A particle is released on a vertical smooth semicircular track from point $X$ so that, $OX$ makes angle $\mathrm{\theta }$  from the vertical (see figure). The normal reaction of the track on the particle vanishes at the point $Y$ where $OY$ makes an angle $\mathrm{\varphi }$ with the horizontal. Then

• Correct answer
  A

  $\mathrm{sin\varphi }=\frac{2}{3}\mathrm{cos\theta }$
• B

  $\mathrm{sin\varphi }=\frac{3}{4}\mathrm{cos\theta }$
• C

  $\mathrm{sin\varphi }=\frac{1}{2}\mathrm{cos\theta }$
• D

  $\mathrm{sin\varphi }=\mathrm{cos\theta }$

Hence, the correct answer is option: 'A'.