Question:
A magnet suspended in the horizontal plane makes 24 oscillations per minute at a place A where the dip angle is \(30^\circ\) and \(n\) oscillations per minute at another place B where the dip angle is \(60^\circ\). If the ratio of the horizontal components of earth's magnetic field at places A and B is \(16:9\), then the value of \(n\) is:
A magnet suspended in the horizontal plane makes 24 oscillations per minute at a place A where the dip angle is \(30^\circ\) and \(n\) oscillations per minute at another place B where the dip angle is \(60^\circ\). If the ratio of the horizontal components of earth's magnetic field at places A and B is \(16:9\), then the value of \(n\) is:
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For a vibrating magnet, frequency varies as the square root of the horizontal component of earth's magnetic field.
Updated On: Jun 12, 2026
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The Correct Option is A
Solution and Explanation
Concept:
Time period of oscillation of a magnet is
\[
T=2\pi\sqrt{\frac{I}{MB_H}}
\]
Therefore,
\[
f\propto\sqrt{B_H}
\]
where \(B_H\) is the horizontal component of earth's magnetic field.
Step 1: Use frequency relation. Given \[ f_A=24 \] and \[ \frac{B_{HA}}{B_{HB}}=\frac{16}{9} \] Therefore, \[ \frac{f_A}{f_B} = \sqrt{\frac{16}{9}} = \frac{4}{3} \] \[ \frac{24}{n} = \frac{4}{3} \]
Step 2: Calculate \(n\). \[ n = 24\times\frac34 \] \[ n=18 \] \[ \boxed{18} \]
Step 1: Use frequency relation. Given \[ f_A=24 \] and \[ \frac{B_{HA}}{B_{HB}}=\frac{16}{9} \] Therefore, \[ \frac{f_A}{f_B} = \sqrt{\frac{16}{9}} = \frac{4}{3} \] \[ \frac{24}{n} = \frac{4}{3} \]
Step 2: Calculate \(n\). \[ n = 24\times\frac34 \] \[ n=18 \] \[ \boxed{18} \]
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