If the angle of dip at places A and B are 30° and 45° respectively… | MHT CET 2020 (16 Oct, Shift 2)

Q7MHT CET 2020 · 16 OCT · SHIFT 2Medium

Earth's Magnetism

If the angle of dip at places $A$ and $B$ are $30^\circ$ and $45^\circ$ respectively, then the ratio of horizontal component of earth's magnetic field at $A$ to that at $B$ will be

$\left[\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}, \sin\frac{\pi}{6} = \frac{1}{2}, \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\right]$

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Answer: (c)

$H = B_E\cos\delta$ ; taking the same total field $B_E$ at both places,
$\frac{H_A}{H_B} = \frac{\cos 30^\circ}{\cos 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$
So the answer is (c) $\sqrt{3} : \sqrt{2}$ .

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Earth's Magnetism question from MHT CET 2020 (16 Oct, Shift 2)

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