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

1. Vertical

2. Horizontal

3. In any direction

4. At an angle of dip to the horizontal

If the angles of dip at two places are 30^o and 45^o respectively, then the ratio of horizontal  components of earth's magnetic field at the two places will be:
(Assume net magnetic field to be equal at the two places)

1. √3 : √2

2. 1 : √2

3. 1 : √3

4. 1 : 2

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

1. Isoclinic line            

2. Agonic line

3. Isogonic line            

4. Aclinic line

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

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

2. ${\sin }^{-1}\left(\frac{3}{4}\right)$

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

4. ${\sin }^{-1}\left(\frac{3}{5}\right)$

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

1. $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$            2. $\theta ={\tan }^{-1}\left(\sqrt{3}\right)$

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

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