A compass needle oscillates 20 times per minute at a place where the dip is $30^{\circ}$ and 30 time

Question

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 $\frac{4}{\sqrt{x}}$. The value of $x$ is\

Professor PC Thomas Classes · Answer

(243) : $\frac{B_{H_{2}}}{B_{H_{1}}}=\frac{T_{1}^{2}}{T_{2}^{2}} \Rightarrow \frac{B_{2} \cos \delta_{2}}{B_{1} \cos \delta_{1}}=\frac{T_{1}^{2}}{T_{2}^{2}}$\ $\frac{B_{2}}{B_{1}}=\frac{T_{1}^{2}}{T_{2}^{2}} \Rightarrow \frac{\cos \delta_{1}}{\cos \delta_{2}}=\left(\frac{3}{2}ight)^{2}$\ $\Rightarrow \frac{\cos 30^{\circ}}{\cos 60^{\circ}}=\left(\frac{3}{2}ight)^{2} \frac{\sqrt{3}}{2} 	imes \frac{2}{1}=\frac{9 \sqrt{3}}{4}$\ Ratio given is $\frac{4}{\sqrt{x}} ; \frac{4}{\sqrt{x}}=\frac{4}{9 \sqrt{3}}=\frac{4}{\sqrt{243}}$\ $x=243$\