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(D) The frequency of oscillation of a magnetic needle in the Earth's magnetic field is given by $f = \frac{1}{2 \pi} \sqrt{\frac{\mu B_H}{I}}$,where $

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$2 \sqrt{2}: 3 \sqrt{3}$

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(D) The frequency of oscillation of a magnetic needle in the Earth's magnetic field is given by $f = \frac{1}{2 \pi} \sqrt{\frac{\mu B_H}{I}}$,where $B_H = B \cos 	heta$ is the horizontal component of the Earth's magnetic field,$\mu$ is the magnetic moment,and $I$ is the moment of inertia.Since the same needle is used,$\mu$ and $I$ are constant. Thus,$f \propto \sqrt{B \cos 	heta}$.Given $f_1 = 20$ oscillations/min at $	heta_1 = 45^{\circ}$ and $f_2 = 30$ oscillations/min at $	heta_2 = 30^{\circ}$.Taking the ratio: $\frac{f_1}{f_2} = \sqrt{\frac{B_1 \cos 45^{\circ}}{B_2 \cos 30^{\circ}}} \Rightarrow \frac{20}{30} = \sqrt{\frac{B_1 (1/\sqrt{2})}{B_2 (\sqrt{3}/2)}}$.Squaring both sides: $\frac{4}{9} = \frac{B_1}{B_2} 	imes \frac{1}{\sqrt{2}} 	imes \frac{2}{\sqrt{3}} = \frac{B_1}{B_2} 	imes \frac{\sqrt{2}}{\sqrt{3}}$.Therefore,$\frac{B_1}{B_2} = \frac{4}{9} 	imes \frac{\sqrt{3}}{\sqrt{2}} = \frac{2 	imes 2 	imes \sqrt{3}}{3 	imes 3 	imes \sqrt{2}} = \frac{2 \sqrt{2} 	imes \sqrt{2} 	imes \sqrt{3}}{3 \sqrt{3} 	imes \sqrt{3} 	imes \sqrt{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}}$.Thus,$B_1: B_2 = 2 \sqrt{2}: 3 \sqrt{3}$.

$A$ compass needle oscillates $20$ times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_1$. The same needle oscillates $30$ times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_2$. Then $B_1: B_2$ is

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