JEE Main 2022 · 27 Jul · Shift 1 · Physics Q43

PhysicsSingle correct+4 / -1Question 43 of 90

A magnet hung at 45° with magnetic meridian makes an angle of 60° with the horizontal. The actual value of the angle of dip is -

\tan^{-1}\left(\sqrt{\dfrac32}\right)

\tan^{-1}\left(\sqrt6\right)

\tan^{-1}\left(\sqrt{\dfrac23}\right)

\tan^{-1}\left(\sqrt{\dfrac12}\right)

Answer and worked solution

Official answer: A

Apparent dip in a plane at θ to the meridian: $\tan\delta'=\frac{\tan\delta}{\cos\theta}$ .

$\tan\delta=\tan60^\circ\cos45^\circ=\sqrt{\frac32}$ . (A)

JEE Main 2022 · 27 Jul · Shift 1, Physics question 43