MCQ
A particle is released on a vertical smooth semicircular track from point $X$ so that $OX$ makes angle $\theta $ from the vertical ( see figure). The normal reaction of the track on the particle vanishes at point $Y$ where $OY$ makes angle $\phi $ with the horizontal. Then
- A$\sin \,\phi = \,\cos \,\phi $
- B$\sin \,\phi = \frac{1}{2}\,\cos \,\theta $
- ✓$\sin \,\phi = \frac{2}{3}\,\cos \,\theta $
- D$\sin \,\phi = \frac{3}{4}\,\cos \,\theta $
