Applying work-energy theorem

Work done by gravity $=$ Change in Kinetic Energy

For part $\mathrm{A}\mathrm{B}$, ( friction will provide torque to generate rotational kinetic energy but work done by friction will be zero because the velocity of the point of action of friction is zero)

${\mathrm{mgy}}_1=\frac{3}{4}{\mathrm{mv}}^2$

For part $\mathrm{BC}$, (there is no friction, hence no change in rotational kinetic energy)

${\mathrm{mgy}}_2=\frac{1}{2}{\mathrm{mv}}^2$

Dividing both equations we get
$\therefore$ ${\mathrm{y}}_2=\frac{2}{3}{\mathrm{y}}_1$