Question

The real angle of dip, if a magnet is suspended at an angle of ${30}^0$ to the magnetic meridian and the dip needle makes an angle of 45${}^0$ with horizontal is :

• A. $ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• B. $ta{n}^{-1}\left(\sqrt{3}\right)$
• C. $ta{n}^{-1}\left(\frac{1}{2}\right)$
• D. $ta{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer 1

The correct option is A $ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Given the apparent dip is ${\delta }_1={45}^o$
Angle made by the magnet with the magnetic meridian is $\theta ={30}^o$
$tan{\delta }_1=\frac{tan\delta }{cos\theta }$
$tan\delta =tan{45}^0\times cos{30}^0$
True dip is $\delta =ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$