ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 5. Magnetism and matter

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

  1. A \(\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)\)
  2. B \(\tan ^{-1}(\sqrt{6})\)
  3. C \(\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)\)
  4. D \(\tan ^{-1}\left(\sqrt{\frac{1}{2}}\right)\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\tan \theta^{\prime}=\frac{\tan \theta}{\cos \alpha}\) \(\theta^{\prime}=60^{\circ}\) \(\alpha=45^{\circ}\) \(\sqrt{3}=\frac{\tan \theta}{\frac{1}{\sqrt{2}}}\) \(\tan \theta=\sqrt{\frac{3}{2}}\) \(\theta=\tan ^{-1} \sqrt{\frac{3}{2}}\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app