At a certain place, the horizontal component B₀ and the vertical component V₀ of the earth's magnetic field are equal in magnitude. The total intensity at the place will be

1. $B_0$                            2. $B_0^2$

3. $2B_0$                          4. $\sqrt{2}{B}_0$

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 $\theta$ with the magnetic meridian such that

(a) $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$            (b) $\theta ={\tan }^{-1}\left(\sqrt{3}\right)$

(c) $\theta ={\tan }^{-1}\left(\frac{1}{2}\right)$               (d) $\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$

The angle of dip at a certain place is 30^o. If the horizontal component of the earth’s magnetic field is H, the intensity of the total magnetic field is

1. $\frac{H}{2}$                            2. $\frac{2H}{\sqrt{3}}$

3. $H\sqrt{2}$                         4. $H\sqrt{3}$

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

1. $T=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{{\mathrm{MB}}_{\mathrm{H}}}}$

2.  $T=2\mathrm{\pi }\sqrt{\frac{M{B}_H}{I}}$

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

4. $T=2\mathrm{\pi }\sqrt{\frac{B_H}{MI}}$

Time period for a magnet is T. If it is divided in four equal parts along its axis and perpendicular to its axis as shown then time period for each part will be

1. 4T

2. T/4

3. T/2

4. T

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) 1 sec                         

(2) 4 sec

(3) 8 sec                         

(4) 0.5 sec

If 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:

1. 2T                            

2. T/2

3. 2/T                           

4. T

A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place where dip angle is 30^o and 15 oscillations per minute at a place where dip angle is 60^o. The ratio of total earth's magnetic field at the two places is:

1. $3\sqrt{3}:8$
2. $16:9\sqrt{3}$
3. 4:9
4. $2\sqrt{3}:9$

A magnet of magnetic moment M oscillating freely in earth's horizontal magnetic field makes n oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled, the number of oscillations made per minute would be

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

3. $2\sqrt{2}n$                        4. $\sqrt{2}n$

Magnets A and B are geometrically similar but the magnetic moment of A is twice that of B. If T₁ and 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. 1/3                             

2. 1/2

3. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.$                         

4. $\sqrt{3}$