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A magnetic needle is free to rotate in a...

A magnetic needle is free to rotate in a vertical plane which makes an angle of `60^@` with the magnetic meridian. If the needle stays in a direction making an angle of `tan^-1(2//sqrt3)` with the horizontal, what would be the true dip at that place?

A

`0^(@)`

B

`45^(@)`

C

`30^(@)`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given `theta = 60^(@)`. Let the apparent dip be `delta` and the real dip be `delta` then , cot `delta = (sqrt(3))/(2)`.
Now , cot `delta = cot delta cos alpha`
`therefore (sqrt(3))/(2) - cot delta . Cos 60^(@) ` and ` cot delta (sqrt(3))/(2) xx 2 = sqrt(3)`
Thus real dip `(delta) = 30^(@)`
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