• # question_answer A dip circle lying initially in the magnetic meridian is rotated through angle $\theta$ is the horizontal plane. The tangent of angle of dip is increased in the ratio  A) $\cos \,\theta :1$                      B)  $\sin \,\theta :1$ C)  $1:\,\cos \,\theta$                   D)  $1:\,\sin \,\theta$

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 $\therefore$