A dip circle is so set that the dip needle moves freely in the magnetic meridian. In this position the angle of dip is `39^(@)`. 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

d

Explanation

(d) `delta_(1)=40^(@), delta_(2)=30^(@), delta=?`
`cotdeltasqrt(cot^(2)delta_(1)+cot^(2)delta_(2))=sqrt(cot^(2)40^(@)+cot^(2)30^(@))`
`cot delta=sqrt(1.19^(2)+3=2.1)`
`therefore delta=25^(@) i.e delta lt 40^(@)`