$A$ small bar magnet $A$ oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^o$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian,its period will be

  • A
    $\frac{T}{\sqrt{2}}$
  • B
    $T$
  • C
    $\sqrt{2} T$
  • D
    $2T$

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