JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • 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)  \[1:\,\sin \,\theta \]

    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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