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
Answer (integer)
243
Solution
$$
\begin{aligned}
& \text {Period of oscillation } \alpha \frac{1}{\sqrt{B_{\mathrm{H}}}} \\\\
& \mathrm{T} \alpha \frac{1}{\sqrt{\mathrm{B} \cos \theta}} \Rightarrow \frac{T_1}{\mathrm{~T}_2}=\sqrt{\frac{\mathrm{B}_2 \cos \theta_2}{\mathrm{~B}_1 \cos \theta_1}} \\\\
& \Rightarrow \frac{60 / 20}{60 / 30}=\sqrt{\frac{\mathrm{B}_2}{\mathrm{~B}_1} \frac{\cos 60^{\circ}}{\cos 30^{\circ}}} \Rightarrow \frac{3}{2}=\sqrt{\frac{\mathrm{B}_2}{\sqrt{3} \mathrm{~B}_1}} \\\\
& \Rightarrow \frac{9}{4}=\frac{\mathrm{B}_2}{\sqrt{3} \mathrm{~B}_1} \Rightarrow \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{4}{9 \sqrt{3}}=\frac{4}{\sqrt{243}}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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