Physics | VITEEE 2014 | Discussion Page

If a magnet is suspened at angle $30^{0}$ to the magnet meridian , the dip of needle makes angle of $45^{0}$ with the horizontal , the real dip is

$\tan^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\tan^{-1}(\sqrt{3})$

$\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$

$\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Option D

Hence,

$tan \delta'=\frac{\tan\delta}{\cos \theta}=\frac{\tan 45^{0}}{\cos 30^{0}}$

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

or $\delta=\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$

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