A compass needle oscillates 20 times per minute at a place where the dip is 45 ° and the magnetic field…

The frequency of oscillations of a magnetic needle in earth’s magnetic field is given as $f=\frac{1}{2\pi }\sqrt{\frac{\mu B_H}{I}}$, where, horizontal component of earth’s magnetic field $B_H=B\cos \theta$, $\mu$ is the magnetic moment of the magnetic needle and $I$ is moment of inertia of the magnetic needle about the axis about which it oscillates. Here, $\theta$ is angle of dip.

As the same magnetic needle is used, so the value of magnetic moment and the moment of inertia will remain the same at the two places.

Now, $f\propto \sqrt{B\cos \theta }$.

Then the ratio, $\frac{f_1}{f_2}=\frac{20}{30}=\sqrt{\frac{B_1\cos 45°}{B_2\cos 30°}}$

$$\begin{gathered} \Rightarrow \frac{400}{900}=\frac{B_1\times {\displaystyle \frac{1}{\sqrt{2}}}}{B_2\times {\displaystyle \frac{\sqrt{3}}{2}}} \\ \Rightarrow \frac{B_1}{B_2}=\frac{2}{3}\times \frac{2}{\sqrt{3}\times \sqrt{2}} \\ \Rightarrow B_1:B_2=2\sqrt{2}:3\sqrt{3} \end{gathered}$$