Magnetism Question 40
Question: If $ {\varphi _{1}} $ and $ {\varphi _{2}} $ be the angles of dip observed in two vertical planes at right angles to each other and f be the true angle of dip, then
Options:
A) $ {{\cos }^{2}}\varphi ={{\cos }^{2}}{\varphi _{1}}+{{\cos }^{2}}{\varphi _{2}} $
B) $ {{\sec }^{2}}\varphi ={{\sec }^{2}}{\varphi _{1}}+{{\sec }^{2}}{\varphi _{2}} $
C) $ {{\tan }^{2}}\varphi ={{\tan }^{2}}{\varphi _{1}}+{{\tan }^{2}}{\varphi _{2}} $
D) $ {{\cot }^{2}}\varphi ={{\cot }^{2}}{\varphi _{1}}+{{\cot }^{2}}{\varphi _{2}} $
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Answer:
Correct Answer: D
Solution:
Let a be the angle which one of the planes make with the magnetic meridian the other plane makes an angle $ (90^{o}-\alpha ) $ with it. The components of H in these planes will be $ H\cos \alpha $ and $ H\sin \alpha $ respectively. If $ {\varphi _{1}} $ and $ {\varphi _{2}} $ are the apparent dips in these two planes, then $ \tan {\varphi _{1}}=\frac{V}{H\cos \alpha } $ i.e. $ \cos \alpha =\frac{V}{H\tan {\varphi _{1}}} $ ….. (i) $ \tan {\varphi _{2}}=\frac{V}{H\sin \alpha } $ i.e. $ \sin \alpha =\frac{V}{H\tan {\varphi _{2}}} $ ….. (ii) Squaring and adding (i) and (ii), we get $ {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha ={{( \frac{V}{H} )}^{2}}( \frac{1}{{{\tan }^{2}}{\varphi _{1}}}+\frac{1}{{{\tan }^{2}}{\varphi _{2}}} ) $ i.e. $ 1=\frac{V^{2}}{H^{2}}( {{\cot }^{2}}{\varphi _{1}}+{{\cot }^{2}}{\varphi _{2}} ) $ or $ \frac{H^{2}}{V^{2}}={{\cot }^{2}}{\varphi _{1}}+{{\cot }^{2}}{\varphi _{2}} $ i.e. $ {{\cot }^{2}}\varphi ={{\cot }^{2}}{\varphi _{1}}+{{\cot }^{2}}{\varphi _{2}} $ This is the required result.
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