Question
A dip needle lies initially in the magnetic meridian when it shows an angle of dip at a place. The dip circle is rotted through an angle $\mathrm{x}$ in the horizontal plane and then it shows an angle of dip $\theta^{\prime}$. Then $\left[\left(\tan \theta^{\prime}\right) /(\tan \theta)\right]$ is(a) $[1 /(\cos x)]$(b) $[1 /(\sin x)]$(c) $[1 /(\tan \mathrm{x})]$(d) $\cos \mathrm{x}$
Step 1
Initially, the dip needle is in the magnetic meridian and shows an angle of dip θ. Let's denote the horizontal component of the Earth's magnetic field as H and the vertical component as V. Then, we have: $\tan \theta = \frac{V}{H}$ Show more…
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A dip needle lies initially in the magnetic meridian when it shows an angle of dip ? at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip ? ? .
The angle of dip, if a dip needle oscillating in a vertical plane makes 40 oscillations per minute in a magnetic meridian and 30 oscillations per minute in a vertical plane at right angle to the magnetic meridian, is: (a) $\theta=\sin ^{-1}(0.5625)$ (b) $\theta=\sin ^{-1}(0.325)$ (c) $\theta=\sin ^{-1}(0.425)$ (d) $\theta=\sin ^{-1}(0.235)$
The real angle of dip, if a magnet is suspended at an angle of $30^{\circ}$ to the magnetic meridian and the dip needle makes an angle of $45^{\circ}$ with horizontal, is : (a) $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ (b) $\tan ^{-1}(\sqrt{3})$ (c) $\tan ^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}}\right)$ (d) $\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
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