A compass needle of oscillation magnetometer oscillates 20 times per… | JEE Main 2022 (27 Jul, Shift 2)

Q10JEE Main 2022 · 27 JUL · SHIFT 2Difficult

Vibration Magnetometer

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

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Answer: (a)

Frequency $n \propto \sqrt{B_H}$ and $B_H = B\cos\delta$ , so $n^2 \propto B\cos\delta$ .
$\dfrac{n_P^2}{n_Q^2} = \dfrac{B_P\cos 30^\circ}{B_Q\cos 60^\circ} \Rightarrow \dfrac{400}{100} = \dfrac{B_P}{B_Q}\cdot\dfrac{\sqrt{3}/2}{1/2}$
$4 = \sqrt{3}\,\dfrac{B_P}{B_Q} \Rightarrow \dfrac{B_Q}{B_P} = \dfrac{\sqrt{3}}{4}$
So the answer is (a) $\sqrt{3} : 4$ .

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