Two solid disks $A \& B$ are released from rest from points $X \& Y$ respectively on an inclined plane as shown in the figure. Which of the following statement(s) is/are incorrect?

Using law of conservation of energy

for wheel A,

 $$\begin{gathered} mgh = \frac{1}{2}I{\omega }^2 + \frac{1}{2}m{v}^2 \\ \Rightarrow mgh = \frac{1}{2}I{\left(\frac{V}{r}\right)}^2 + \frac{1}{2}m{v}^2 \\ \therefore mgh = \frac{1}{2}\left[\frac{I}{r^2}+m\right] v^2 \\ \therefore v^2 = \frac{mg}{\frac{1}{2}\left[\frac{I}{r^2}+m\right]}. h \\ \Rightarrow v \alpha \sqrt{h} \\ \therefore V_A \alpha \sqrt{h} \& V_B \alpha \sqrt{4h} \\ \Rightarrow V_B= 2V_A \\ Also V_A^2 = 2a_A \frac{h}{\sin \theta } \\ \Rightarrow a_A = \frac{g \sin \theta }{1+{\displaystyle \frac{1}{m{r}^2}}} \\ \therefore Acceleration of disk is independent of h \\ \therefore a_A = a_B \\ also \frac{h}{\sin \theta } = \frac{1}{2}{a}_A{t}_A^2 \\ \Rightarrow t_{A }\alpha \sqrt{h} \\ \therefore \frac{t_B}{t_A} = \sqrt{\frac{4h}{h}} = 2 \end{gathered}$$