Q 12-05-077JEE MainJEE Main 2019 (12 Apr, Shift 1)Medium
A magnetic compass needle oscillates $30$ times per minute at a place where the dip is $45^\circ$ and $40$ times per minute where the dip is $30^\circ$. If $B_1$ and $B_2$ are the net magnetic fields due to the earth at the two places respectively, then the ratio $B_1/B_2$ is approximately equal to
Answer: (D) $0.7$
A compass needle oscillates in the horizontal plane, so $f \propto \sqrt{B_H} = \sqrt{B\cos\delta}$.
$$\left(\frac{30}{40}\right)^2 = \frac{B_1\cos45^\circ}{B_2\cos30^\circ} \Rightarrow \frac{B_1}{B_2} = \frac{9}{16}\times\frac{0.866}{0.707} \approx 0.69 \approx 0.7$$
Solution by Sreeraj P, M.Sc Physics