#
5th no

A wedge of height 'h'is released from rest with a light particles P placed on it as shown.the wedge slides down an incline which makes an angle $\theta $ with horizontal.All the surface are smooth ,P will reach the surface of the surface of the incline in time.$\left(a\right)\sqrt{\frac{2h}{g{\mathrm{sin}}^{2}\theta}}\left(b\right)\sqrt{\frac{2h}{g\mathrm{sin}\theta \mathrm{cos}\theta}}\left(c\right)\sqrt{\frac{2h}{g\mathrm{tan}\theta}}\left(d\right)\sqrt{\frac{2h}{g{\mathrm{cos}}^{2}\theta}}$

I think the answer is A) root 2h/g sin2theta
If its correct letme know
HOPE IT HELPS)