Q 12-05-042JEE MainJEE Main 2023 (12 Apr, Shift 1)Medium
A compass needle oscillates $20$ times per minute at a place where the dip is $30^\circ$ and $30$ times per minute where the dip is $60^\circ$. The ratio of total magnetic field due to the earth at two places respectively is $\dfrac4{\sqrt x}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 243
A horizontal compass needle responds to $B_H=B\cos\delta$, with frequency $f\propto\sqrt{B_H}$.
$$\left(\frac{20}{30}\right)^2=\frac{B_1\cos30^\circ}{B_2\cos60^\circ}=\sqrt3\,\frac{B_1}{B_2}\ \Rightarrow\ \frac{B_1}{B_2}=\frac{4}{9\sqrt3}=\frac{4}{\sqrt{243}}$$
So $x=243$.
Solution by Sreeraj P, M.Sc Physics