Figure shows the two inclined tracks of different lengths but of same heights, such that the angle of inclination $\theta_2$ is greater than $\theta_1$ . The stones along OX and OY will slide down with accelerations $a_1 = g\sin\theta_1$ and $a_2 = g\sin\theta_2$ respectively. As $\theta_2 > \theta_1$ therefore, $a_2 > a_1$. Now, for motion of the two objects : K.E. at the bottom = P.E. at the top or $Mgh = \frac{1}{2} M v^{2}$ or $v=\sqrt{2 g h}$ As heights of two tracks is same, both the objects will reach the bottom withthe same speed. Now $, v=u+a t=0+a t$ or $t= \frac{v}{a}$ As $v$ is same for two objects, $t \propto \frac{1}{a}$. Since $a_2 > a_1$, the object sliding on the inclined track OY will reach the bottom earlier. In other words, the two objects will reach the bottom at different times.