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(D) The frequency of oscillation $f$ of a magnetic needle in a magnetic field is given by $f \propto \sqrt{B_H}$,where $B_H$ is the horizontal compone

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$243$

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(D) The frequency of oscillation $f$ of a magnetic needle in a magnetic field is given by $f \propto \sqrt{B_H}$,where $B_H$ is the horizontal component of the Earth's magnetic field.Given $B_H = B \cos 	heta$,where $B$ is the total magnetic field and $	heta$ is the angle of dip.Thus,$f \propto \sqrt{B \cos 	heta}$.Let $f_1 = 20 	ext{ oscillations/min}$ at $	heta_1 = 30^{\circ}$ and $f_2 = 30 	ext{ oscillations/min}$ at $	heta_2 = 60^{\circ}$.$\frac{f_1}{f_2} = \sqrt{\frac{B_1 \cos 	heta_1}{B_2 \cos 	heta_2}}$$\frac{20}{30} = \sqrt{\frac{B_1 \cos 30^{\circ}}{B_2 \cos 60^{\circ}}}$$\frac{2}{3} = \sqrt{\frac{B_1 (\sqrt{3}/2)}{B_2 (1/2)}} = \sqrt{\frac{B_1 \sqrt{3}}{B_2}}$Squaring both sides: $\frac{4}{9} = \frac{B_1 \sqrt{3}}{B_2} \Rightarrow \frac{B_1}{B_2} = \frac{4}{9\sqrt{3}} = \frac{4}{\sqrt{81 	imes 3}} = \frac{4}{\sqrt{243}}$.Comparing this with $\frac{4}{\sqrt{x}}$,we get $x = 243$.

$A$ 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 the two places respectively is $\frac{4}{\sqrt{x}}$. The value of $x$ is $............$.

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