A small bar magnet is placed with its north pole facing the magnetic north pole. The neutral points are located at a distance r from its centre. If the magnet is rotated by 180^o, the neutral point shall be obtained at a distance of:
1. \(2r\)
2. \(\sqrt{2}r\)
3. \(2^{\frac{1}{3}}r\)
4. \(\frac{r}{2\sqrt{2}}\)

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 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. $2\sqrt{3}:9$

Given below are two statements: 

Given below are two statements: 

If \(\phi_1\) and \(\phi_2\) are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip \(\phi\) is given by:
1. \(cos^2{\phi}=cos^2{\phi_1}+cos^2{\phi_2}\)
2. \(sec^2{\phi}=sec^2{\phi_1}+sec^2{\phi_2}\)
3. \(tan^2{\phi}=tan^2{\phi_1}+tan^2{\phi_2}\)
4. \(cot^2{\phi}=cot^2{\phi_1}+cot^2{\phi_2}\)

A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It:

1. will become rigid showing no movement

2. will stay in any position

3. will stay in north-south direction only

4. will stay in east-west direction only

At point A on the earth's surface, the angle of dip is, $\delta =+25°$. At a point B on the earth's surface, the angle of dip is, $\delta =-25°$. We can interpret that:

The current of 30 A is flowing in a vertical straight wire. If the horizontal component of the earth's magnetic field is 2 x 10^-5 Tesla, then the position of the null point will be:

1. 0.9 m 

2. 0.3 mm

3. 0.3 cm 

4. 0.3 m