Question

At a place, a freely suspended magnet shows a dip angle of ${30}^0$. If the plane of the magnet is at an angle of ${45}^0$ from the magnetic meridian, then the real dip of the magnet will be

• A. $ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• B. $ta{n}^{-1}\sqrt{3}$
• C. $ta{n}^{-1}\left(\frac{1}{\sqrt{6}}\right)$
• D. $co{t}^{-1}\sqrt{6}$

Answer 1

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

$\tan \delta =\frac{V}{H}$ ----eq.1

$\tan {45}^{\circ }=\frac{V}{H\cos {30}^{\circ }}$----eq.2

eq.1 / eq.2

$\frac{\tan \delta }{\tan {45}^{\circ }}=cos{30}^{\circ }$

$\tan \delta =\frac{\sqrt{3}}{2}$

$\delta ={\tan }^{-1}\frac{\sqrt{3}}{2}$

Hence option 'A' is correct.