Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are
allowed to slide down from rest, one on each track as shown in figure, which of the following statement(s) is/are correct?

As the given tracks are frictionless, hence mechanical energy will be conserved, as both the tracks having common height h. 

From conservation of mechanical energy,

$\frac{1}{2}m{v}^2=mgh$                  (For both tracks I and II)

$\Rightarrow v=\sqrt{2gh}$

Hence, speed is same for both stones.

For stone I, a₁ = acceleration along inclined plane = g sin ${\theta }_1$ Similarly, for stone II, a₂ = g sin ${\theta }_2$  as  ${\theta }_2$> ${\theta }_1$, hence a₂ > a₁,
And length for track II is also less, hence stone II reaches earlier than stone I.