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If a magnet is suspended at angle $30^{\circ}$ to the magnet meridian, the dip of needle makes angle of $45^{\circ}$ with the horizontal, the real dip is
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The correct answer is:
$\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
Here, $\tan \delta^{\prime}=\frac{\tan \delta}{\cos \theta}=\frac{\tan 45^{\circ}}{\cos 30^{\circ}}$ $\tan \delta=\frac{1}{\sqrt{3} / 2}=\frac{2}{\sqrt{3}}$ or $\delta=\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
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