A compass needle of oscillation magnetometer oscillates 20 times per minute at a place $P$ of $\oper

Question

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place $P$ of $\operatorname{dip} 30^{\circ}$. The number of oscillations per minute become 10 at another place $Q$ of $60^{\circ} \mathrm{dip}$. The ratio of the total magnetic field at the two places ( $B_{Q}: B_{P}$ ) is\

Professor PC Thomas Classes · Answer

(a) : Time period of oscillations of the needle is,\ $T=2 \pi \sqrt{\frac{I}{M(B \cos 	heta)}}=\frac{1}{f}$\ $\Rightarrow \sqrt{\frac{B_{Q} \cos 60^{\circ}}{B_{P} \cos 30^{\circ}}}=\frac{f_{Q}}{f_{P}}=\frac{10}{20}=\frac{1}{2}$\ $\Rightarrow \frac{B_{Q}}{B_{P}}=\frac{1}{4} 	imes \sqrt{3}=\frac{\sqrt{3}}{4}$\