Question

A dip circle is so set that its needle 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 an angle.

• A. ${40}^o$
• B. ${30}^o$
• C. more than ${40}^o$
• D. less than ${40}^o$

Answer 1

The correct option is C more than ${40}^o$

${\delta }_1={40}^o,{\delta }_2={30}^o$

$\cot \delta =\sqrt{{\cot }^2{\delta }_1+{\cot }^2{\delta }_2}$

$=\sqrt{{\cot }^2{40}^o+{\cot }^2{30}^o}$

$\cot \delta =\sqrt{{1.19}^2+3}=2.1$

$\therefore \delta ={25}^{o}i.e.\delta <{40}^o$