question_answer

The earth is considered as a short magnet with its centre coinciding with the geometric centre of earth. The angle of dip \[\phi \] related to the magnetic latitude \[\lambda \] as

A) \[\tan \phi =\frac{1}{2\tan \alpha }\]      

B) \[\tan \lambda =2\tan \phi \]

C) \[\tan \lambda =2\tan \phi \]       

D) \[\tan \phi =2\tan \lambda \]

Correct Answer: D

Solution :

For a dipole at position (R, Q) We have                \[{{\text{p}}_{\text{1}}}\text{=10mm, }{{\text{p}}_{\text{2}}}\text{=50mm}\]                  ...(1) and          \[{{C}_{6}}{{H}_{6}}\]                 ...(2) Also          \[=\frac{10}{60}=\frac{1}{6}\]              ...(3)                  Dividing Eq. (i) by Eq. (ii)                   \[=8\times \frac{1}{8}+6\times \frac{1}{2}=4\]         ....(iv) From  Eq. (iii) and Eq. (iv)                        \[=12\times \frac{1}{4}=3\] From figure,    \[\text{C}{{\text{u}}_{\text{4}}}\text{A}{{\text{g}}_{\text{3}}}\text{Au}\text{.}\] So      \[-nF{{E}^{\circ }}_{cell}=-RT\] \[{{E}^{\circ }}_{cell}=\frac{RT}{nF}\]     \[{{E}^{\circ }}_{cell}\]