Q 12-05-052JEE MainJEE Main 2022 (27 Jul, Shift 2)Medium
A compass needle of oscillation magnetometer oscillates $20$ times per minute at a place $P$ of dip $30^\circ$. The number of oscillations per minute become $10$ at another place $Q$ of $60^\circ$ dip. The ratio of the total magnetic field at the two places $(B_Q : B_P)$ is:
Answer: (A) $\sqrt3 : 4$
The needle oscillates in the horizontal plane, so $f \propto \sqrt{B_H} = \sqrt{B\cos\delta}$.
$$\left(\frac{f_P}{f_Q}\right)^2 = \frac{B_P\cos30^\circ}{B_Q\cos60^\circ} \Rightarrow 4 = \frac{B_P(\sqrt3/2)}{B_Q(1/2)} = \sqrt3\frac{B_P}{B_Q}$$
$$\frac{B_Q}{B_P} = \frac{\sqrt3}{4}$$
Solution by Sreeraj P, M.Sc Physics