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A magnet hung at \mathrm{ 45^{\circ}} with magnetic meridian makes an angle of \mathrm{ 60^{\circ}} with the horizontal. The actual value of the angle of dip is -

Option: 1

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


Option: 2

\mathrm{\tan ^{-1}(\sqrt{6})}


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Apparent angle of dip \mathrm{=60^{\circ}=\delta^{\prime}}
Angle of declination \mathrm{=\theta=60^{\circ}}
Actual angle of dip \mathrm{=\tan \delta^{\circ}}
                            \mathrm{=\tan \delta^{\prime} \times \cos \theta}
                            \mathrm{=(\sqrt{3}) \times \frac{1}{2}}
                            \mathrm{\tan \delta=\frac{\sqrt{3}}{2}}
\mathrm{Actual \: angle \: \mathrm{=\delta=\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)} of \: dip}

Hence 1 is correct option.
 

Posted by

manish painkra

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