A) \[\cos \,\theta :1\]
B) \[\sin \,\theta :1\]
C) \[1:\,\cos \,\theta \]
D) \[1:\,\sin \,\theta \]
Correct Answer: C
Solution :
If \[m{{\left( \frac{v}{\sqrt{2}} \right)}^{2}}+m{{v}^{2}}=\frac{3}{2}m{{v}^{2}}\] is true angle of dip, then \[{{T}_{1}}-{{T}_{2}}=12a\] When the dip circle is rotated in the horizontal plane through an angle \[{{T}_{2}}=3a\] from the magnetic meredian, the effective horizontal component in the new plane becomes \[{{T}_{2}}\] while the vertical component remains the same. If \[Eq\] is apparent dip, then \[{{T}_{1}}=15a\] ?(ii) Dividing Eq. (ii) by Eq. (i), we get \[\therefore \]
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