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The real angle of dip, if a magnet is su...

The real angle of dip, if a magnet is suspended at an angle of `30^(@)` to the magnetic meridian and the dip needle makes an angle of `45^(@)` with horizontal, is:

A

`tan^(-1)(3//sqrt2)`

B

`tan^(-1)(sqrt3)`

C

`tan^(-1)(3//sqrt2)`

D

`tan^(-1)(2//sqrt3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`tan delta'=(tan delta)/(cos theta)=(tan 45^(@))/(cos 30^(@)), tan delta' =(1)/(sqrt3//2)=(2)/(sqrt3)`
`therefore " " delta'=tan^(-1)((2)/sqrt3)`
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