A dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position, the angle of dip ia `40^(@)`. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of `30^(@)` with the magnetic meridian. In this position the needle will dip by an angle

Options

A. `40^(@)`
B. `30^(@)`
C. More than `40^(@)`
D. Less than `40^(@)`

Answer

C

Explanation

`tan delta = (B_(V))/(B_(H))`, If apparent dip is `delta`
`tan delta. = (B_(V))/(B_(H))= (B_(V))/(B_(H) cos 30^(@))= (B_(V))/(B_(H) xx (sqrt3)/(2))`
`rArr tan delta.= ((2)/(sqrt3)) tan delta rArr tan delta. gt tan delta`