A rolling wheel of 12kg is on an inclined plane at position P and...

If we take the wheel as a ring then its moment of inertia will be, $I = M{r}^{2}$.

Using mechanical energy conservation,

Mgh - mgh = \dfrac{1}{2}m{v}^{2} + \dfrac{1}{2}M{v}^{2} + \dfrac{1}{2}I{\omega}^{2} ⇒ 12gh - 3gh = \dfrac{1}{2}3{v}^{2} + \dfrac{1}{2}12{v}^{2} + \dfrac{1}{2}12{r}^{2}\dfrac{{v}^{2}}{{r}^{2}}$</p><p>$⇒ 9gh = \dfrac{15{v}^{2} + 12{v}^{2}}{2}$</p><p>$⇒ v = \sqrt{\dfrac{2}{3}gh} = \dfrac{1}{2}\sqrt{\dfrac{8}{3}gh}$</p><p>Therefore,$x = \dfrac{8}{3} ≃ 3$