A magnet is suspended in such a way that it oscillates in the horizontal plane. If it makes 20 oscillations per minute at a place where dip angle is 30° and 15 oscillations per minute at a place where dip angle is 60° . The ratio of total earth magnetic field at the two places is $\frac{x}{9 \sqrt{3}}$ then x is

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answer is 16.

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Detailed Solution

At Place I, $T_{1} = \frac{60}{20} = 2 π \sqrt{\frac{I}{{MB}_{H_{1}}}}$
At Place II, $T_{2} = \frac{60}{15} = 2 π \sqrt{\frac{I}{{MB}_{H_{2}}}}$
From equations (i) and (ii) , we have
$\begin{array}{l} {(\frac{4}{3})}^{2} = \frac{B_{H_{1}}}{B_{H_{2}}} \\ or \frac{B_{H_{1}}}{B_{H_{2}}} = \frac{16}{9} \\ B = B_{H} / \cos θ \\ \frac{B_{1}}{B_{2}} = \frac{B_{H_{1}}}{B_{H_{2}}} \times \frac{\cos 60^{\circ}}{\cos 30^{\circ}} \\ = \frac{16}{9} \times \frac{1 / 2}{\sqrt{3} / 2} = \frac{16}{9 \sqrt{3}} \end{array}$