A dip circle is so set that its needle moves freely in the magnetic meridian. In this position, the angle of dip is 40º.Now the dipcircle is rotated so that the plane in which the needle moves makes an angle of 30º with the magnetic meridian.In this position, the needle will dip by an angle

40º
30º
more than 40º
less than 40º

Solution

A dip needle lies initially in the magnetic meridian when it shows an angle of dip θ at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip θ’.Then \(\frac{tan\, \theta ‘}{tan\, \theta }\) is

\(\frac{1}{cos\, x}\)
\(\frac{1}{sin\, x}\)
\(\frac{1}{sin\, x}\)
cos x

Solution

Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magnet is placed 20cms in tan A position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to

20 cms
20 (2)^1/3cms
20 (2)^2/3cms
20 (2)^3/3cms

Solution

A compass needle placed at a distance r from a short magnet in Tan A position shows a deflection of 60º. If the distance is increased to r (3)^1/3, then deflection of compass needle is

1/3
60 × 3^1/3
60 × 3^2/3
60 × 3^3/3

Solution

Two bar magnets of the same mass, same length and breadth but having magnetic moments M and 2M are joined together pole for pole and suspended by a string. The time period of assembly in a magnetic field of strength H is 3 seconds. If now the polarity of one of the magnets is reversed and combination is again made to oscillate in the same field, the time of oscillation is

√3 sec
3√3 sec
3 sec
6 sec

Solution

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

tan^-1(\(\sqrt{3/2}\))
tan^-1(√3)
tan^-1(√3 / 2)
tan^-1(2 / √3)

Solution

Two points A and B are situated at a distance x and 2x respectively from the nearer pole of a magnet 2cm long. The ratio of magnetic field at A and B is

4 : 1 exactly
4 : 1 approximately
8 : 1 approximately
1 : 1 approximately

Solution

In the above question, magnetic moment of each part is

Solution

As length of each part also becomes half, therefore magnetic moment M = pole strength × length

\(\Rightarrow \frac{1}{2}\times \frac{1}{2}=\frac{1}{4}th\: i.e. M/4.\)

A thin bar magnet of length 2 and breadth 2 b pole strength m and magnetic moment M is divided into four equal parts with length and breadth of each part being half of original magnet. Then the pole strength of each part is

Solution

As breadth of each part is half the original breadth,therefore, pole strength becomes half (i.e. m/2).

The B – H curve (i) and (ii) shown in fig.associated with

(i) diamagnetic and (ii) paramagnetic substance
(i) paramagnetic and (ii) ferromagnetic substance
(i) soft iron and (ii) steel respectively
(i) steel and (ii) soft iron respectively

Solution

The loop (i) is for soft iron and the loop (ii) is for steel in Fig.

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