{\sqrt{3}}$

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Explanation

T $α \frac{1}{\sqrt{B_{H}}} = \frac{1}{\sqrt{B \cos \emptyset}} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{B_{2} \cos \emptyset_{2}}{B_{1} \cos \emptyset_{1}}}$

$\Rightarrow \frac{B_{1}}{B_{2}} = \frac{{T^{2}}_{2}}{{T^{2}}_{1}} \times \frac{\cos \emptyset 2}{\cos \emptyset 1} = {(\frac{3}{2})}^{2} \times \frac{\cos 60^{\circ}}{\cos 30^{\circ}} \Rightarrow \frac{B_{1}}{B_{2}} = \frac{9}{4 \sqrt{3}}$

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A superconductor exhibits perfect :

Ferrimagnetism

Ferromagnetism

Paramagnetism

Diamagnetism

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Explanation

(d)

A superconducting material kept in a magnetic field expels the magnetic flux out its body when cooled below the critical temperature and exhibits perfect diamagnetism.

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Among the following properties describing diamagnetism identify the property that is wrongly stated

Diamagnetic material do not have permanent magnetic moment

Diamagnetism is explained in terms of electromagnetic induction

Diamagnetic materials have a small positive susceptibility

The magnetic moment of individual electrons neutralize each other

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Explanation

(c)

Susceptibility of diamagnetic substance is negative and it does not change with temperature.

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If a magnet is suspended at an angle 30^o to the magnetic meridian, it makes an angle of 45^owith the horizontal. The real dip is

$\tan^{- 1} (\frac{\sqrt{3}}{2})$

$\tan^{- 1} (\sqrt{3})$

$\tan^{- 1} (\sqrt{\frac{3}{2}})$

$\tan^{- 1} (\frac{2}{\sqrt{3}})$

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The true value of angle of dip at a place is 60^o, the apparent dip in a plane inclined at an angle of 30^o with magnetic meridian is

$\tan^{- 1} (\frac{1}{2})$

$\tan^{- 1} (2)$

$\tan^{- 1} (\frac{2}{3})$

None of these

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