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The real angle of dip, if a magnet is suspended at an angle of `30^(@)` to the magnetic meridian and the dip needle makes an angle of `45^(@)` with horizontal, is:A. `tan^(-1)(sqrt(3)/2)`B. `tan^(-1)(sqrt(3))`C. `tan^(-1)(sqrt(3)/sqrt(2))`D. `tan^(-1)(2/sqrt(3))` |
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Answer» Correct Answer - A `:. tam delta=V/H` `tan 45^(@)=V/(H cos 30^(@))` (Divide `(1)` and `(2)`) `delta=tan^(-1)(sqrt(3)/2)` |
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