JEE Mains · Physics · STD 12 - 5. Magnetism and matter
At an angle of \(30^{\circ}\) to the magnetic meridian, the apparent dip is \(45^{\circ} .\) Find the true dip :
- A \(\tan ^{-1} \frac{1}{\sqrt{3}}\)
- B \(\tan ^{-1} \frac{\sqrt{3}}{2}\)
- C \(\tan ^{-1} \sqrt{3}\)
- D \(\tan ^{-1} \frac{2}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1} \frac{\sqrt{3}}{2}\)
Step-by-step Solution
Detailed explanation
\(A \tan \delta=\tan \delta^{\prime} \cos \theta\) \(=\tan 45^{\circ} \cos 30^{\circ}\) \(\tan \delta=1 \times \frac{\sqrt{3}}{2}\) \(\delta=\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
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