First slide

Earth's Magnetism

Question

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

Difficult

A

40^o
B

30^o
C

more than 40^o
D

less than 40^o

Solution

Let new angle of dip be $θ$ '. Then
$\tan θ^{'} = \frac{\tan 40}{\cos 30} = \frac{\tan 40 \times 2}{\sqrt{3}}$
This is more than 40^o.

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